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Norm, Trace, Characteristic Polynomial and Discriminant
Norm,Trace and Characteristic Polynomial
Consider a finite field extension $L/k$ of degree $n=[L:k]$, then $L$ is a $k$ vector space. For an element $\alpha\in L$ we can define a multiplication map $m_\alpha:L\rightarrow L$, which is a $k$ linear map. Note that $m_\alpha$ is a $n\times n$ matrix and the trace is Trace of $m_\alpha$ (sum of diagonal entries) denoted by $\tr_{L/k}(\alpha)$, Norm is $\det m_\alpha $ denoted by $\nm_{L/k}(\alpha)$ and the Characteristic polynomial is $~^{L/k}\chp_\alpha(X)=\det(X\cdot I_n-m_{\alpha})$.Example Let $L=\mathbb{Q}(\ci)$ and $k=\mathbb{Q}$ then $[L:k]=2$ and multiplication by $\alpha=a+\mathrm{i}b$ for a given basis $\\{1 , \ci\\}$ is $$ \begin{align*} \alpha\cdot 1 &=a\cdot 1+ b\ci\\ \alpha\cdot \ci &=a\cdot \ci-b = -b+a\cdot \ci\\ \end{align*} $$ The above can be rewritten as $$ \begin{alignat}{1} \alpha\cdot\text{basis}&= \alpha \cdot \begin{pmatrix} 1 & \ci \end{pmatrix}\\ &= \begin{pmatrix} 1 & \ci \end{pmatrix}\cdot \begin{pmatrix} a & -b \\ b & a \end{pmatrix}\\ &=\text{ basis }\cdot m_\alpha\label{basis1} \end{alignat} $$ Hence, $\tr(\alpha)=\tr(m_\alpha)=2a$ and $\nm(\alpha)=\nm(m_\alpha)=a^2+b^2$. The Characteristic polynomial corresponding to the matrix $m_\alpha$ is $$\det \begin{pmatrix} X-a & b\\ -b & X-a \end{pmatrix}= (X-a)^2+b^2=X^2-2aX+a^2+b^2 $$ Example Let $L=\mathbb{Q}(\sqrt{d})$ where $d$ is a square free integer and $k=\mathbb{Q}$ then $[L:k]=2$ and multiplication by $\alpha=a+b\sqrt{d}$ for a given basis $\\{1 , \sqrt{d}\\}$ is $$ \begin{align*} \alpha\cdot 1 &=a\cdot 1+ b\sqrt{d}\\ \alpha\cdot \sqrt{d} &=a\cdot \sqrt{d}+db = db+a\cdot \sqrt{d}\\ \end{align*} $$ The above can be rewritten as $$ \begin{align*} \alpha\cdot\text{basis}&= \alpha \cdot \begin{pmatrix} 1 & \sqrt{d} \end{pmatrix}\\ & = \begin{pmatrix} 1 & \sqrt{d} \end{pmatrix}\cdot \begin{pmatrix} a & db \\ b & a \end{pmatrix}\\ &=\text{ basis }\cdot m_\alpha\label{basis2} \end{align*} $$ Hence, $\tr(\alpha)=\tr(m_\alpha)=2a$ and $\nm(\alpha)=\nm(m_\alpha)=a^2-db^2$. The Characteristic polynomial corresponding to the matrix $m_\alpha$ is $$\det \begin{pmatrix} X-a & -db\\ -b & X-a \end{pmatrix}= (X-a)^2-db^2=X^2-2aX+a^2-db^2 $$ Example Let $a\in k$, and $n=[L:k]$, then $m_a$ is the matrix $a\cdot I_n$ for example if $n=3$ $$m_a= \begin{pmatrix} a& 0& 0\\ 0 &a& 0\\ 0 &0& a \end{pmatrix},\qquad \qquad\tr(a)=3a\qquad\nm(a)=a^3$$ If $\alpha\in L$ then $m_a\cdot m_{\alpha}=a\cdot I_n m_{\alpha}=m_{a\alpha}$. One can consider $a$ as a scalar $a\cdot\alpha=\alpha\cdot a$ which commutes with $\alpha$.
Let $\mathbf{B}=\{e_0,\ldots, e_n\}$ denote the basis, then we have $\alpha\cdot \mathbf{B}=\mathbf{B}\cdot m_\alpha$ as in \eqref{basis1}. In particular note the following relations
$$
\begin{aligned}
(\alpha+\beta)\cdot\mathbf{B}&=\mathbf{B}\cdot m_{\alpha+\beta}\nonumber\\
\alpha\cdot\mathbf{B}+\beta\cdot\mathbf{B}&=\mathbf{B}\cdot(m_\alpha+m_\beta)\nonumber\\
m_\alpha+m_\beta&=m_{\alpha+\beta}\label{addtrace}\\
(\alpha\cdot\beta)\cdot\mathbf{B}&=\mathbf{B}\cdot m_{\alpha\beta}\nonumber\\
(\alpha\cdot\beta)\cdot\mathbf{B}&=\alpha\cdot \mathbf{B}\cdot m_{\beta}\nonumber\\
&=\mathbf{B}\cdot m_{\alpha} \cdot m_{\beta}\nonumber\\
m_{\alpha}\cdot m_{\beta}&=m_{\alpha\beta}\label{multrace}
\end{aligned}
$$
If $c\in k$ then $c\alpha=\alpha c $ (since, $c$ is in $k$ it is just considered a scalar), hence $c\alpha\cdot\mathbf{B}=\mathbf{B}\cdot c m_{\alpha}$, but $c\alpha\cdot\mathbf{B}=\mathbf{B}\cdot m_{c\alpha}$ which gives $cm_{\alpha}=m_{c\alpha}$ (or see example in the previous section). Also, notice that $m_\alpha\cdot(1,0,\ldots, 0)^T$ gives the first column of $m_{\alpha}$ which is $\alpha$. Thus, we can recover $\alpha$ from $m_\alpha$ by looking at the first column. Also, $m_1$ is just the identity matrix.
Lemma
Additivity: $\tr(\alpha+\beta) =\tr(m_{\alpha+\beta})=\tr(m_{\alpha})+\tr(m_{\beta})=\tr(\alpha)+\tr(\beta)$ and $k$ linearity follows from the fact that $cm_\alpha=m_{c\alpha}$, which gives $c\tr(m_\alpha)=\tr(cm_\alpha)=\tr(m_{c\alpha})$ or $c\tr(\alpha)=\tr(c\alpha)$.
Multiplicativity: As $\nm(\alpha\cdot\beta) =\det(m_{\alpha\beta})=\det(m_{\alpha})\cdot\det(m_{\beta})=\nm(\alpha)\cdot\nm(\beta)$
Discriminant
Let $L=k(\alpha)$ and $f(X)$ be the minimal polynomial of $\deg n$ corresponding to $\alpha$, then the discriminant is given as $$ ~~~\text{disc}_{L/k}(1,\alpha,\ldots,\alpha^n)=(-1)^{n(n-1)/2}N_{L/k}f'(\alpha).$$Example Let $f(X)=X^3+pX+q$ then $f'(\alpha)=3\alpha^{2}+p$ the matrix for multiplication in the basis $\\{1,\alpha,\alpha^2\\}$ is computed as follows $$ \begin{align} 3\alpha^{2}+p \cdot 1&= 3\alpha^{2}+p\\ 3\alpha^{2}+p \cdot \alpha&= 3\alpha^{3}+p\alpha=-2p\alpha -3q \label{eq2}\\ 3\alpha^{2}+p \cdot \alpha^{2}&= 3\alpha^{4}+p\alpha^{2}=-2p\alpha^2 -3q\alpha\label{eq3} \end{align} $$ Use the relation $\alpha^3+p\alpha +q=0$ to get $3\alpha^{3}=-3p\alpha -3q $ in \eqref{eq2} and similarly, use the relation $\alpha^3+p\alpha +q=0$ to get $3\alpha^4=-3p\alpha^2 -3q\alpha$ in \eqref{eq3}. Rewrite the above in matrix form $$ (1,\alpha,\alpha^2)\begin{pmatrix} p&-3q&0\\ 0&-2p&-3q\\ 3&0&-2p\\ \end{pmatrix} $$ Hence the $N_{L/k}f'(\alpha)=4p^3+27q^2$ and $(-1)^{n(n-1)/2}=-1$ since $n=3$. $$ \text{disc}_{L/k}(1,\alpha,\alpha^2)=(-1)^{n(n-1)/2}N_{L/k}f'(\alpha)=-4p^3-27q^2 $$ Example Let $f(X)=X^2+bX+C$ then $f'(\alpha)=2\alpha +b$ the matrix for multiplication in the basis $\\{1,\alpha\\}$ is computed as follows $$ \begin{align} 2\alpha+b \cdot 1&=2\alpha+b\\ 2\alpha+b\cdot \alpha&=2\alpha^2+b\alpha=-b\alpha-2c\label{discex2} \end{align} $$ and we use the relation $\alpha^2+b\alpha+C=0$ to get $\alpha^2=-b\alpha -c$ which is then used to get \eqref{discex2} Rewrite the above in matrix form $$ (1,\alpha)\begin{pmatrix} b&-2c\\ 2&-b\\ \end{pmatrix} $$ which gives $N_{L/k}f'(\alpha)=-b^2+4c$ and $(-1)^{n(n-1)/2}=-1$ since $n=2$. $$ \text{disc}_{L/k}(1,\alpha)=(-1)^{n(n-1)/2}N_{L/k}f'(\alpha)=b^2-4c. $$ In case $L=\mathbb{Q}(\ci)$ and $k=\mathbb{Q}$ the minimal polynomial is $X^2+1$ thus, $b=0$ and $c=1$ which gives the discriminant $-4$.
Minimal polynomial and Characteristic polynomial
The Cayley Hamilton theorem states that for a square matrix $A$ if its Characteristic polynomial is given as $\chp(X)=\det(XI-A)$ then $\chp(A)=\bz$ (zero matrix). For $\alpha\in L$ the corresponding matrix is $m_\alpha$, and now Cayley Hamilton implies $\chp(m_{\alpha})=\bz$ for $\chp_\alpha(X)=\det(XI-m_\alpha)$.For any polynomial $h(X)$ (Lemma below)$ h(m_\alpha)=m_{h(\alpha)}$ replacing $h$ by $\chp$ we get $\chp(m_{\alpha})=m_{\chp(\alpha)}=\bz$ (Cayley Hamilton above). But, matrix $m_\beta=\bz$ is precisely when $\beta =0$, now setting $0=\beta=\chp(\alpha)$, we see $\alpha$ is a root of $\chp_\alpha(X)$.
Lemma
$$ \begin{align*} h(m_\alpha)&=m_\alpha^n+a_1m_\alpha^{n-1}+\ldots +a_0\\ &=m_{\alpha^n}+a_1m_{\alpha^{n-1}}+\ldots +a_0\text{ using } m_\alpha\cdot m_\beta=m_{\alpha\beta}\\ &=m_{\alpha^n}+m_{a_1\alpha^{n-1}}+\ldots +a_0\text{ using } c\cdot m_\beta=m_{c\beta}, c\in k\\ &=m_{\alpha^n+a_1\alpha^{n-1}+\ldots +a_0}\text{ using }m_\alpha+ m_\beta=m_{\alpha+\beta}\\ &=m_{h(\alpha)} \qquad \hfill \qquad \blacksquare \end{align*} $$ See the properties of $m_\alpha$.
Consider the extension $k(\alpha)/k$ and $f(X)$ is minimal polynomial corresponding to $\alpha$ then $\chp_\alpha(X)=f(X)$, since the degree of $\chp_\alpha(X)=\deg[k(\alpha):k]$ (minimal) and it is a monic polynomial with $\alpha$ as a root. This result is not helpful, since we need the mimimal polynomial first to compute $m_\alpha$ which in turn gives Characteristic polynomial. But in some cases we might obtain the minimal polynomial of $\alpha^2$ via Characteristic polynomial corresponding to $m_\alpha$.
Transitivity of Trace and Norm
Theorem
Let $V$ be a vector space over $L$ and $\mathbf{A}:V\rightarrow V$ a linear map. Then from linear algebra we can write it in a rational canonical form for some basis $\\{e_i\\}$ $$ \mathbf{A} = \begin{pmatrix} \mathbf{A}_{1} & 0 & \cdots & 0 \\ 0 & \mathbf{A}_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf{A}_{n} \end{pmatrix} $$ where each diagonal element looks like $$ \begin{equation}\label{0919NT2} \mathbf{A}_{i}= \begin{pmatrix} 0&0&\ldots&0&a_0\\ 1&0&\ldots&0&a_1\\ 0&1&\ldots&0&a_2\\ \vdots&\vdots&\ddots&\vdots&a_0\\ 0&0&\ldots&1&a_{r-1} \end{pmatrix} \end{equation} $$ From above $\det \mathbf{A}=\prod\det\mathbf{A}_i$ and $\tr \mathbf{A}=\sum\tr\mathbf{A}_i$, hence, without loss of generality look only at $\mathbf{A}_i$. $$ \tr\ \mathbf{A}_i=a_{r-1} \text{ and }\nm\ \mathbf{A}_i=\det\ \mathbf{A}_i =(1)^{r-1}a_0 $$ Consider the field extension $L/k$ and pick the basis $\\{f_j\\}$, thus $\\{e_if_j\\}$ is a basis for $V$ over $k$. In this basis we can write $\mathbf{A}_i$ as follows $$ \mathbf{B}_i= \begin{pmatrix} \bz&\bz&\ldots&\bz&m_{a_0}\\ \bon&\bz&\ldots&\bz&m_{a_1}\\ \bz&\bon&\ldots&\bz&m_{a_2}\\ \vdots&\vdots&\ddots&\vdots&m_{a_0}\\ \bz&\bz&\ldots&\bon&m_{a_{r-1}} \end{pmatrix} $$ Let us compute the trace of the above matrix, and note that $m_{a_{r-1}}$ just becomes $a_{r-1}$ when we go back to bigger field $L$ as in \eqref{0919NT2}. $$ \begin{equation}\label{0919NT3} \tr_k\ \mathbf{B}_i =\tr_k(m_{a_{r-1}})=\tr_{L/k}(a_{r-1})=\tr_{L/k}(\tr_L\mathbf{A}_i) \end{equation} $$ In a similar vein $$\begin{equation}\label{0919NT4} \det_k\ \mathbf{B}_i =\det_k((-1)^{r-1}m_{a_{0}})=\nm_{L/k}((-1)^{r-1}a_{0})=\nm_{L/k}(\nm_L\mathbf{A}_i) \end{equation} $$ Taking the vector space $V$ to be $E$ we get the finite field extensions $E/L/k$ and \eqref{0919NT3} and \eqref{0919NT4} give the Transitivity of trace and Norm. $$ \tr_{E/k}=\tr_{L/k}\circ\tr_{E/L} \text{ and } \nm_{E/k}=\nm_{L/k}\circ\nm_{E/L} $$
Proposition
Case 1 First, consider a simpler case of $L=k(\alpha)$, (that is, $r=1$), and immediately below this lemma it is shown that minimal and characteristic polynomial coincide. Given a square matrix $A$ we have $$ \begin{align*} \det(XI_n-A)&= X^n-\tr(A)X^{n-1}+\cdots+(-1)^n\det A\\ \chp_\alpha(X)=\det(XI_n-m_\alpha)&=X^n-\tr(\alpha)X^{n-1}+\cdots +(-1)^n\nm(\alpha) \end{align*} $$ Comparing minimal polynomial and characteristic polynomial we get the result.
Case 2 If $r \not= 1$, consider the tower of extensions $L/k(\alpha)/k$ and note that $\nm_{L/k(\alpha)}(\alpha)=\alpha^r$ since it lies in the base field $k(\alpha)$. $$\begin{align*} \nm_{L/k}(\alpha) &= \nm_{k(\alpha)/k} (\nm_{L/k(\alpha)}(\alpha))\\ &= \nm_{K(\alpha)/K}(\alpha^r) = (\nm_{k(\alpha)/k} (\alpha))^r\\ &= (-1)^{nr} a_0^r. \end{align*}$$ In the above we use, Transitivity, multiplicativity of the norm, and the last equality comes from Case 1. $$\begin{align*} \tr_{L/k}(\alpha) &= \tr_{k(\alpha)/k} (\tr_{L/k(\alpha)}(\alpha)) \\ &= \tr_{K(\alpha)/K}(r\alpha) = r(\tr_{k(\alpha)/k} (\alpha)\\ & = -ra_{n-1}. \end{align*}$$ In the above we use, Transitivity, additivity of the Trace, and the last equality comes from Case 1. For a finite extension $L/k$ with char $ k = 0$, $$\tr_{L/k}(1) = [L:K] \not= 0.$$ implying $\tr_{L/k} \not= 0$. It is in fact true that all separable extensions have $\tr_{L/K} \not= 0$, but for inseparable extensions the following Proposition holds.
Let $\alpha$ be the root of an irreducible polynomial $X^3-X-1\in k[X]$. Setting $L=k(\alpha)$ we have $r=1$ in the proposition and $[L:k]=3$. Thus, $\tr(\alpha)=0$ (no term corresponding to $a_2$) and $\nm(\alpha)=-a_0=-(-1)=1$.
Proposition
Let $f_\beta(X)$ be the minimal polynomial of $\beta\in L$ over $k$, and this polynomial is not separable, which implies that $f_\beta(X)=g(X^p)$ for $g\in k[X]$ and $p=$ char $k$. We have a series of inclusions $$k \subseteq k(\beta^p) \subseteq k(\beta) \subseteq L.$$ Thus, to show $\tr_{L/k} = 0$ it suffices to show $\tr_{k(\beta)/k(\beta^p)} = 0$ (using Transitivity of trace). Note that the minimal polynomial of $\beta^p$ over $k$ is $g(X)$ because $g(\beta^p) = 0$ and $g$ is irreducible. Then $[k(\beta):k] = \deg f_\beta = p \deg g$ and $\deg[k(\beta^p):k] = \deg g$ implying $[k(\beta):k(\beta^p)] = p$. Let $\{1, \beta, \beta^2, \cdots, \beta^{p - 1}\}$ be a basis of $k(\beta)$ over $k(\beta^p)$. Since, $p$ is prime this implies that $k(\beta^p,\beta^i)=k(\beta)$ for $0<i<p$ giving us $[K(\beta^p,\beta^i):k(\beta^p)]=p$. Hence, the minimal polynomial is $$ \begin{cases} X - 1 & i = 0\\ X^p - \beta^{i} & i \not=0 \end{cases}, $$ Looking at the minimal polynomial we see $\tr_{k(\beta)/k(\beta^p)} (\beta^i) = 0$ for all $i$.
Proposition
Let $f_\alpha(X)\in k[X]$ be the minimal polynomial of $\alpha\in L$ of degree $n$. There is one-to-one correspondence between $$\Hom_k(k(\alpha), E)\longleftrightarrow \Root_{f_\alpha}(E) = \{\alpha_1, \cdots, \alpha_n\}\text{ with }\alpha_1=\alpha$$ Since $|\Hom_k(L, E)| = [L:k],$ setting $L=k(\alpha)$ $$|\Hom_k(k(\alpha), E)| = [k(\alpha): k] = \deg f_\alpha.$$ Consider the extension $L/k(\alpha)$ with the surjective restriction map $$\Hom_k(L, E) \to \Hom_k(k(\alpha), E)\qquad\varphi \mapsto \varphi\vert_{k(\alpha)}.$$ Since, the extension $E/L/k(\alpha)/k$ is being collapsed to $E/k(\alpha)/k$ above, thus exactly $[L:k(\alpha)]$ elements map to any particular element in $\Hom_k(k(\alpha), E)$. Let $\psi_j \in \Hom_k(k(\alpha), E)$ then $$\sum_i \varphi_i(\alpha) = [L:k(\alpha)] \sum_j \psi_j(\alpha) = [L:k(\alpha)] \sum_{i = 1}^n \alpha_i$$ The sum of roots of a polynomial is the (negative of the) coefficient of $X^{n - 1}$, and product is the constant term. $$f_\alpha(X) = X^n + a_{n - 1} X^{n - 1} + \cdots + a_0.$$ Hence, $$\sum \varphi_i(\alpha) = [L:k(\alpha)] (- a_{n - 1}) = \tr_{L/K}(\alpha).$$ Similarly, setting $r=[L:k(\alpha)]$ we have $$ \prod_i \varphi_i(\alpha) = \left(\prod_j\psi_j(\alpha)\right)^{r} = \left(\prod_{i = 1}^n \alpha_i\right)^{r} = ((- 1)^n a_0)^{r}= \nm_{L/K}(\alpha). $$
Let $L/k$ be a Galois extension with $k = \F_q $ and $L = \F_{q^n}$, the $\gal(L/k) = {\Z}/{n\Z}$ is generated by Frobenius $\varphi$ (given as $\varphi^i(\alpha)=\alpha^i$), then by the proposition above $$\begin{aligned} \tr_{L/K}(\alpha) &= \sum_{i = 0}^{n - 1} \varphi^i (\alpha)= \alpha + \alpha^q + \alpha^{q^2} + \cdots + \alpha^{q^{n - 1}}.\\ \nm_{L/K}(\alpha) &= \prod_{i = 0}^{n - 1} \varphi^i (\alpha) = \alpha\cdot \alpha^q \cdot \cdots \cdot \alpha^{q^{n - 1}} \end{aligned}$$
Corollary
Notice that $\varphi_1, \cdots, \varphi_n$ are linearly independent over $E$, and hence $\sum_i \varphi_i$ cannot be identically zero. Hence there is some $\alpha$ such that $$\tr_{L/K}(\alpha) = \sum \varphi_i(\alpha) \not= 0.$$
Discriminant again
Let $\alpha_1,\ldots,\alpha_n$ be distinct roots of $f(X)\in k[X]$, then discriminant of $f$ is defined as $$\disc_f = \prod_{i<j}(\alpha_i-\alpha_j)^2 .$$Lemma
Let $f(X)=(X-\alpha_1)\cdots(X-\alpha_n)$, then taking derivative using the product rule gives $f'(\alpha_i)=\prod_{j\neq i}(\alpha_j-\alpha_i)$. Now multiply over all $i$ to get $$\prod_if'(\alpha_i)=\prod_{i=1}^n\prod_{j\neq i}(\alpha_j-\alpha_i) $$ The right hand side can be rewritten as over index pairs $i<j$ by collecting $(\alpha_i-\alpha_j)$ and $\alpha_j-\alpha_i$ as $-(\alpha_j-\alpha_i)^2$ and there are $\binom{n}{2}=n(n-1)/2$ such pairs. $$\prod_{i<j}(\alpha_i-\alpha_j)^2= (-1)^{n(n-1)/2}\prod_if'(\alpha_i)= (-1)^{n(n-1)/2}\nm(f'(\alpha_i)). $$ The last equality follows from proposition above.