Add field theory and bilinear form enhancements

Expanded field theory content with definitions and propositions on algebraic and separable extensions, minimal polynomials, separable degree, and trace pairing. Improved vector space notes with bilinear map, bilinear form, and nondegeneracy definitions and characterizations. Updated .gitignore for additional LaTeX build artifacts. Refined example box styling in preamble.tex. Minor corrections and clarifications in associative_algebra.tex and field.tex.

Author: hooyuser

.gitignore

@@ -8,6 +8,10 @@
 *.thm
 *.auxlock
 *.mmz
+*.idx
+*.ilg
+*.ind
+*.bbl
 
 .history/*
 .idea/*

associative_algebra.tex

@@ -402,18 +402,6 @@ \subsection{Exterior Algebra and Symmetric Algebra}
     \]  
 \end{definition}
 
-\[
-    \begin{tikzcd}[ampersand replacement=\&]
-        \mathsf{PoFin}_I^{\mathrm{op}}\&[-25pt]\&[+10pt]\&[-30pt] \mathsf{Mble}\&[-30pt]\&[-30pt] \\ [-15pt] 
-        J_2  \arrow[dd, ""{name=L, left}] 
-        \&[-25pt] \& [+10pt] 
-        \& [-30pt]\prod\limits_{i \in J_2}(\Omega_i,\mathcal{F}_i)\arrow[dd, "\mathrm{pr}_{J_1\subseteq J_2}"{name=R}] \&[-20pt]\ni\& [+20pt](\omega_i)_{i\in J_2} \arrow[dd, mapsto, ""{name=L, right}] 
-        \\ [-10pt] 
-        \&  \phantom{.}\arrow[r, "U\circ F", squigarrow]\&\phantom{.}  \&   \\[-10pt] 
-        J_1\& \& \& \prod\limits_{i \in J_1}(\Omega_i,\mathcal{F}_i)\&[-0pt]\ni\& (\omega_i)_{i\in J_1}
-    \end{tikzcd}
-\]  
-
 
 \begin{proposition}{Adjunction $\Largewedge^{\bullet}\dashv U_{\mathsf{R\text{-}\mathsf{Mod}}}$}{}
     Let $R$ be a commutative ring. Suppose $U:R\text{-}\mathsf{AcAlg_{\mathbb{Z}}}\to R\text{-}\mathsf{Mod}$ is the forgetful functor. Then the exterior algebra functor $\Largewedge^{\bullet}:R\text{-}\mathsf{Mod}\to R\text{-}\mathsf{AcAlg_{\mathbb{Z}}}$ is left adjoint to $U$.
@@ -591,7 +579,7 @@ \section{Algebra over Field}
 \end{prf}
 
 \begin{definition}{Algebraic Element and Transcendental Element}{algebraic_element_and_transcendental_element}
-    Let $K$ be a field and $A$ be a $K$-algebra. Consider the evaluation ring homomorphism
+    Let $K$ be a field, $A$ be a $K$-algebra, and $a \in A$. Consider the evaluation ring homomorphism
     \begin{align*}
         \mathrm{ev}_a:K[X] &\longrightarrow A\\
         f &\longmapsto f(a).

field.tex

@@ -43,7 +43,7 @@ \section{Field Extension}
 
 
 \begin{definition}{$K$-embedding and $K$-isomorphism}{}
- A \textbf{$K$-embedding} is a morphism in the \hyperref[th:coslice_category]{coslice category} $(K/\mathsf{Field})$. A \textbf{$K$-isomorphism} is an isomorphism in $(K/\mathsf{Field})$. We say two field extensions $L_1/K$ and $L_2/K$ are \textbf{$K$-isomorphic} if there exists a $K$-isomorphism between them.
+ A \textbf{$K$-embedding} is a morphism in the \hyperref[th:coslice_category]{coslice category} $(\mathsf{Field}/K)$. A \textbf{$K$-isomorphism} is an isomorphism in $(\mathsf{Field}/K)$. We say two field extensions $L_1/K$ and $L_2/K$ are \textbf{$K$-isomorphic} if there exists a $K$-isomorphism between them.
 \end{definition}
 
 \begin{proposition}{Equivalent Characterizations of $K$-embedding}{equivalent_characterizations_of_K_embedding}
@@ -61,7 +61,7 @@ \section{Field Extension}
     \end{enumerate}
 \end{proposition}
 \begin{prf}
-Note $(K/\mathsf{Field})$ is a full subcategory of $(K/\mathsf{CRing})\cong K\text{-}\mathsf{CAlg}$. 
+Note $(\mathsf{Field}/K)$ is a full subcategory of $(K/\mathsf{CRing})\cong K\text{-}\mathsf{CAlg}$. 
 \end{prf}
 
 \begin{corollary}{$K$-endomorphism Fixes Base Field}{}
@@ -78,10 +78,10 @@ \section{Field Extension}
     \]
     be the set of roots of $f$ in $L$. Then $\phi|_{S_f}:S_f\to S_f$ is a bijection and we can define a monoid homomorphism
     \begin{align*}
-        \mathrm{End}_{(K/\mathsf{Field})}(L/K)&\longrightarrow \mathrm{Aut}_{\mathsf{Set}}\left(S_f\right)\\
+        \mathrm{End}_{(\mathsf{Field}/K)}(L/K)&\longrightarrow \mathrm{Aut}_{\mathsf{Set}}\left(S_f\right)\\
         \phi&\longmapsto \phi|_{S_f}
     \end{align*}
-    Moreover, $\mathrm{Aut}_{(K/\mathsf{Field})}(L/K)$ acts on $S_f$ through this map.
+    Moreover, $\mathrm{Aut}_{(\mathsf{Field}/K)}(L/K)$ acts on $S_f$ through this map.
 \end{proposition}
 \begin{prf}
     Let $x\in S_f$. Since
@@ -90,7 +90,7 @@ \section{Field Extension}
     \]
     we have $\phi(x)\in S_f$. Hence $\phi(S_f)\subseteq S_f$. Since $S_f$ is finite set and $\phi$ is injective, we see $\phi|_{S_f}:S_f\to S_f$ is a bijection. It is easy to check
     \begin{align*}
-        \mathrm{Aut}_{(K/\mathsf{Field})}(L/K)&\longrightarrow \mathrm{Aut}_{\mathsf{Set}}\left(S_f\right)\\
+        \mathrm{Aut}_{(\mathsf{Field}/K)}(L/K)&\longrightarrow \mathrm{Aut}_{\mathsf{Set}}\left(S_f\right)\\
         \phi&\longmapsto \phi|_{S_f}
     \end{align*}
     is a group homomorphism. 
@@ -251,19 +251,67 @@ \section{Field Extension}
     If $\Bbbk$ is any field and $\mathfrak{m}$ is a maximal ideal of $\Bbbk\left[x_1, \ldots, x_n\right]$, then $\Bbbk\left[x_1, \ldots, x_n\right]/\mathfrak{m}$ is a finite extension of $\Bbbk$.
 \end{theorem}
 
+\begin{proposition}{}{}
+    If $K$ is a field, then any finite subgroup of $K^\times$ is cyclic.
+\end{proposition}
 
-
-
+\begin{prf}
+    Let $G$ be any subgroup of $K^\times$. By the structure theorem for finite abelian groups, we can write
+    \[
+     G\cong C_{n_1}\times C_{n_2}\times\cdots \times C_{n_r}
+    \]
+    with $n_1\divides n_2 \divides\cdots\divides n_r$. Let $n=G=n_1n_2\cdots n_r$. It is sufficient to show that $r=1$. Since for any $g=(g_1,\cdots, g_r)\in C_{n_1}\times C_{n_2}\times\cdots \times C_{n_r}$, we have
+    \[
+      g^{n_r}-1=(g_1^{n_r},g_2^{n_r},\cdots, g_r^{n_r})-(1,1,\cdots, 1)=0,
+    \]
+    we know the polynomial $f(X)=X^{n_r}-1$ has $n$ distinct roots in $K$. Since 
+    \[
+    n\le \deg f= n_r,
+    \]
+    there must be $n=n_r$ and $r=1$.
+\end{prf}
 
 
 
 
 \subsection{Algebraic Extension}
 
+\begin{definition}{Algebraic Element and Transcendental Element}{algebraic_element_and_transcendental_element_for_field_extension}
+    Let $L/K$ be a field extension and $\alpha \in L$. Consider the evaluation ring homomorphism
+    \begin{align*}
+        \mathrm{ev}_\alpha:K[X] &\longrightarrow L\\
+        f &\longmapsto f(\alpha).
+    \end{align*}
+    and $\ker \mathrm{ev}_\alpha=(P_\alpha)$ for some $P_\alpha\in K[X]$. Polynomials in $\ker \mathrm{ev}_\alpha$ are called \textbf{annihilating polynomials} of $\alpha$ over $K$.
+    \begin{itemize}
+        \item If $P_\alpha=0$, then $\alpha$ is called a \textbf{transcendental element} over $K$. In this case, zero polynomial is the only annihilating polynomial of $\alpha$ over $K$.
+        \item If $P_\alpha\ne 0$, then $\alpha$ is called an \textbf{algebraic element} over $K$. Suppose $P_\alpha(X)=\sum_{i=0}^n a_iX^i$ with $a_n\in K^\times$. Then the monic polynomial $m_\alpha(X)=P_\alpha(X)/a_n$ is called the \textbf{minimal polynomial}\index{minimal polynomial} of $\alpha$ over $K$, which is irreducible in $K[X]$.
+    \end{itemize}
+\end{definition}
+\begin{remark}
+    This definition is a special case of \Cref{th:algebraic_element_and_transcendental_element}.
+\end{remark}
+
 \begin{definition}{Algebraic Extension}{}
     A field extension $L/K$ is \textbf{algebraic} if every element of $L$ is \hyperref[th:algebraic_element_and_transcendental_element]{algebraic} over $K$. That is, for any $\alpha\in L$, there exists a nonzero polynomial $f\in K[X]$ such that $f(\alpha)=0$.
 \end{definition}
 
+\begin{proposition}{}{}
+    Let $L/K$ be a field extension and $\alpha\in L$ be an algebraic element over $K$. Suppose $P(X)\in K[X]$ is an irreducible annihilating polynomial of $\alpha$, or equivalently $P(X)\in K[X]-\{0\}$ is a constant multiple of the minimal polynomial $m_\alpha(X)$ of $\alpha$ over $K$. Then
+    \[
+    K(\alpha)\cong K[X]/(P(X))
+    \]
+    and $[K(\alpha):K]=\deg P(X)$. 
+\end{proposition}
+\begin{prf}
+    The equivalence between irreducible annihilating polynomial and constant multiple of minimal polynomial follows from \Cref{th:irreducible_annihilating_polynomial_is_minimal_polynomial}. According to \Cref{th:structure_of_K_a}, we see
+    \[
+    K[X]/(P(X))\cong K[\alpha]=\left\{ f(\alpha) \in L\midv f\in K[X]
+    \right\},
+    \]
+    $K[\alpha]$ is a field and $[K[\alpha]:K]=\deg P(X)$. It is sufficient to show $K(\alpha)=K[\alpha]$. It is clear that $K[\alpha]\subseteq K(\alpha)$. Since $K(\alpha)$ is the smallest subfield of $L$ containing $K$ and $\alpha$, we have $K(\alpha)\subseteq K[\alpha]$. Therefore, $K(\alpha)=K[\alpha]$.
+\end{prf}
+
 
 \begin{proposition}{}{K_embedding_is_K_isomorphism_for_algebraic_extension}
     Let $L/K$ be an algebraic extension. Then any $K$-embedding $L\hookrightarrow L$ is an $K$-isomorphism, namely
@@ -387,6 +435,9 @@ \section{Algebraic Closure}
     Let $K$ be a field and $f\in K[X]$ be a irreducible polynomial. Denote $L:=K[X]/(f)$ and $\alpha:=X+(f(X))\in L$. Then
     \begin{enumerate}[(i)]
         \item $u:K\hookrightarrow K[X]\twoheadrightarrow L$ is a field extension of degree $\deg f$.
+        \[
+         [L:K]=\deg f.
+        \]
         \item $L=K(\alpha)$.
         \item $\alpha$ is algebraic over $K$ with minimal polynomial $f\in K[X]$.
     \end{enumerate}  
@@ -491,24 +542,24 @@ \section{Algebraic Closure}
 \end{prf}
 
 \begin{corollary}{}{}
-    Let $L/K$ be a simple extension and $L=K(\alpha)$ for some $\alpha\in L$. Suppose $m(X)\in K[X]$ is the minimal polynomial of $\alpha$ over $K$. Then 
+    Let $L/K$ be a simple extension and $L=K(\alpha)$ for some $\alpha\in L$. Suppose $m_\alpha(X)\in K[X]$ is the minimal polynomial of $\alpha$ over $K$. Then 
     \begin{align*}
-        \mathrm{ev}_{\alpha}:\mathrm{Aut}_{(K/\mathsf{Field})}(K(\alpha)/K)&\longrightarrow \left\{ x\in K(\alpha)\midv m(x)=0\right\}\\
+        \mathrm{ev}_{\alpha}:\mathrm{Aut}_{(\mathsf{Field}/K)}(K(\alpha)/K)&\longrightarrow \left\{ x\in K(\alpha)\midv m_\alpha(x)=0\right\}\\
         \sigma&\longmapsto \sigma(\alpha)
     \end{align*}
     is a bijection and we have
     \[
-    |\mathrm{Aut}_{(K/\mathsf{Field})}(K(\alpha)/K)|=\left|\left\{ x\in K(\alpha)\midv m(x)=0\right\}\right|\le \deg m(X),
+    |\mathrm{Aut}_{(\mathsf{Field}/K)}(K(\alpha)/K)|=\left|\left\{ x\in K(\alpha)\midv m_\alpha(x)=0\right\}\right|\le \deg m_\alpha(X),
     \]
-    with equality if and only if $m(X)$ splits over $K(\alpha)$ into $\deg m(X)$ distinct linear factors.
+    with equality if and only if $m_\alpha(X)$ splits over $K(\alpha)$ into $\deg m_\alpha(X)$ distinct linear factors.
 \end{corollary}
 \begin{prf}
     Suppose $\sigma:K(\alpha)\hookrightarrow K(\alpha)$ is a $K$-automorphism. By \Cref{th:K_embedding_preserves_algebraic_element_and_minimal_polynomial}, $u(\alpha)$ is algebraic over $K$ with minimal polynomial $m(X)$. Thus we can define a map
     \begin{align*}
-        \mathrm{ev}_{\alpha}:\mathrm{Aut}_{(K/\mathsf{Field})}(K(\alpha)/K)&\longrightarrow \left\{ x\in L\midv m(x)=0\right\}\\
+        \mathrm{ev}_{\alpha}:\mathrm{Aut}_{(\mathsf{Field}/K)}(K(\alpha)/K)&\longrightarrow \left\{ x\in L\midv m(x)=0\right\}\\
         \sigma&\longmapsto \sigma(\alpha)
     \end{align*}
-    Since $\sigma$ is totally determined by $\sigma(\alpha)$, for any $\sigma_1, \sigma_2\in \mathrm{Aut}_{(K/\mathsf{Field})}(K(\alpha)/K)$, we have 
+    Since $\sigma$ is totally determined by $\sigma(\alpha)$, for any $\sigma_1, \sigma_2\in \mathrm{Aut}_{(\mathsf{Field}/K)}(K(\alpha)/K)$, we have 
     \[
     \sigma_1(\alpha)=\sigma_2(\alpha)\implies \sigma_1=\sigma_2.
     \]
@@ -566,11 +617,11 @@ \section{Algebraic Closure}
 \begin{proposition}{Embed Algebraic Extension into Algebraic Closure}{embed_algebraic_extension_into_algebraic_closure}
     Let $L/K$ be an algebraic extension and $\overline{K}/K$ be an algebraic closure of $K$. Then there exists a $K$-embedding $\gamma:L\hookrightarrow \overline{K}$, namely
     \[
-        \mathrm{Hom}_{(\mathsf{Field}/K)}(L/K,\overline{K}/K)\ne \varnothing.
+        \mathrm{Hom}_{(\mathsf{Field}/K)}(L,\overline{K})\ne \varnothing.
     \]
     If $L/K$ is a finite extension, then 
     \[
-    |\mathrm{Hom}_{(\mathsf{Field}/K)}(L/K,\overline{K}/K)|\le \left[L:K\right].
+    |\mathrm{Hom}_{(\mathsf{Field}/K)}(L,\overline{K})|\le \left[L:K\right].
     \]
 \end{proposition}
 \begin{prf}
@@ -593,11 +644,11 @@ \section{Algebraic Closure}
     Thus we get a chain of restrictions of $\gamma$:
     \[
         \begin{tikzcd}
-            {\mathrm{Hom}_{(K/\mathsf{Field})}\left(K(\alpha_1,\cdots,\alpha_n)/K,\overline{K}/K\right)} \arrow[d, "\mathrm{res}_n"']         & \gamma \arrow[d, maps to]                                      \\
-            {\mathrm{Hom}_{(K/\mathsf{Field})}\left(K(\alpha_1,\cdots,\alpha_{n-1})/K,\overline{K}/K\right)} \arrow[d, "\mathrm{res}_{n-1}"'] & {\gamma|_{K(\alpha_1,\cdots,\alpha_{n-1})}} \arrow[d, maps to] \\
+            {\mathrm{Hom}_{(\mathsf{Field}/K)}\left(K(\alpha_1,\cdots,\alpha_n)/K,\overline{K}/K\right)} \arrow[d, "\mathrm{res}_n"']         & \gamma \arrow[d, maps to]                                      \\
+            {\mathrm{Hom}_{(\mathsf{Field}/K)}\left(K(\alpha_1,\cdots,\alpha_{n-1})/K,\overline{K}/K\right)} \arrow[d, "\mathrm{res}_{n-1}"'] & {\gamma|_{K(\alpha_1,\cdots,\alpha_{n-1})}} \arrow[d, maps to] \\
             {\raisebox{1.5ex}{$\vdots$}}  \arrow[d, "\mathrm{res}_{2}"']                                                                                                 & {\raisebox{1.5ex}{$\vdots$}}  \arrow[d,maps to]                                      \\
-            {\mathrm{Hom}_{(K/\mathsf{Field})}\left(K(\alpha_1)/K,\overline{K}/K\right)} \arrow[d, "\mathrm{res}_1"']                         & \gamma|_{K(\alpha_1)} \arrow[d, maps to]                       \\
-            {\mathrm{Hom}_{(K/\mathsf{Field})}\left(K/K,\overline{K}/K\right)}                                                                & \gamma|_{K}                                                   
+            {\mathrm{Hom}_{(\mathsf{Field}/K)}\left(K(\alpha_1)/K,\overline{K}/K\right)} \arrow[d, "\mathrm{res}_1"']                         & \gamma|_{K(\alpha_1)} \arrow[d, maps to]                       \\
+            {\mathrm{Hom}_{(\mathsf{Field}/K)}\left(K/K,\overline{K}/K\right)}                                                                & \gamma|_{K}                                                   
             \end{tikzcd}
     \]
 \end{prf}
@@ -729,6 +780,13 @@ \section{Separable Extension}
     Let $K$ be a field and $f\in K[X]$ be a nonzero polynomial. We say $f$ is \textbf{separable} if $f$ has no multiple roots in an algebraic closure of $K$. We say $f$ is \textbf{inseparable} if $f$ is not separable.
 \end{definition}
 
+\begin{definition}{Separable Degree of Irreducible Polynomial}{separable_degree_of_irreducible_polynomial}
+ Let $K$ be a field and $f\in K[x]$ be an irreducible polynomial. The \textbf{separable degree} of $f$ is the cardinality of the set of roots of $f$ in any algebraic closure $\overline{K}$ of $K$, which is denoted by
+ \[
+ \deg_s(f):= \left|\left\{\alpha \in \overline{K} : f(\alpha) = 0\right\}\right|
+ \]
+\end{definition}
+
 \begin{proposition}{Equivalent Characterization of Separable Polynomial}{equivalent_characterization_of_separable_polynomial}
     Let $K$ be a field and $f\in K[X]$ be a nonzero polynomial. The following are equivalent:
     \begin{enumerate}[(i)]
@@ -801,6 +859,50 @@ \section{Separable Extension}
     A algebraic extension $L/K$ is \textbf{separable} if every element of $L$ is separable over $K$.
 \end{definition}
 
+
+\begin{lemma}{}{}
+    Let $L/K$ be a finite extension such that $L=K(\alpha_1,\cdots,\alpha_n)$ for some $\alpha_1,\cdots,\alpha_n\in L$. Suppose $\overline{K}/K$ is an algebraic closure of $K$. Define $K_0:=K$ and $K_i:=K_{i-1}(\alpha_i)$ for $1\le i\le n$. Denote the minimal polynomial of $\alpha_i$ over $K_{i-1}$ by $m_{\alpha_i}\in K_{i-1}[X]$. 
+    \begin{enumerate}[(i)]
+        \item If $m_{\alpha_i}$ is separable for any $1\le i\le n$, then 
+        \[
+        \left|\mathrm{Hom}_{\left(\mathsf{Field}/K\right)}(L,\overline{K})\right| = [L:K]
+        \]
+        and $L/K$ is separable.
+        \item If there exists $1\le j\le n$ such that $m_{\alpha_j}$ is inseparable, then
+        \[
+        \left|\mathrm{Hom}_{\left(\mathsf{Field}/K\right)}(L,\overline{K})\right| < [L:K].
+        \]
+    \end{enumerate}
+\end{lemma}
+\begin{prf}
+    If $\alpha_i$ is separable over $K_{i-1}$, then the minimal polynomial $m_{\alpha_i}$ is separable, which implies 
+    \[
+    \operatorname{deg}_s\left(m_{\alpha_i}\right)=\operatorname{deg}\left(m_{\alpha_i}\right)=\left[K_i: K_{i-1}\right]
+    \] 
+    (last equality by Lemma 9.9.2). By multiplicativity (Lemma 9.7.7) we have
+
+    $$
+    [K: F]=\prod\left[K_i: K_{i-1}\right]=\prod \operatorname{deg}\left(m_{\alpha_i}\right)=\prod \operatorname{deg}_s\left(m_{\alpha_i}\right)=\left|\operatorname{Mor}_F(K, \bar{F})\right|
+    $$
+
+    where the last equality is Lemma 9.12.9. By the exact same argument we get the strict inequality $\left|\operatorname{Mor}_F(K, \bar{F})\right|<[K: F]$ if one of the $\alpha_i$ is not separable over $K_{i-1}$.
+
+    Finally, assume again that each $\alpha_i$ is separable over $K_{i-1}$. We will show $K / F$ is separable. Let $\gamma=\gamma_1 \in K$ be arbitrary. Then we can find additional elements $\gamma_2, \ldots, \gamma_m$ such that $K=F\left(\gamma_1, \ldots, \gamma_m\right)$ (for example we could take $\gamma_2=\alpha_1, \ldots, \gamma_{n+1}=\alpha_n$ ). Then we see by the last part of the lemma (already proven above) that if $\gamma$ is not separable over $F$ we would have the strict inequality $\left|\operatorname{Mor}_F(K, \bar{F})\right|<[K: F]$ contradicting the very first part of the lemma (already prove above as well).
+\end{prf}
+\begin{proposition}{Equivalent Characterization of Finite Separable Extension}{}
+    Let $L/K$ be a finite field extension. The following are equivalent:
+    \begin{enumerate}[(i)]
+        \item $L/K$ is a separable extension.
+        \item $L=K(\alpha_1,\cdots,\alpha_n)$ for some separable elements $\alpha_1,\cdots,\alpha_n\in L$ over $K$.
+        \item Let $\overline{K}/K$ be an algebraic closure of $K$.  
+        \[
+        \left|\mathrm{Hom}_{\left(\mathsf{Field}/K\right)}(L,\overline{K})\right| = [L:K]
+        \]
+        \item The trace pairing is \hyperref[th:nondegenerate_bilinear_form]{nondegenerate}.
+        \item $L \otimes_K \overline{K} \;\cong\; \overline{K} \times \cdots \times \overline{K}$
+    \end{enumerate}
+\end{proposition}
+
 \begin{definition}{Perfect Field}{perfect_field}
     A field $K$ is \textbf{perfect} if every finite extension of $K$ is separable.
 \end{definition}
@@ -835,7 +937,7 @@ \section{Separable Extension}
 \section{Trace and Norm of Field Extension}
 
 
-\begin{definition}{}{}
+\begin{definition}{Trace and Norm of Finite Extension}{}
     Let $L/K$ be a finite field extension and $\alpha\in L$ be an algebraic element over $K$. We can define a $K$-linear map
     \begin{align*}
         l_{\alpha}:L&\longrightarrow L\\
@@ -857,6 +959,19 @@ \section{Trace and Norm of Field Extension}
 \begin{remark}
     This definition is a special case of \Cref{th:trace_norm_and_characteristic_polynomial}.
 \end{remark}
+
+
+\begin{definition}{Trace Pairing}{}
+    Let $L/K$ be a finite extension. The \textbf{trace pairing} is the symmetric $K$-bilinear form
+    \begin{align*}
+        \mathrm{Tr}_{L/K}:L\times L&\longrightarrow K\\
+        (x,y)&\longmapsto \mathrm{Tr}_{L/K}(xy).
+    \end{align*}
+\end{definition}
+
+
+
+
 \section{Finite Field}
 
 \begin{definition}{Finite Field}{}