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\chapter{Associative Algebra}
\section{Basic Properties}
\begin{definition}{Associative Algebra over Commutative Ring}{}
Let $R$ be a commutative ring. An \textbf{associative $R$-algebra} is a ring $A$ together with a ring homomorphism $\varphi:R\to Z(A)$, which makes $A$ an $R$-module by defining the scalar multiplication as
\begin{align*}
R\times A &\longrightarrow A\\
(r,a) &\longmapsto r\cdot a:=\varphi(r)a.
\end{align*}
$\varphi:R\to Z(A)$ is called the \textbf{structure homomorphism} of $A$.
\end{definition}
\begin{remark}
We can check that
\[
r\cdot (ab)=\sigma(r)ab=(r\cdot a)b=\sigma(r)ab=a\left(\sigma(r)b\right) =a(r\cdot b),
\]
which justifies the naming ``associative".
\end{remark}
We usually call associative $R$-algebra as $R$-algebra for short.
\begin{proposition}{Commutative Ring homomorphism $R\to S$ induces functor $S\text{-}\mathsf{Alg}\to R\text{-}\mathsf{Alg}$}{}
Let $R$ and $S$ be commutative rings with a ring homomorphism $f: R\to S$. Then every $S$-algebra $A$ is an $R$-algebra by defining $ra = f(r)a$, or equivalently through $R\to S\to Z(A)$. This defines a functor $F: S\text{-}\mathsf{Alg}\to R\text{-}\mathsf{Alg}$, which is identify map on objects and morphisms.
\[
\begin{tikzcd}[ampersand replacement=\&]
S\text{-}\mathsf{Alg}\&[-25pt]\&[+10pt]\&[-30pt] R\text{-}\mathsf{Alg}\&[-30pt]\&[-30pt] \\ [-15pt]
A \arrow[dd, "g"{name=L, left}]
\&[-25pt] \& [+10pt]
\& [-30pt] A\arrow[dd, "g"{name=R}] \&[-30pt]\\ [-10pt]
\& \phantom{.}\arrow[r, "F", squigarrow]\&\phantom{.} \& \\[-10pt]
B \& \& \& B\&
\end{tikzcd}
\]
\end{proposition}
In particular, commutative ring homomorphism $R\to S$ makes $S$ an $R$-algebra.
\section{Construction}
\subsection{Quotient Object}
\begin{definition}{Quotient Algebra}{quotient_algebra}
Let $A$ be an $R$-algebra and $\mathfrak{a}$ be a two-sided ideal of $A$. Since $\mathfrak{a}$ is an $R$-submodule of $A$, the quotient ring $A/\mathfrak{a}$ can also be endowed with an $R$-module structure, which makes $A/\mathfrak{a}$ an $R$-algebra. We call $A/\mathfrak{a}$ the \textbf{quotient algebra} of $A$ by $\mathfrak{a}$.
\end{definition}
\begin{proposition}{Universal Property of Quotient Algebra}{universal_property_of_quotient_algebra}
Let $A$ be an $R$-algebra and $\mathfrak{a}$ be a two-sided ideal of $A$. Let $\pi:A\to A/\mathfrak{a}$ be the canonical projection. For any $R$-algebra homomorphism $f:A\to B$ such that $\mathfrak{a}\subseteq \ker(f)$
or equivalently $f(\mathfrak{a})=\{0\}$, there exists a unique $R$-algebra homomorphism $\widetilde{f}:A/\mathfrak{a}\to B$ such that the following diagram commutes
\begin{center}
\begin{tikzcd}
& A \arrow[ld, "\pi"'] \arrow[rd, "f"] & \\
A/\mathfrak{a} \arrow[rr, "{\exists !\,\widetilde{f}}"', dashed] & & B
\end{tikzcd}
\end{center}
The unique $R$-algebra homomorphism $\widetilde{f}:A/\mathfrak{a}\to B$ is defined as
\begin{align*}
\widetilde{f}: A/\mathfrak{a} &\longrightarrow B\\
a + \mathfrak{a} &\longmapsto f(a).
\end{align*}
\end{proposition}
\subsection{Free Object}
\begin{definition}{Free $R$-Algebra}{}
Let $X$ be a set and $R$ be a commutative ring. The \textbf{free $R$-algebra} on $X$, denoted by $\mathrm{Free}_{R\text{-}\mathsf{Alg}}(X)$, together with a function $\iota:X\to \mathrm{Free}_{R\text{-}\mathsf{Alg}}(X)$, is defined by the following universal property: for any $R$-algebra $A$ and any function $f:X\to A$, there exists a unique $R$-algebra homomorphism $\widetilde{f}:\mathrm{Free}_{R\text{-}\mathsf{Alg}}(X)\to A$ such that the following diagram commutes
\begin{center}
\begin{tikzcd}[ampersand replacement=\&]
\mathrm{Free}_{R\text{-}\mathsf{Alg}}(X)\arrow[r, dashed, "\exists !\,\widetilde{f}"] \& A \\[0.3cm]
X\arrow[u, "\iota"] \arrow[ru, "f"'] \&
\end{tikzcd}
\end{center}
The free $R$-algebra $\mathrm{Free}_{R\text{-}\mathsf{Alg}}(X)$ can be contructed by direct sum of copies of $R$
\[
\mathrm{Free}_{R\text{-}\mathsf{Alg}}(X)\cong\bigoplus_{w\in\mathrm{Free}_{\mathsf{Mon}}(X)}Rw.
\]
\end{definition}
\subsection{Graded Object}
\begin{definition}{$I$-Graded Algebra over an Graded Commutative Ring}{}
Let $(I,+)$ be a monoid and $R$ be a \hyperref[th:I_graded_ring]{$I$-graded commutative ring} with grading $(R_i)_{i\in I}$. An \textbf{$I$-graded algebra over graded ring $R$} is an $R$-algebra $A$ together with a family of subalgebras $\left(A_i\right)_{i\in I}$ such that
\begin{enumerate}[(i)]
\item $A=\bigoplus_{i\in I}A_i$.
\item $A_iA_j\subseteq A_{i+j}$ for all $i, j\in I$.
\item $R_iA_j\subseteq A_{i+j}$ for all $i, j\in I$.
\end{enumerate}
Elements in $A_i$ are called \textbf{homogeneous elements of degree $i$}.
\end{definition}
\begin{definition}{Degree-preserving $R$-algebra homomorphism}{degree_preserving_R_algebra_homomorphism}
Let $(I,+)$ be a monoid and let $R=\bigoplus_{i\in I} R_i$ be an $I$-graded commutative ring. Let $A=\bigoplus_{i\in I} A_i$ and $B=\bigoplus_{i\in I} B_i$ be $I$-graded $R$-algebras.
An $R$-algebra homomorphism $f:A\to B$ is called \textbf{degree-preserving} (or \textbf{graded}) if for every $i\in I$,
\[
f(A_i)\subseteq B_i.
\]
We write $\mathsf{GrAlg}^I_R$ for the category whose objects are $I$-graded $R$-algebras and whose morphisms are degree-preserving $R$-algebra homomorphisms.
\end{definition}
\begin{proposition}{Graded Algebra Quotients out Graded Ideal}{graded_algebra_quotients_out_graded_ideal}
Let $A$ be an $I$-graded algebra over graded ring $R$ with grading $(A_i)_{i\in I}$ and $\mathfrak{a}$ be a \hyperref[th:graded_ideal]{$I$-graded two-sided ideal} of $A$. Then we have an $R$-module isomorphism
\begin{align*}
A_i/\left(\mathfrak{a}\cap A_i\right)&\xlongrightarrow{\sim} \left( A_i+\mathfrak{a}\right)/\mathfrak{a}\\
a + \left(\mathfrak{a}\cap A_i\right) &\longmapsto a + \mathfrak{a},
\end{align*}
and the \hyperref[th:quotient_algebra]{quotient algebra} $A/\mathfrak{a}$ has a decomposition
\[
A/ \mathfrak{a}=\bigoplus_{i\in I}\left( A_i+\mathfrak{a}\right)/\mathfrak{a}\cong \bigoplus_{i\in I}A_i/\left(\mathfrak{a}\cap A_i\right),
\]
which makes $A/\mathfrak{a}$ an $I$-graded $R$-algebra.
\end{proposition}
\begin{prf}
By the definition of graded ideal, we have
\[
\mathfrak{a}=\bigoplus_{i\in I}\mathfrak{a}_i,\quad \mathfrak{a}_i:=\mathfrak{a}\cap A_i.
\]
Let $\pi:A\to A/\mathfrak{a}$ be the canonical projection. Restrict $\pi$ to $A_i$, we have an $R$-module homomorphism
\begin{align*}
\pi|_{A_i}: A_i &\longrightarrow \pi(A_i)\\
\end{align*}
with kernel
\[
\ker\left(\pi|_{A_i}\right)=\left\{ x\in A_i \midv \pi(x)=0 + \mathfrak{a} \right\}=\left\{ x\in A_i \midv x\in \mathfrak{a} \right\} = A_i \cap \mathfrak{a} = \mathfrak{a}_i.
\]
Thus we have an $R$-module isomorphism
\begin{align*}
A_i/\mathfrak{a}_i&\xlongrightarrow{\sim} \pi(A_i)\\
a + \mathfrak{a}_i &\longmapsto a + \mathfrak{a}.
\end{align*}
Take any $\pi(a)\in A/\mathfrak{a}$, where $a\in A$. Since $A=\bigoplus_{i\in I} A_i$, we can write $a=\sum_{i\in I} a_i$ with $a_i\in A_i$ and only finitely many $a_i$ being nonzero. Thus
\[
\pi(a)=\pi\left(\sum_{i\in I} a_i\right)=\sum_{i\in I} \pi(a_i)\in \sum_{i\in I}\pi(A_i),
\]
which means $A/\mathfrak{a}\subseteq \sum_{i\in I}\pi(A_i)$. On the other hand, we have $\sum_{i\in I}\pi(A_i)\subseteq A/\mathfrak{a}$. Therefore, we have $A/\mathfrak{a}=\sum_{i\in I}\pi(A_i)$.
Moreover, if $\sum_{i\in I}\pi(a_i)=0$ with $a_i\in A_i$, then $\pi\left(\sum_{i\in I} a_i\right)=0$, which means $\sum_{i\in I} a_i\in \mathfrak{a}$. According to \Cref{th:membership_criterion_for_graded_ideals}, this implies $a_i\in \mathfrak{a}\cap A_i $ for all $i\in I$. Thus we have $\pi(a_i)=0$ for all $i\in I$. Therefore, the sum $A/\mathfrak{a}=\sum_{i\in I}\pi(A_i)$ is direct, which means
\[
A/\mathfrak{a}=\bigoplus_{i\in I}\pi(A_i).
\]
Note
\[
\pi(A_i) =\left\{x+\mathfrak{a}\midv x\in A_i\right\} = \left\{x+a +\mathfrak{a}\midv x\in A_i, a\in \mathfrak{a}\right\}= \left( A_i+\mathfrak{a}\right)/\mathfrak{a}.
\]
Combining these results, we have the $R$-module isomorphism
\[
A/ \mathfrak{a}=\bigoplus_{i\in I}\pi(A_i)=\bigoplus_{i\in I}\left( A_i+\mathfrak{a}\right)/\mathfrak{a}\cong \bigoplus_{i\in I}A_i/\left(\mathfrak{a}\cap A_i\right).
\]
Finally, we can check that the multiplication and scalar multiplication respect the grading:
\[
\left(a_i + \mathfrak{a}\right)\left(a_j + \mathfrak{a}\right)=a_ia_j + \mathfrak{a}\in \left(A_{i+j}+\mathfrak{a}\right)/\mathfrak{a},\quad \forall a_i\in A_i, a_j\in A_j,
\]
\[
r\cdot\left(a_i + \mathfrak{a}\right)=ra_i + \mathfrak{a}\in \left(A_{i+j}+\mathfrak{a}\right)/\mathfrak{a},\quad \forall r\in R_j, a_i\in A_i.
\]
\end{prf}
\begin{example}{Polynomial Algebra $R[X_1,\cdots,X_n]$}{}
Let $R$ be a commutative ring and $X_1,\cdots,X_n$ be indeterminates. Then $R[X_1,\cdots,X_n]$ is an $\mathbb{N}$-graded $R$-algebra with grading $R[X_1,\cdots,X_n]_i$ being the set of homogeneous polynomials of degree $i$.
\end{example}
\subsection{Tensor Product}
\begin{definition}{Tensor Product of Algebras}{tensor_product_of_algebras}
Let $R$ be a commutative ring and $A$, $B$ be $R$-algebras. The \textbf{tensor product of $R$-algebras $A$ and $B$} is defined by the following universal property: for any triple $(C, f_A, f_B)$, where $C$ is an $R$-algebra and $f_A:A\to C$, $f_B:B\to C$ are $R$-algebra homomorphisms which satisfy
\[
f_A(a)f_B(b)=f_B(b)f_A(a),\quad \forall a\in A, b\in B,
\]
the tensor product
\[
(A\otimes_R B, \iota_A: A \times B \to A\otimes_R B, \iota_B: A \times B \to A\otimes_R B)
\]
is initial among such triples, i.e. there exists a unique $R$-algebra homomorphism \begin{align*}
\phi: A\otimes_R B &\longrightarrow C
\end{align*}
such that the following diagram commutes
\[
\begin{tikzcd}
A \arrow[r, "\iota_A"] \arrow[rd, "f_A"'] & A\otimes_R B \arrow[d, "\exists! \phi", dashed] & B \arrow[l, "\iota_B"'] \arrow[ld, "f_B"] \\[0.5em]
& C &
\end{tikzcd}
\]
Concretely, $A\otimes_R B$ can be constructed as the tensor product of $R$-modules $A\otimes_R B$ together with multiplication defined as
\[
(a_1\otimes b_1)(a_2\otimes b_2):=(a_1a_2)\otimes(b_1b_2),\quad \forall a_1,a_2\in A, b_1,b_2\in B
\]
and unity
\[
1_{A\otimes_R B}:=1_A\otimes 1_B.
\]
And the $R$-algebra homomorphisms $\iota_A$, $\iota_B$ are defined as
\begin{align*}
\iota_A : A &\longrightarrow A\otimes_R B\\
a &\longmapsto a\otimes 1_B,
\end{align*}
\begin{align*}
\iota_B : B &\longrightarrow A\otimes_R B\\
b &\longmapsto 1_A\otimes b.
\end{align*}
The unique $R$-algebra homomorphism $\phi:A\otimes_R B\to C$ is defined as
\begin{align*}
\phi: A\otimes_R B &\longrightarrow C\\
a\otimes b &\longmapsto f_A(a)f_B(b).
\end{align*}
\end{definition}
\begin{remark}
It is straightforward to check that the multiplication defined above is well-defined and makes $A\otimes_R B$ an $R$-algebra. According to \Cref{th:pure_tensors_generate_tensor_product}, since $(a,b)\mapsto f_A(a)f_B(b)$ is $\mathbb{Z}$-bilinear, $\phi$ is a well-defined abelian group homomorphism. We can further check that $\phi$ is an $R$-algebra homomorphism:
\[
\phi(r(a\otimes b))=\phi((r a)\otimes b)=f_A(r a)f_B(b)=r f_A(a)f_B(b)=r \phi(a\otimes b)
\]
\[
\phi\left((a_1\otimes b_1)(a_2\otimes b_2)\right)=\phi\left((a_1a_2)\otimes(b_1b_2)\right)= f_A(a_1a_2)f_B(b_1b_2)=f_A(a_1)f_A(a_2)f_B(b_1)f_B(b_2)=\phi(a_1\otimes b_1)\phi(a_2\otimes b_2)
\]
\[
\phi\left(1_{A\otimes_R B}\right)=\phi\left(1_A\otimes 1_B\right)=f_A(1_A)f_B(1_B)=1_C
\]
\end{remark}
\begin{definition}{Tensor Product of $R$-algebra Homomorphisms}{tensor_product_of_R_algebra_homomorphisms}
Let $R$ be a commutative ring and $A_1$, $A_2$, $B_1$, $B_2$ be $R$-algebras. Given two $R$-algebra homomorphisms $f:A_1\to A_2$ and $g:B_1\to B_2$, the \textbf{tensor product of $R$-algebra homomorphisms} is defined as the $R$-algebra homomorphism
\begin{align*}
f\otimes_R g: A_1\otimes_R B_1 &\longrightarrow A_2\otimes_R B_2\\
a\otimes b &\longmapsto f(a)\otimes g(b).
\end{align*}
which is induced by the universal property of tensor product $A_1\otimes_R B_1$ through the following commutative diagram:
\[
\begin{tikzcd}
A_1 \arrow[r, "\iota_{A_1}"] \arrow[d, "f"'] & A_1\otimes_R B_1 \arrow[d, "f\otimes_R g", dashed] & B_2 \arrow[l, "\iota_{B_1}"'] \arrow[d, "g"] \\[0.7em]
A_2 \arrow[r, "\iota_{A_2}"'] & A_2\otimes_R B_2 & B_2 \arrow[l, "\iota_{B_2}"]
\end{tikzcd}
\]
\end{definition}
\begin{proposition}{Symmetric Monoidal Structure on $R$-$\mathsf{Alg}$}{}
Let $R$ be a commutative ring. The tensor product $\otimes_R$ defines a symmetric monoidal structure on the category $R$-$\mathsf{Alg}$, with unit object $R$.
\begin{enumerate}[(i)]
\item Tensor product: the tensor product functor is
\[
\begin{tikzcd}[ampersand replacement=\&]
R\text{-}\mathsf{Alg}\times R\text{-}\mathsf{Alg}\&[-25pt]\&[+10pt]\&[-30pt] R\text{-}\mathsf{Alg}\&[-30pt]\&[-30pt] \\ [-15pt]
(A_1, B_1) \arrow[dd, "f\times g"{name=L, left}]
\&[-25pt] \& [+10pt]
\& [-30pt] A_1\otimes_R B_1\arrow[dd, "f\otimes_R g"{name=R}] \&[-30pt]\\ [-10pt]
\& \phantom{.}\arrow[r, "\otimes_R", squigarrow]\&\phantom{.} \& \\[-10pt]
(A_2, B_2) \& \& \& A_2\otimes_R B_2\&
\end{tikzcd}
\]
\item Associator: for any $R$-algebras $A$, $B$, $C$, there is a natural isomorphism
\begin{align*}
\alpha_{A,B,C}:(A\otimes_R B)\otimes_R C &\xlongrightarrow{\sim}A\otimes_R (B\otimes_R C)\\
(a\otimes b)\otimes c &\longmapsto a\otimes (b\otimes c)
\end{align*}
\item Unit object: $R$.
\item An isomorphism in $R$-$\mathsf{Alg}$:
\begin{align*}
\iota: R \otimes_R R &\xlongrightarrow{\sim} R\\
r\otimes r' &\longmapsto rr'
\end{align*}
\item Symmetry: for any $R$-algebras $A$, $B$, there is a natural isomorphism
\begin{align*}
\gamma_{A,B}: A\otimes_R B &\xlongrightarrow{\sim} B\otimes_R A\\
a\otimes b &\longmapsto b\otimes a
\end{align*}
\end{enumerate}
\end{proposition}
% \begin{proposition}{}{}
% Let $R$ be a commutative ring, and $A$, $B$, $C$ be $R$-algebras. Suppose $\otimes_R:A\times B\to A\otimes_R B$ is the tensor product map. Let $f:\mathrm{im}(\otimes_R)\to C$ be a map. Define
% \[
% \beta:A\times B\longrightarrow C,\qquad \beta(a,b)\coloneqq f(a\otimes b).
% \]
% then find equivalent consitions for that there exists a $R$-algebra homomorphism $\tilde{f}:A\otimes_R B\to C$ such that the following diagram commutes
% \[
% \begin{tikzcd}[ampersand replacement=\&] \mathrm{im}(\otimes) \arrow[r, "i", hook] \arrow[rd, "f"'] \& A\otimes_R B \arrow[d, dashed, "\tilde{f}"] \\[0.3cm] \& C
% \end{tikzcd}
% \]
% \end{proposition}
\begin{proposition}{Tensor Product of Quotient Algebras}{tensor_product_of_quotient_algebras}
Let $R$ be a commutative ring and $A_1$, $A_2$ be $R$-algebras. Let $I_1\subseteq A_1$, $I_2\subseteq A_2$ be two-sided ideals of $A_1$, $A_2$ respectively. Then we have an $R$-algebra isomorphism
\begin{align*}
(A_1/I_1)\otimes_R (A_2/I_2) &\xlongrightarrow{\sim}(A_1\otimes_R A_2)/(I_1\otimes_R A_2 + A_1\otimes_R I_2)\\
\overline{a_1}\otimes \overline{a_2} &\longmapsto \overline{a_1\otimes a_2}
\end{align*}
Here, given inclusion $i_1: I_1\hookrightarrow A_1$, the $R$-module $I_1\otimes_R A_2$ is identified as the image of $i_1\otimes_R \mathrm{id}_{A_2}:I_1\otimes_R A_2 \to A_1\otimes_R A_2$
\[
\mathrm{im}\left(i_1\otimes_R \mathrm{id}_{A_2}\right)=\left\{\sum_{n=1}^m x_n\otimes y_n\in A_1\otimes_R A_2\midv m\in\mathbb{Z}_{\ge 1},\,x_n\in I_1,\, y_n\in A_2\right\},
\]
which is a two-sided ideal of $A_1\otimes_R A_2$. The similar identification applies to $A_1\otimes_R I_2$.
\end{proposition}
\begin{prf}
Let $J:=I_1\otimes_R A_2 + A_1\otimes_R I_2$. Define
\begin{align*}
\iota_1: A_1/I_1 &\longrightarrow (A_1\otimes_R A_2)/J\\
a_1 + I_1 &\longmapsto (a_1\otimes 1_{A_2}) + J,
\end{align*}
If $a_1, a_1'\in A_1$ satisfy $a_1-a_1'\in I_1$, then
\[
(a_1\otimes 1_{A_2}) - (a_1'\otimes 1_{A_2}) = (a_1 - a_1')\otimes 1_{A_2} \in I_1\otimes_R A_2 \subseteq J\implies (a_1\otimes 1_{A_2}) + J = (a_1'\otimes 1_{A_2}) + J,
\]
which shows that $\iota_1$ is well-defined. And we can check that $\iota_1$ is an $R$-algebra homomorphism. Similarly, we can define an $R$-algebra homomorphism
\begin{align*}
\iota_2: A_2/I_2 &\longrightarrow (A_1\otimes_R A_2)/J\\
a_2 + I_2 &\longmapsto (1_{A_1}\otimes a_2) + J.
\end{align*}
Moreover, the images of $\iota_1$ and $\iota_2$ commute in $(A_1\otimes_R A_2)/J$: for any $a_1\in A_1$, $a_2\in A_2$,
\begin{align*}
\iota_1(a_1 + I_1)\iota_2(a_2 + I_2) &= \left((a_1\otimes 1_{A_2}) + J\right)\left((1_{A_1}\otimes a_2) + J\right)\\
&= (a_1\otimes a_2) + J\\
&= \left((1_{A_1}\otimes a_2) + J\right)\left((a_1\otimes 1_{A_2}) + J\right) \\
&= \iota_2(a_2 + I_2)\iota_1(a_1 + I_1).
\end{align*}
Thus by the \hyperref[th:tensor_product_of_algebras]{universal property of tensor product}, there exists a unique $R$-algebra homomorphism
\begin{align*}
\varphi: (A_1/I_1)\otimes_R (A_2/I_2) &\longrightarrow (A_1\otimes_R A_2)/J\\
(a_1 + I_1)\otimes (a_2 + I_2) &\longmapsto (a_1\otimes a_2) + J.
\end{align*}
Next, we construct the inverse of $\varphi$. Given the quotient maps $\pi_1:A_1\to A_1/I_1$ and $\pi_2:A_2\to A_2/I_2$, we can define an $R$-algebra homomorphism $\psi:=\pi_1\otimes_R \pi_2$ as
\begin{align*}
\psi: A_1\otimes_R A_2 &\longrightarrow (A_1/I_1)\otimes_R (A_2/I_2)\\
a_1\otimes a_2 &\longmapsto (a_1 + I_1)\otimes (a_2 + I_2).
\end{align*}
Since for any $x\in I_1$, $y\in A_2$, we have
\[
\psi(x\otimes y)=(0 + I_1)\otimes (y + I_2)=0
\]
and for any $x\in A_1$, $y\in I_2$, we have
\[
\psi(x\otimes y)=(x + I_1)\otimes (0 + I_2)=0,
\]
we have $J\subseteq \ker(\psi)$. Thus, by the universal property of quotient algebra, there exists a unique $R$-algebra homomorphism
\begin{align*}
\widetilde{\psi}: (A_1\otimes_R A_2)/J &\longrightarrow (A_1/I_1)\otimes_R (A_2/I_2)\\
(a_1\otimes a_2) + J &\longmapsto (a_1 + I_1)\otimes (a_2 + I_2).
\end{align*}
We can check that $\widetilde{\psi}$ is the inverse of $\varphi$:
\begin{align*}
\widetilde{\psi}\circ \varphi\left((a_1 + I_1)\otimes (a_2 + I_2)\right) &= \widetilde{\psi}\left((a_1\otimes a_2) + J\right) = (a_1 + I_1)\otimes (a_2 + I_2),\\
\varphi\circ \widetilde{\psi}\left((a_1\otimes a_2) + J\right) &= \varphi\left((a_1 + I_1)\otimes (a_2 + I_2)\right) = (a_1\otimes a_2) + J.
\end{align*}
Therefore, we show that $\varphi$ is an $R$-algebra isomorphism.
\end{prf}
\begin{corollary}{$mod\; I$ Reduction of $R$-Algebras}{mod_I_reduction_of_R_algebras}
Let $R$ be a commutative ring and $I\subseteq R$ be an ideal of $R$. For any $R$-algebra $A$, there is an isomorphism of $R$-algebras
\begin{align*}
A/IA &\xlongrightarrow{\sim} A\otimes_R (R/I) \\
\overline{a} &\longmapsto a\otimes \overline{1_R}\\
\overline{ra}&\longmapsfrom a\otimes \overline{r}
\end{align*}
where
\[
IA:= \left\{ r a \in A\midv r\in I,\, a\in A\right\}
\]
is the two-sided ideal of $A$ generated by $I$.
\end{corollary}
\begin{prf}
Apply \Cref{th:tensor_product_of_quotient_algebras} with $A_1=A$, $A_2=R$, $I_1=\{0\}$, $I_2=I$. We obtain an $R$-algebra isomorphism
\begin{align*}
A \otimes_R (R/I) &\xlongrightarrow{\sim} (A\otimes_R R)/( 0\otimes_R R+A\otimes_R I)\\
a \otimes (r+I) &\longmapsto (a\otimes r) + (A\otimes_R I).
\end{align*}
Under the canonical isomorphism
\begin{align*}
\phi: A\otimes_R R &\xlongrightarrow{\sim} A\\
a\otimes r &\longmapsto r a,
\end{align*}
$A \otimes_R I$ is mapped to $IA$. Thus we have an $R$-algebra isomorphism
\begin{align*}
A \otimes_R (R/I) &\xlongrightarrow{\sim} A/IA\\
a \otimes (r+I) &\longmapsto r a + IA.
\end{align*}
\end{prf}
\begin{corollary}{}{}
Let $R$ be a commutative ring and $I\subseteq R$ be an ideal of $R$. We have a cononical isomorphism of $R$-algebras
\begin{align*}
(R/I)\otimes_R (R/I) &\xlongrightarrow{\sim} R/I\\
(r+I)\otimes (r'+I) &\longmapsto rr' + I.
\end{align*}
\end{corollary}
\begin{prf}
By \Cref{th:mod_I_reduction_of_R_algebras}, we have an $R$-algebra isomorphism
\begin{align*}
(R/I)\otimes_R (R/I) &\xlongrightarrow{\sim} (R/I)/ (I(R/I))\\
(r+I)\otimes (r'+I) &\longmapsto rr' + I(R/I).
\end{align*}
Since for any $r\in I$, $r'+I\in R/I$, we have
\[
r(r'+I)=rr' + I = 0 + I,
\]
which shows that $I(R/I)=\{0\}$. Thus we obtain the desired isomorphism.
\end{prf}
\subsection{Tensor Algebra}
\begin{definition}{Tensor Algebra $T^{\bullet}(M)$}{}
Given a $R$-module $M$, the \textbf{$k$-th tensor power of $M$} is defined as
\begin{align*}
T^k(M)&:=M^{\otimes k}=\underbrace{M\otimes_R\cdots\otimes_R M}_{k\text{ times}},\\
T^0(M)&:=R.
\end{align*}
The \textbf{tensor algebra} of $M$ is defined as
\[
T^{\bullet}(M):=\bigoplus_{k=0}^{\infty}T^k(M)
\]
with multiplication $\otimes$ defined as
\[
(m_1\otimes\cdots\otimes m_k)\otimes(m_{k+1}\otimes\cdots\otimes m_{k+l})=m_1\otimes\cdots\otimes m_{k+l}
\]
$T^{\bullet}(M)$ is an $\mathbb{N}$-graded $R$-algebra with grading $(T^k(M))_{k\ge 0}$.
\end{definition}
\begin{proposition}{Tensor Algebra Functor $T^{\bullet}:R\text{-}\mathsf{Mod}\to \mathsf{GrAlg}^{\mathbb{N}}_R$}{}
Let $R$ be a commutative ring. The tensor algebra construction $T^{\bullet}:R\text{-}\mathsf{Mod}\to \mathsf{GrAlg}^{\mathbb{N}}_R$ is a functor defined as follows
\[
\begin{tikzcd}[ampersand replacement=\&]
R\text{-}\mathsf{Mod}\&[-25pt]\&[+10pt]\&[-30pt] \mathsf{GrAlg}^{\mathbb{N}}_R\&[-30pt]\&[-30pt] \\ [-15pt]
M \arrow[dd, "g"{name=L, left}]
\&[-25pt] \& [+10pt]
\& [-30pt] T^{\bullet}(M)\arrow[dd, "T^{\bullet}(g)"{name=R}] \&[-20pt]\ni\& [+10pt]m_1\otimes\cdots\otimes m_k \arrow[dd, mapsto, "g\otimes g\cdots\otimes g"{name=L, right}]
\\ [-10pt]
\& \phantom{.}\arrow[r, "T^{\bullet}", squigarrow]\&\phantom{.} \& \\[-10pt]
N \& \& \& T^{\bullet}(N)\&[-0pt]\ni\& g(m_1)\otimes\cdots\otimes g(m_k)
\end{tikzcd}
\]
\end{proposition}
\begin{prf}
According to \Cref{th:tensor_product_of_R_algebra_homomorphisms}, for each $k\ge 0$, we can define an $R$-module homomorphism $T^k(g):=g^{\otimes k}$ on degree-$k$ component:
\begin{align*}
T^k(g): T^k(M) &\longrightarrow T^k(N)\\
m_1\otimes\cdots\otimes m_k &\longmapsto g(m_1)\otimes\cdots\otimes g(m_k).
\end{align*}
Then we can define a \hyperref[th:degree_preserving_R_algebra_homomorphism]{degree-preserving $R$-algebra homomorphism} $T^{\bullet}(g):=\bigoplus_{k=0}^{\infty} T^k(g)$ as follows:
\begin{align*}
T^{\bullet}(g): T^{\bullet}(M) &\longrightarrow T^{\bullet}(N)\\
(x_0, x_1, x_2, \cdots) &\longmapsto \left(T^0(g)(x_0),\, T^1(g)(x_1),\, T^2(g)(x_2), \cdots\right).
\end{align*}
It is straightforward to check that $T^{\bullet}(\mathrm{id}_M)=\mathrm{id}_{T^{\bullet}(M)}$ and $T^{\bullet}(g_2\circ g_1)=T^{\bullet}(g_2)\circ T^{\bullet}(g_1)$ for any $R$-module homomorphisms $g_1:M\to N$, $g_2:N\to P$. Thus, $T^{\bullet}$ is a functor.
\end{prf}
\begin{proposition}{Adjunction $T^{\bullet}\dashv U_{\mathsf{R\text{-}\mathsf{Mod}}}$}{}
Let $R$ be a commutative ring. Suppose $U:R\text{-}\mathsf{Alg}\to R\text{-}\mathsf{Mod}$ is the forgetful functor. Then the tensor algebra functor $T^{\bullet}:R\text{-}\mathsf{Mod}\to R\text{-}\mathsf{Alg}$ is left adjoint to $U$.
\end{proposition}
\subsection{Exterior Algebra and Symmetric Algebra}
\begin{definition}{Exterior Algebra $\Largewedge^{\bullet} (M)$}{}
Given an $R$-module $M$,
\begin{align*}
I_{\largewedge}(M) & :=\langle x \otimes x: x \in M\rangle=\left\{\sum_{i=1}^m a_i (x_i \otimes x_i) b_i \midv m\in\mathbb{Z}_{\ge 1},\, a_i,b_i\in T^{\bullet}(M),\, x_i\in M\right\}
\end{align*}
is a graded two-sided ideal of $T^{\bullet}(M)$. The \textbf{exterior algebra} of $M$ is defined as
\[
\largewedge^{\bullet} (M)=T^{\bullet}(M)/I_{\largewedge}(M).
\]
According to \Cref{th:graded_algebra_quotients_out_graded_ideal}, $\largewedge^{\bullet} (M)$ is a graded $R$-algebra with grading
\[
\largewedge^{\bullet} (M)\cong \bigoplus_{k=0}^{\infty} \largewedge^k(M)
\]
where
\[
\largewedge^k(M):=T^k(M)/\left(I_{\largewedge}(M)\cap T^k(M)\right)
\]
is an $R$-module and is called the \textbf{$k$-th exterior power of $M$}. Especially, we have $\Largewedge^0(M)\cong R$ and $\Largewedge^1(M)\cong M$ as $R$-modules, and we identify them directly.
The multiplication of $\Largewedge^{\bullet} (M)$ is denoted by
\begin{align*}
\wedge: \Largewedge^{\bullet} (M)\times \Largewedge^{\bullet} (M) &\longrightarrow \Largewedge^\bullet (M)\\
(a + I_{\wedge}(M), b + I_{\wedge}(M)) &\longmapsto (a\otimes b) + I_{\wedge}(M).
\end{align*}
and is called the \textbf{wedge product}. The graded version of the wedge product for degree-1 elements is given by
\begin{align*}
\wedge: \Largewedge^1 (M)\times \Largewedge^1 (M) &\longrightarrow \Largewedge^{2} (M)\\
(m_1, m_2) &\longmapsto m_1\wedge m_2 := (m_1\otimes m_2) + I_{\wedge}(M)\cap T^{2}(M)
\end{align*}
We can prove $\Largewedge^{k} (M)$ is an $R$-module generated by the elements of the form
\[
m_1\wedge m_2\wedge \cdots \wedge m_k = (m_1\otimes m_2\otimes \cdots \otimes m_k) + I_{\wedge}(M)\cap T^k(M)
\]
for $m_1, m_2, \cdots, m_k \in M$. The wedge product for degree-$k$ and degree-$l$ elements is given by
\begin{align*}
\wedge: \Largewedge^k (M)\times \Largewedge^l (M) &\longrightarrow \Largewedge^{k+l} (M)\\
(m_1\wedge\cdots\wedge m_k,\, m_{k+1}\wedge\cdots\wedge m_{k+l}) &\longmapsto m_1\wedge\cdots\wedge m_k \wedge m_{k+1}\wedge\cdots\wedge m_{k+l}
\end{align*}
where $m_i\in M$ for $1\le i \le k+l$.
\end{definition}
\begin{remark}
Since $T^k(M)$ is an $R$-module generated by the pure tensors of the form $m_1\otimes m_2\otimes \cdots \otimes m_k$ for $m_i\in M$, the quotient module $\Largewedge^k(M)=T^k(M)/\left(I_{\largewedge}(M)\cap T^k(M)\right)$ is generated by the elements of the form $(m_1\otimes m_2\otimes \cdots \otimes m_k) + I_{\wedge}(M)\cap T^k(M)$. And by induction on $k$, we can show that
\[
m_1 \wedge m_2 \wedge \cdots m_k = (m_1\otimes m_2\otimes \cdots \otimes m_k) + I_{\wedge}(M)\cap T^k(M).
\]
\end{remark}
\begin{proposition}{}{}
Given an $R$-module $M$ and $m_1, m_2 \in M$, we have
\[
m_1 \wedge m_2 = - m_2 \wedge m_1.
\]
For any homogeneous elements $x, y \in \Largewedge^{\bullet}(M)$ , we have
\[
x \wedge y = (-1)^{\deg(x)\deg(y)} y \wedge x.
\]
\end{proposition}
\begin{definition}{Exterior Algebra Functor: $\Largewedge^{\bullet}:R\text{-}\mathsf{Mod}\to \mathsf{GrAlg}^{\mathbb{N}}_R$}{}
The exterior algebra construction $\Largewedge^{\bullet}:R\text{-}\mathsf{Mod}\to \mathsf{GrAlg}^{\mathbb{N}}_R$ is a functor defined as follows
\[
\begin{tikzcd}[ampersand replacement=\&]
R\text{-}\mathsf{Mod}\&[-25pt]\&[+10pt]\&[-30pt] \mathsf{GrAlg}^{\mathbb{N}}_R\&[-30pt]\&[-30pt] \\ [-15pt]
M \arrow[dd, "f"{name=L, left}]
\&[-25pt] \& [+10pt]
\& [-30pt] \Largewedge^{\bullet}(M)\arrow[dd, "\Largewedge^{\bullet}(f)"{name=R}] \&[-20pt]\ni\& [+20pt]m_1\wedge\cdots\wedge m_k \arrow[dd, mapsto, "f\wedge f\wedge\cdots\wedge f"{name=L, right}]
\\ [-10pt]
\& \phantom{.}\arrow[r, "\Largewedge^{\bullet}", squigarrow]\&\phantom{.} \& \\[-10pt]
N \& \& \& \Largewedge^{\bullet}(N)\&[-0pt]\ni\& f(m_1)\wedge\cdots\wedge f(m_k)
\end{tikzcd}
\]
where $\Largewedge^{\bullet}(f)$ is induced by the \hyperref[th:universal_property_of_quotient_algebra]{universal property of the quotient algebra} $T^{\bullet}(M)/I_\wedge(M)$ through the following commutative diagram
\[
\begin{tikzcd}
T^{\bullet}(M) \arrow[r, "T^{\bullet}(f)"] \arrow[d, "\pi_M"'] &[1em] T^{\bullet}(N) \arrow[d, "\pi_N"] \\[2em]
T^{\bullet}(M)/I_\wedge(M) \arrow[r, "\wedge^\bullet(f)"',dashed] & T^{\bullet}(N)/I_\wedge(N)
\end{tikzcd}
\]
\end{definition}
\begin{remark}
Since for any $x\in M$,
\begin{align*}
\pi_N\circ T^{\bullet}(f)(x\otimes x) &= \pi_N(f(x)\otimes f(x))=f(x)\wedge f(x) + I_\wedge(N)=0+ I_\wedge(N),
\end{align*}
we see each generator of $I_\wedge(M)$ is mapped to $0$ in $\pi_N\circ T^{\bullet}(f)$. Thus, we have $I_\wedge(M)\subseteq \ker(\pi_N\circ T^{\bullet}(f))$, which guarantees that there exists a unique $R$-algebra homomorphism
\begin{align*}
\Largewedge^{\bullet}(f): \Largewedge^{\bullet}(M) &\longrightarrow \Largewedge^{\bullet}(N)\\
m_1\wedge\cdots\wedge m_k &\longmapsto f(m_1)\wedge\cdots\wedge f(m_k).
\end{align*}
such that $\Largewedge^{\bullet}(f)\circ \pi_M = \pi_N \circ T^{\bullet}(f)$.
\end{remark}
\begin{proposition}{Adjunction $\Largewedge^{\bullet}\dashv U_{\mathsf{R\text{-}\mathsf{Mod}}}$}{}
Let $R$ be a commutative ring. Suppose $U_{\mathsf{R\text{-}\mathsf{Mod}}}:\mathsf{GrAlg}^{\mathbb{N}}_R\to R\text{-}\mathsf{Mod}$ is the forgetful functor. Then the exterior algebra functor $\Largewedge^{\bullet}:R\text{-}\mathsf{Mod}\to \mathsf{GrAlg}^{\mathbb{N}}_R$ is left adjoint to $U$.
\end{proposition}
\begin{example}{Take Degree $k$ Functor}{take_degree_k_functor}
Let $R$ be a commutative ring and $M$ be an $R$-module. The \textbf{take degree $k$ functor} is defined as
\[
\begin{tikzcd}[ampersand replacement=\&]
\mathsf{GrAlg}^{\mathbb{N}}_R\&[-25pt]\&[+10pt]\&[-30pt]R\text{-}\mathsf{Mod} \\ [-15pt]
A=\bigoplus\limits_{i=0}^{\infty} A_i\arrow[dd, "g"{name=L, left}]
\&[-25pt] \& [+10pt]
\& [-30pt] A_k\arrow[dd, "g|_{A_k}"{name=R}] \\ [-10pt]
\& \phantom{.}\arrow[r, "\left(-\right)_k", squigarrow]\&\phantom{.} \& \\[-10pt]
B=\bigoplus\limits_{i=0}^{\infty} B_i \& \& \& B_k
\end{tikzcd}
\]
In particular, we have the composition functor $\Largewedge^{k} := (-)_k \circ \bigwedge^\bullet : R\text{-}\mathsf{Mod}\to R\text{-}\mathsf{Mod}$ defined as follows
\[
\begin{tikzcd}[ampersand replacement=\&]
R\text{-}\mathsf{Mod}\&[-25pt]\&[+10pt]\&[-30pt]R\text{-}\mathsf{Mod} \&[-30pt]\&[-30pt] \\ [-15pt]
M \arrow[dd, "f"{name=L, left}]
\&[-25pt] \& [+10pt]
\& [-30pt] \Largewedge^{k}(M)\arrow[dd, "\Largewedge^{k}(f)"{name=R}] \&[-20pt]\ni\& [+20pt]m_1\wedge\cdots\wedge m_k \arrow[dd, mapsto, "f\wedge f\cdots\wedge f"{name=L, right}]
\\ [-10pt]
\& \phantom{.}\arrow[r, "\Largewedge^{k}", squigarrow]\&\phantom{.} \& \\[-10pt]
N \& \& \& \Largewedge^{k}(N)\&[-0pt]\ni\& f(m_1)\wedge\cdots\wedge f(m_k)
\end{tikzcd}
\]
\end{example}
\begin{proposition}{}{}
Suppose $R$ is a commutative ring and $M=\bigoplus_{x\in X}Rx$ is a free $R$-module. Then
\begin{enumerate}[(i)]
\item $\Largewedge^{\bullet}(M)$ has a basis $\{x_1\wedge\cdots\wedge x_k: x_1,\cdots,x_k\in X, x_i\ne x_j\text{ for all }i\ne j\}$.
\item If $M$ has a basis $\{x_1,\cdots,x_n\}$, then we have an $R$-linear isomorphism
\begin{align*}
\largewedge^{n}(M)&\xlongrightarrow{\sim} R\\
x_1 \wedge \cdots \wedge x_n& \longmapsto 1_R.
\end{align*}
Moreover, we have $\Largewedge^{m}(M)=0$ for all $m>n$.
\end{enumerate}
\end{proposition}
\section{Integral Element}
\begin{definition}{Integral Element}{}
Let $R$ be a commutative ring and $A$ be an $R$-algebra with structure homomorphism $\varphi:R\to Z(A)$. An element $x\in A$ is called \textbf{integral} over $R$ if there exists a monic polynomial $f\in R[T]$ such that $\leftindex^{\varphi}\!f(x)=0$.
\end{definition}
\begin{definition}{Generated Subalgebra}{generated_subalgebra}
Let $R$ be a commutative ring and $A$ be an $R$-algebra. By the universal property of $R\langle T\rangle$, there exists a unique $R$-algebra homomorphism $\psi:R\langle T\rangle\to A$ such that
$\psi(T)=x$.
\begin{center}
\begin{tikzcd}[ampersand replacement=\&]
R\langle T\rangle\arrow[r, dashed, "\exists !\,\psi"] \& A\\[0.3cm]
\{T\}\arrow[u, "\iota"] \arrow[ru, "\mathrm{const}_x"'] \&
\end{tikzcd}
\end{center}
The \textbf{$R$-subalgebra of $A$ generated by $x$} is defined as
\[
R[x]:=\psi\left(R\langle T\rangle\right)=\left\{\sum_{k=0}^n r_k x^k \in A\;\middle|\; r_k\in R\right\}.
\]
\end{definition}
\begin{proposition}{Equivalent Definition of Integral Element}{equivalent_definition_of_integral_element}
Let $R$ be a commutative ring and $A$ be an $R$-algebra. Let $R[x]$ be the \hyperref[th:generated_subalgebra]{$R$-subalgebra of $A$ generated by $x$}. Then $A$ is an $R[x]$-module. And the following statements are equivalent:
\begin{enumerate}[(i)]
\item $x$ is integral over $R$.
\item $R[x]$ is a finitely generated $R$-module.
\item There exists a faithful $R[x]$-submodule of $A$ that is finitely generated as an $R$-module and contains $x$.
\end{enumerate}
\end{proposition}
\section{Trace and Norm}
\begin{lemma}{Left Multiplication Endomorphism}{}
Let $R$ be a commutative ring and $A$ be an $R$-algebra. For any $a\in A$, we can define the left multiplication endomorphism $l_a\in\mathrm{End}_{R\text{-}\mathsf{Mod}}(A)$ by
\begin{align*}
l_a:A &\longrightarrow A\\
x &\longmapsto ax.
\end{align*}
Moreover,
\begin{align*}
l_{-}:A &\longrightarrow \mathrm{End}_{R\text{-}\mathsf{Mod}}(A)\\
a &\longmapsto l_a
\end{align*}
is an $R$-algebra homomorphism.
\end{lemma}
\begin{prf}
For any $r\in R$, $a,b\in A$ and $x\in A$, we have
\begin{align*}
l_{ra+b}(x)&=(ra+b)x=r(ax)+bx=(ra)x+bx=l_{ra}(x)+l_b(x),\\
l_{ab}(x)&=(ab)x=a(bx)=l_a(l_b(x)),\\
l_{1_A}(x)&=1_Ax=x.
\end{align*}
Hence $l_{-}$ is an $R$-algebra homomorphism.
\end{prf}
\begin{definition}{Trace, Norm and Characteristic Polynomial}{trace_norm_and_characteristic_polynomial}
Let $R$ be a commutative ring and $A$ be an $R$-algebra. Suppose $A$ as an $R$-module is free of finite rank. For any $a\in A$, we can define the left multiplication endomorphism $l_a\in\mathrm{End}_{R\text{-}\mathsf{Mod}}(A)$ by
\begin{align*}
l_a:A &\longrightarrow A\\
x &\longmapsto ax.
\end{align*}
\begin{itemize}
\item The \textbf{trace} of $a\in A$ is defined as the trace of $l_a$, denoted by
\[
\mathrm{Tr}_{A|R}(a):=\mathrm{Tr}(l_a) \in R.
\]
That is, $\mathrm{Tr}_{A|R}:A\to R$ is an $R$-module homomorphism through the following composition
\[
\mathrm{Tr}_{A|R}: A\xrightarrow{l_{-}} \mathrm{End}_{R\text{-}\mathsf{Mod}}(A)\xrightarrow{\mathrm{Tr}} R.
\]
\item The \textbf{norm} of $a\in A$ is defined as the determinant of $l_a$, denoted by
\[
\mathrm{N}_{A|R}(a):=\det(l_a) \in R.
\]
That is, $\mathrm{N}_{A|R}:A\to R$ is a multiplicative monoid homomorphism through the following composition
\[
\mathrm{N}_{A|R}: A\xrightarrow{l_{-}} \mathrm{End}_{R\text{-}\mathsf{Mod}}(A)\xrightarrow{\det} R.
\]
\item The \textbf{characteristic polynomial} of $a\in A$ is defined as the characteristic polynomial of $l_a$, denoted by
\[
\mathrm{char}_{A|R}(a; X):=\mathrm{char}(l_a; X)=\det(X\cdot\mathrm{id}_A-l_a)= \mathrm{N}_{A[X]|R[X]}(X-a)\in R[X]
\]
\end{itemize}
\end{definition}
\begin{proposition}{Trace, Norm, and Characteristic Polynomial under Change of Base Ring}{trace_norm_and_characteristic_polynomial_under_change_of_base_ring}
Let $R$ be a commutative ring, $A$ be an commutative $R$-algebra and $M$ be a free $R$-module of finite rank. Suppose $A$ as an $R$-module is free of finite rank. Given any $A$-linear transformation $\varphi\in\mathrm{End}_{A\text{-}\mathsf{Mod}}(M)$, by applying the functor of restriction of scalars to $R\to A$, we can regard $\varphi$ as an $R$-linear transformation on $M$ through $\mathrm{End}_{A\text{-}\mathsf{Mod}}(M)\hookrightarrow\mathrm{End}_{R\text{-}\mathsf{Mod}}(M)$. And we have
\begin{enumerate}[(i)]
\item $\mathrm{Tr}_{R}(\varphi)=\mathrm{Tr}_{A|R}(\mathrm{Tr}_{A}(\varphi))$.
\item $\mathrm{N}_{R}(\varphi)=\mathrm{N}_{A|R}(\mathrm{N}_{A}(\varphi))$.
\item $\mathrm{char}_{R}(\varphi;X)=\mathrm{N}_{A[X]|R[X]}(\mathrm{char}_{A}(\varphi;X))$.
\end{enumerate}
\end{proposition}
\begin{corollary}{}{}
Let $R$ be a commutative ring, $A$ be an commutative $R$-algebra and $B$ be an $A$-algebra. Suppose $A$ as an $R$-module is free of finite rank and $B$ as an $A$-module is free of finite rank. Then for any $b\in B$, we have
\begin{enumerate}[(i)]
\item $\mathrm{Tr}_{B|R}(b)=\mathrm{Tr}_{A|R}(\mathrm{Tr}_{B|A}(b))$.
\item $\mathrm{N}_{B|R}(b)=\mathrm{N}_{A|R}(\mathrm{N}_{B|A}(b))$.
\item $\mathrm{char}_{B|R}(b;X)=\mathrm{N}_{A[X]|R[X]}(\mathrm{char}_{B|A}(b;X))$.
\end{enumerate}
\end{corollary}
\begin{prf}
This is a direct consequence of \Cref{th:trace_norm_and_characteristic_polynomial_under_change_of_base_ring} by taking $M=A$ and $\varphi=l_b$.
\end{prf}
\begin{definition}{Trace Pairing}{trace_pairing}
Let $R$ be a commutative ring and $A$ be an $R$-algebra. Suppose $A$ as an $R$-module is free of finite rank. The \textbf{trace pairing} is the symmetric $R$-bilinear form
\begin{align*}
\langle \cdot,\cdot \rangle_{A|R}:A\times A&\longrightarrow R\\
(x,y)&\longmapsto \mathrm{Tr}_{A|R}(xy).
\end{align*}
\end{definition}
\subsection{Discriminant}
We first define the discriminant of a polynomial.
\begin{definition}{Resultant}{}
Let $R$ be a commutative ring and
\begin{align*}
f(X)&=a_n X^n + a_{n-1}X^{n-1} + \cdots + a_1 X + a_0 \in R[X],\\
g(X)&=b_m X^m + b_{m-1}X^{m-1} + \cdots + b_1 X + b_0 \in R[X]
\end{align*}
be two polynomials of degree $n$ and $m$ respectively. The \textbf{resultant} of $f(X)$ and $g(X)$ is defined as
\[
\mathrm{Res}(f,g):= \det \left[
\begin{array}{c@{\;}c}
\underbrace{
\begin{matrix}
a_n & 0 & \cdots & 0 \\
a_{n-1} & a_n & \cdots & 0 \\
a_{n-2} & a_{n-1} & \ddots & \vdots \\
\vdots & \vdots & \ddots & a_n \\
a_0 & a_1 & \cdots & a_{n-1} \\
0 & a_0 & \cdots & a_{n-2} \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_0
\end{matrix}
}_{m \text{ columns}}
&
\underbrace{
\begin{matrix}
b_m & 0 & \cdots & 0 \\
b_{m-1} & b_m & \cdots & 0 \\
b_{m-2} & b_{m-1} & \ddots & \vdots \\
\vdots & \vdots & \ddots & b_m \\
b_0 & b_1 & \cdots & b_{m-1} \\
0 & b_0 & \cdots & b_{m-2} \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & b_0
\end{matrix}
}_{n \text{ columns}}
\end{array}
\right]_{(m+n) \times (m+n)}.
\]
If $R$ is an integral domain with field of fractions $K$ and $f(X),g(X)$ have roots $\alpha_1,\alpha_2,\cdots,\alpha_n$ and $\beta_1,\beta_2,\cdots,\beta_m$ in some algebraic closure $\overline{K}$ of $K$ respectively, then
\[
\mathrm{Res}(f,g)=a_n^m b_m^n \prod_{i=1}^n \prod_{j=1}^m (\alpha_i - \beta_j)=a_n^m \prod_{i=1}^n g(\alpha_i)=(-1)^{mn} b_m^n \prod_{j=1}^m f(\beta_j).
\]
\end{definition}
\begin{remark}
Suppose $R$ is an integral domain with field of fractions $K$ and $f(X),g(X)$ have roots $\alpha_1,\alpha_2,\cdots,\alpha_n$ and $\beta_1,\beta_2,\cdots,\beta_m$ in some algebraic closure $\overline{K}$ of $K$ respectively. Then we have
\begin{align*}
\frac{1}{a_n}f(X)&=(X-\alpha_1)(X-\alpha_2)\cdots (X-\alpha_n),\\
\frac{1}{b_m}g(X)&=(X-\beta_1)(X-\beta_2)\cdots (X-\beta_m).
\end{align*}
Let
\[
V =
\begin{bmatrix}
\alpha_1^{m+n-1} & \alpha_1^{m+n-2} & \cdots & \alpha_1 & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
\alpha_n^{m+n-1} & \alpha_n^{m+n-2} & \cdots & \alpha_n & 1 \\
\beta_1^{m+n-1} & \beta_1^{m+n-2} & \cdots & \beta_1 & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
\beta_m^{m+n-1} & \beta_m^{m+n-2} & \cdots & \beta_m & 1
\end{bmatrix},\quad S=\begin{bmatrix}
a_n & 0 & \cdots & 0 & b_m & 0 & \cdots & 0 \\
a_{n-1} & a_n & \cdots & 0 & b_{m-1} & b_m & \cdots & 0 \\
a_{n-2} & a_{n-1} & \ddots & \vdots & b_{m-2} & b_{m-1} & \ddots & \vdots \\
\vdots & \vdots & \ddots & a_n & \vdots & \vdots & \ddots & b_m \\
a_0 & a_1 & \cdots & a_{n-1} & b_0 & b_1 & \cdots & b_{m-1} \\
0 & a_0 & \cdots & a_{n-2} & 0 & b_0 & \cdots & b_{m-2} \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_0 & 0 & 0 & \cdots & b_0
\end{bmatrix}.
\]
Then
\[
{\renewcommand{\arraystretch}{1.5}}
V S = \begin{bmatrix}
\alpha_1^{m-1}f(\alpha_1) & \alpha_1^{m-2}f(\alpha_1) & \cdots & f(\alpha_1) & \alpha_1^{n-1}g(\alpha_1) & \alpha_1^{n-2}g(\alpha_1) &\cdots & g(\alpha_1) \\
\alpha_2^{m-1}f(\alpha_2) & \alpha_2^{m-2}f(\alpha_2) & \cdots & f(\alpha_2) & \alpha_2^{n-1}g(\alpha_2) & \alpha_2^{n-2}g(\alpha_2) &\cdots & g(\alpha_2) \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
\alpha_n^{m-1}f(\alpha_n) & \alpha_n^{m-2}f(\alpha_n) & \cdots & f(\alpha_n) & \alpha_n^{n-1}g(\alpha_n) & \alpha_n^{n-2}g(\alpha_n) &\cdots & g(\alpha_n) \\
\beta_1^{m-1}f(\beta_1) & \beta_1^{m-2}f(\beta_1) & \cdots & f(\beta_1) & \beta_1^{n-1}g(\beta_1) & \beta_1^{n-2}g(\beta_1) &\cdots & g(\beta_1) \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
\beta_m^{m-1}f(\beta_m) & \beta_m^{m-2}f(\beta_m) & \cdots & f(\beta_m) & \beta_m^{n-1}g(\beta_m) & \beta_m^{n-2}g(\beta_m) &\cdots & g(\beta_m)
\end{bmatrix}=\begin{bmatrix}
\mathbf{0}_{n\times m} & D_g \\
D_f & \mathbf{0}_{m\times n}.
\end{bmatrix}
\]
This implies that
\begin{align*}
\det(V)\det(S)&=\det(V S)\\
&=(-1)^{mn}\det(D_g)\det(D_f)\\
&=(-1)^{mn} \left(\prod_{i=1}^ng(\alpha_i)\prod_{1\le i<j\le n}(\alpha_i-\alpha_j)\right) \left(\prod_{j=1}^m f(\beta_j)\prod_{1\le i<j\le m}(\beta_i-\beta_j)\right).
\end{align*}
Since $\det(V)$ is a Vandermonde determinant, we have
\[
\det(V)=\prod_{1\le i<j\le n}(\alpha_i-\alpha_j)\prod_{1\le i<j\le m}(\beta_i-\beta_j)\prod_{i=1}^n\prod_{j=1}^m(\alpha_i-\beta_j)
\]
Thus
\[
\det(S)=\frac{(-1)^{mn}\displaystyle\prod_{i=1}^n g(\alpha_i)\prod_{j=1}^m f(\beta_j)}{\displaystyle\prod_{i=1}^n\prod_{j=1}^m(\alpha_i-\beta_j)}.
\]
Note
\begin{align*}
&\;\prod_{i=1}^n\prod_{j=1}^m(\alpha_i-\beta_j)=\prod_{i=1}^n \frac{g(\alpha_i)}{b_m}=\frac{1}{b_m^n} \prod_{i=1}^n g(\alpha_i)\\
=&\;(-1)^{mn}\prod_{j=1}^m\prod_{i=1}^n(\beta_j-\alpha_i)=(-1)^{mn}\prod_{j=1}^m \frac{f(\beta_j)}{a_n}=\frac{(-1)^{mn}}{a_n^m} \prod_{j=1}^m f(\beta_j).
\end{align*}
We can conclude that
\[
\det(S)=a_n^m b_m^n \prod_{i=1}^n \prod_{j=1}^m (\alpha_i - \beta_j)=a_n^m \prod_{i=1}^n g(\alpha_i)=(-1)^{mn} b_m^n \prod_{j=1}^m f(\beta_j).
\]
\end{remark}
\begin{definition}{Discriminant of Polynomial}{discriminant_of_polynomial}
Let $R$ be an integral domain and $f(X)\in R[X]$ be a polynomial. The \textbf{discriminant} of $f(X)$ is defined as
\[
\mathrm{Disc}(f):=(-1)^{\frac{n(n-1)}{2}}\frac{\mathrm{Res}(f,f')}{a_n}.
\]
\end{definition}
\begin{proposition}{}{}
Let $K$ be a field and $f(X)\in K[X]$ be a polynomial of degree $n$ with roots $\alpha_1,\alpha_2,\cdots,\alpha_n$ in some algebraic closure $\overline{K}$ of $K$. Then
\[
\mathrm{Disc}(f)=(-1)^{\frac{n(n-1)}{2}} a_n^{n-2} \prod_{i=1}^n f'(\alpha_i) = (-1)^{\frac{n(n-1)}{2}} a_n^{2n-2} \prod_{\substack{i,j \\ i \ne j}}(\alpha_i-\alpha_j)=a_n^{2n-2}\prod_{1\le i<j\le n}(\alpha_i-\alpha_j)^2.
\]
\end{proposition}
\begin{prf}
By definition of resultant, we have
\[
\mathrm{Disc}(f)=(-1)^{\frac{n(n-1)}{2}}\frac{\mathrm{Res}(f,f')}{a_n}=(-1)^{\frac{n(n-1)}{2}}\frac{a_n^{n-1} \prod\limits_{i=1}^n f'(\alpha_i)}{a_n}=(-1)^{\frac{n(n-1)}{2}} a_n^{n-2} \prod_{i=1}^n f'(\alpha_i).
\]
Since
\begin{align*}
f'(X)&=\frac{\mathrm{d}}{\mathrm{d}X}\left(a_n \prod_{i=1}^n (X - \alpha_i) \right)= a_n \sum_{i=1}^n \prod_{j\ne i} (X - \alpha_j)\implies f'(\alpha_i)= a_n \prod_{j\ne i} (\alpha_i - \alpha_j),
\end{align*}
we have
\[
\prod_{i=1}^n f'(\alpha_i)= \prod_{i=1}^n \left( a_n \prod_{j\ne i} (\alpha_i - \alpha_j) \right)= a_n^{n} \prod_{\substack{i,j \\ i \ne j}} (\alpha_i - \alpha_j).
\]
Thus
\[
\mathrm{Disc}(f)=(-1)^{\frac{n(n-1)}{2}} a_n^{n-2} \prod_{i=1}^n f'(\alpha_i)=(-1)^{\frac{n(n-1)}{2}} a_n^{2n-2} \prod_{\substack{i,j \\ i \ne j}} (\alpha_i - \alpha_j).
\]
Note
\begin{align*}
\prod_{\substack{i,j \\ i \ne j}} (\alpha_i - \alpha_j)&=\prod_{1\le i<j\le n} (\alpha_i - \alpha_j)(\alpha_j - \alpha_i)\\
&=\prod_{1\le i<j\le n} -(\alpha_i - \alpha_j)^2\\
&=(-1)^{\frac{n(n-1)}{2}} \prod_{1\le i<j\le n} (\alpha_i - \alpha_j)^2.
\end{align*}
We get
\[
\mathrm{Disc}(f)=(-1)^{\frac{n(n-1)}{2}} a_n^{2n-2} \prod_{\substack{i,j \\ i \ne j}} (\alpha_i - \alpha_j)= a_n^{2n-2}\prod_{1\le i<j\le n}(\alpha_i-\alpha_j)^2.
\]
\end{prf}
\begin{definition}{Discriminant of Bilinear Form $M\times M\to R$}{}
Let $R$ be a commutative ring and $M$ be an free $R$-module of rank $n$. Let $Q:M\times M\to R$ be a $R$-bilinear form. Then
\begin{align*}
\varphi_Q:M &\longrightarrow M^*\\
x &\longmapsto Q(x,-)
\end{align*}
is an $R$-module homomorphism. Taking top exterior powers gives an $R$-module homomorphism
\begin{align*}
\Largewedge^{n}(\varphi_Q):\Largewedge^{n}(M) &\longrightarrow \Largewedge^{n}(M^*)\\
x_1\wedge\cdots\wedge x_n &\longmapsto \varphi_Q(x_1)\wedge\cdots\wedge \varphi_Q(x_n).
\end{align*}
Note we have natural isomorphisms
\begin{align*}
\psi_M:\Largewedge^{n}(M^*)&\xlongrightarrow{\sim} (\Largewedge^{n}(M))^*\\
f_1\wedge\cdots\wedge f_n &\longmapsto \left( x_1\wedge\cdots\wedge x_n \mapsto \det\left( [f_i(x_j)]_{1\le i,j\le n} \right) \right),
\end{align*}
By composing $\Largewedge^{n}(\varphi_Q)$ with $\psi_M$, we get an $R$-module endomorphism of the rank-$1$ free $R$-module $\Largewedge^{n}(M)$
\begin{align*}
\theta:=\psi_M \circ \Largewedge^{n}(\varphi_Q):\Largewedge^{n}(M) &\longrightarrow (\Largewedge^{n}(M))^*\\
x_1\wedge\cdots\wedge x_n &\longmapsto \left(x_1\wedge\cdots\wedge x_n\mapsto \det\left( [Q(x_i,x_j)]_{1\le i,j\le n} \right) \right).
\end{align*}
Let $\varepsilon$ be a basis of $\Largewedge^{n}(M)$ and
\begin{align*}
\varepsilon^*:\Largewedge^{n}(M) &\longrightarrow R\\
\varepsilon &\longmapsto 1_R.
\end{align*}
be the dual basis of $\varepsilon$. Then there exists a unique $\lambda\in R$ such that
\[
\theta(\varepsilon) = \lambda \cdot \varepsilon^*.
\]
Then the \textbf{discriminantof $Q$} is defined as the image of $\lambda$ under the canonical projection $R\to R/(R^{\times})^2$
\[
\mathrm{Disc}(Q):=\lambda \mod (R^{\times})^2 \in R/(R^{\times})^2,
\]
where $R/(R^{\times})^2$ is the quotient monoid of the multiplicative monoid $R$ by modulo the congruence relation
\[
a \sim b \iff \exists u\in R^{\times}, a = u^2 b.
\]
We can check that this definition is independent of the choice of basis $\varepsilon$. If we choose another basis $\tilde{\varepsilon} = c \varepsilon$ for some $c\in R^{\times}$, then $\tilde{\varepsilon}^* = c^{-1} \varepsilon^*$ and
\[
\theta(\tilde{\varepsilon}) = \theta(c \varepsilon) = c \theta(\varepsilon) = c \lambda \varepsilon^* =c \lambda (c \tilde{\varepsilon}^*) = (c^2 \lambda)\cdot \tilde{\varepsilon}^*.
\]
If $E=\{e_1,e_2,\cdots,e_n\}$ is a basis of $M$, then the \textbf{discriminant of $Q$ with respect to the basis $\{e_1,e_2,\cdots,e_n\}$} is defined as
\[
\mathrm{Disc}(Q; e_1,e_2,\cdots,e_n):=\det\left( [Q(e_i,e_j)]_{1\le i,j\le n} \right)\in R.
\]
The relation between $\mathrm{Disc}(Q)$ and $\mathrm{Disc}(Q; e_1,e_2,\cdots,e_n)$ is given by
\[
\mathrm{Disc}(Q)=\mathrm{Disc}(Q; e_1,e_2,\cdots,e_n) \mod (R^{\times})^2 \in R/(R^{\times})^2.
\]
The ideal of $R$ generated by $\mathrm{Disc}(Q; e_1,e_2,\cdots,e_n)$ is called the \textbf{discriminant ideal of $Q$}, which is independent of the choice of basis $\{e_1,e_2,\cdots,e_n\}$.
\end{definition}
\begin{remark}
Here we check the relation between $\mathrm{Disc}(Q)$ and $\mathrm{Disc}(Q; e_1,e_2,\cdots,e_n)$. Let $\langle \cdot,\cdot \rangle :M^*\times M \to R$ be the dual pairing defined by $\langle f,x \rangle := f(x)$. Note $\varepsilon:=e_1\wedge e_2 \wedge \cdots \wedge e_n$ is a basis of $\Largewedge^{n}(M)$. Then we have
\[
\langle\theta(\varepsilon),\varepsilon \rangle= \langle \lambda\varepsilon^*,\varepsilon \rangle =\lambda \langle \varepsilon^*,\varepsilon \rangle = \lambda.
\]
Denote
\[
f_i:=\varphi_Q(e_i) \in M^*.
\]
We have
\begin{align*}
\langle\theta(\varepsilon),\varepsilon \rangle &= \langle \psi_M(f_1\wedge f_2 \wedge \cdots \wedge f_n), e_1\wedge e_2 \wedge \cdots \wedge e_n \rangle \\
&= \det\left( [f_i(e_j)]_{1\le i,j\le n} \right) \\
&= \det\left( [\varphi_Q(e_i)(e_j)]_{1\le i,j\le n} \right) \\
&= \det\left( [Q(e_i,e_j)]_{1\le i,j\le n} \right).
\end{align*}
\end{remark}
\begin{definition}{Discriminant of Algebra}{discriminant_of_algebra}
Let $R$ be a commutative ring and $A$ be an $R$-algebra. Suppose $A$ is a free $R$-module of rank $n$. Then the \textbf{discriminant} of $A$ over $R$ is defined as
\[
d_A=\mathrm{Disc}\left(\langle \cdot ,\cdot \rangle_{A|R}\right)
\]
where $\langle \cdot ,\cdot \rangle_{A|R}$ is the \hyperref[th:trace_pairing]{trace pairing} of $A$ over $R$.
Assume $\{e_1,e_2,\cdots,e_n\}$ is a basis of $A$ over $R$. The \textbf{discriminant} of $A$ with respect to the basis $\{e_1,e_2,\cdots,e_n\}$ is defined as
\begin{align*}
d_A(e_1,e_2,\cdots,e_n):&=\det\left( \left[ \langle e_i,e_j \rangle_{A|R} \right]_{1\le i,j\le n} \right)\\
&= \det \begin{bmatrix}
\mathrm{Tr}_{A|R}(e_1 e_1) & \mathrm{Tr}_{A|R}(e_1 e_2) & \cdots & \mathrm{Tr}_{A|R}(e_1 e_n) \\
\mathrm{Tr}_{A|R}(e_2 e_1) & \mathrm{Tr}_{A|R}(e_2 e_2) & \cdots & \mathrm{Tr}_{A|R}(e_2 e_n) \\
\vdots & \vdots & \ddots & \vdots \\
\mathrm{Tr}_{A|R}(e_n e_1) & \mathrm{Tr}_{A|R}(e_n e_2) & \cdots & \mathrm{Tr}_{A|R}(e_n e_n)
\end{bmatrix}\in R.
\end{align*}
The relation between $d_A$ and $d_A(e_1,e_2,\cdots,e_n)$ is given by
\begin{align*}
d_A=d_A(e_1,e_2,\cdots,e_n)\mod (R^{\times})^2.
\end{align*}
The ideal of $R$ generated by $d_A(e_1,e_2,\cdots,e_n)$ is called the \textbf{discriminant ideal
of $A$ over $R$} and is denoted by $\mathfrak{d}_A$.
\end{definition}
\begin{definition}{Order}{}
Let $R$ be an integral domain with field of fractions $K$ and $A$ be a finite-dimensional $K$-algebra. An $R$-\textbf{order} in $A$ is a unital $R$-subalgebra $\mathcal{O}\subseteq A$ which is finitely generated as an $R$-module and satisfies $\mathcal{O}\otimes_R K \cong A$.
\end{definition}
\begin{remark}
Since the $K$-algebra $A$ can be regarded as an $R$-algebra through the restriction of scalars along $R\hookrightarrow K$, it makes sense to talk about $R$-subalgebras of $A$.
\end{remark}
\begin{lemma}{}{discriminant_relation}
Let $R$ be a commutative ring, $M$ be a free $R$-module of rank $n$ and $N\subseteq M$ be a submodule which is also a free $R$-module of rank $n$. Let $Q:M\times M\to R$ be a $R$-bilinear form.
\begin{enumerate}[(i)]
\item Given any basis $\{e_1,e_2,\cdots,e_n\}$ of $M$ over $R$ and any basis $\{f_1,f_2,\cdots,f_n\}$ of $N$ over $R$, there exists a matrix $P\in M_n(R)$ such that
\[
\begin{bmatrix}
f_1 & f_2 & \cdots & f_n
\end{bmatrix} = \begin{bmatrix}
e_1 & e_2 & \cdots & e_n
\end{bmatrix} P.
\]
The discriminants with respect to the bases $\{e_1,e_2,\cdots,e_n\}$ and $\{f_1,f_2,\cdots,f_n\}$ satisfy
\[
\mathrm{Disc}\left(Q|_{N\times N};f_1,f_2,\cdots,f_n\right)=\det(P)^2\; \mathrm{Disc}\left(Q;e_1,e_2,\cdots,e_n\right).
\]
\item Let $\mathrm{Fitt}_0(M/N)$ be the 0-th Fitting ideal of the $R$-module $M/N$. Then we have equality of ideals
\[
\left(\mathrm{Disc}(Q|_{N\times N})\right)= \mathrm{Fitt}_0(M/N)^2 \left(\mathrm{Disc}(Q)\right).
\]
\end{enumerate}
\end{lemma}
\begin{prf}
\begin{enumerate}[(i)]
\item Let $\{e_1,e_2,\cdots,e_n\}$ be a basis of $M$ over $R$. Since $N$ is a unital $R$-submodule of $M$ which is also a free $R$-module of rank $n$, we can assume that $\{f_1,f_2,\cdots,f_n\}$ is a basis of $N$ over $R$. Note each $f_i$ can be expressed as a linear combination of $e_1,e_2,\cdots,e_n$. Thus there exists a matrix $M\in M_n(R)$ such that
\[
\begin{bmatrix}
f_1 & f_2 & \cdots & f_n
\end{bmatrix} = \begin{bmatrix}
e_1 & e_2 & \cdots & e_n
\end{bmatrix} P.
\]
Thus we have
\begin{align*}
\mathrm{Disc}\left(Q|_{N\times N};f_1,f_2,\cdots,f_n\right)&=\det\left( \left[ \langle f_i,f_j \rangle_{Q|_{N\times N}} \right]_{1\le i,j\le n} \right)\\