1. Primitive elements, trace and norms. Let $K$ be a finite field, or $char(K)=0$, and $L$ is finite extension field of $K$. We know that there exists exactly $[L:K]$ embeddings that fix $K$ from $L$ to $\overline{K}$-the algebraic closure of $K$. Also, there exists $\alpha\in L$ such that $K(\alpha)=K$, and such $\alpha$ is called the primitive element of the extension field $L/K$.
We also recall that if $L/K$ is a finite extension, then for any $x\in L$, we have the map $x_l: L\to L$ that sends $y$ to $xy$, i.e. the left multiplication map by $x$. It can be seen that $x_l$ is a $K$-linear map. The trace of $x_l$ is called the trace of $x$, and denoted $tr_{L/K}(x)$. Also, the determinant of $x_l$ is called the norm of $x$, and denoted $N_{L/K}(x)$. If $L,K$ are understood, we can write shortly as $tr(x)$ and $N(x)$. We have the following
Proposition 1.1. If $K$ is a finite field or $char(K)=0$, and $L/K$ is finite extension, with $n:=[L:K]$. Let us denote $\sigma_i(i=1,...,n)$ the $n$ embeddings from $L$ to $\overline{K}$. For any $x\in L$, we have $tr(x)=\sum_{i=1}^n \sigma_i(x)$, and $N(x)=\prod_{i=1}^n \sigma_i(x)$. In particular, $tr(x), N(x)\in K$.
Proof sketch. If $x$ is the primitive element of $L$ over $K$, then the basis of $L$ as $K$-vector space is $(1,x,...,x^{n-1})$, and the minimum polynomial of $x$ is of the form $f_x(X)=a_0+a_1X+...+a_{n-1}X^{n-1}+X^n\in K[X]$. From this, one can easily write the the matrix for $x_l$, it is just the companion matrix of $f_x(X)$, and then compute the trace and determinant of $x_l$. It is exact $-a_{n-1}$ and $(-1)^na_0$. On the other hand, we have $f_x(X) = \prod_{i=1}^n(X-\sigma_i(x))$. Comparing the coefficient gives us the result.
If $x$ is not the primitive element, we can consider $x_l$ as the matrix whose blocks are exactly the form the companion matrix of $f_x(X)$. And from this, we will get the same result.
(Q.E.D)
Using this proposition, one can obtain
Corollary 1.2. If $A$ is a domain of characteristic zero and $K$ is the quotient field of $K$. $L/K$ is a field extension of degree n, and $x\in L$ such that $x$ is integral over $A$, then $tr(x)$, $N(x)$ are also integral over $K$. In particular, if $A$ is integrally closed in K, then $tr(x), N(x)\in A$.
Proof. Let $\sigma_1,...,\sigma_n$ be $n$ embeddings from $L$ to $\overline{K}$, it can be seen that $\sigma_i(x)$ is also integral over $K$. And hence, their sum and their product are also integral over $K$. From this, if $A$ is integrally closed in $K$, then from Proposition 1.1, we have $N(x),tr(x)$ is in $K$, and integral over $A$. This implies, both of them lie in $A$.
(Q.E.D)
2. Trace form and discriminant. We first fix the notions in this section, $K$ is a field and $char(K)=0$, $L/K$ is a field extension of degree $n$. Let $\langle.,.\rangle: L\times L\to K$ defined by $\langle x,y\rangle = tr(xy)$. One can see that $\langle.,.\rangle$ is a symmetric bilinear form, which is called the trace form. We will prove
Lemma 2.1. The trace form defined above is non-degenerate.
Proof. Assume that there exists $x\in L$, such that $tr(xy)=0$ for all $y\in L$. If $x\ne 0$, then $tr(xx^{-1})=tr(1)=n=0$, a contradiction. Hence, $tr(xy)=0$ for all $y\in L$ implies that $x=0$. This yields the trace form is non-degenrate.
(Q.E.D)
We are now ready to define the discriminant. Let $x_1,...,x_n\in L$, we define $D(x_1,...,x_n):=\det(tr(x_ix_j))_{ij}$. From Proposition 1.1, we have $D(x_1,...,x_n)\in K$, since any $tr(x_ix_j)\in K$. Also, by Lemma 2.1, we easily get
Proposition 2.2. Let $x_1,...,x_n\in L$, then $(x_1,...,x_n)$ is a base for $L$ as $K$-vector space iff $D(x_1,...,x_n)\ne 0$.
Proof. Because the trace form is non-degenerate. One obtains from this that the matrix of the bilinear form $tr(x_ix_j)_{ij}$ is non-degenrate iff $(x_1,...,x_n)$ is a basis for $L$ as $K$-vector space.
(Q.E.D)
Due to Proposition 1.1, if $\sigma_1,...,\sigma_n$ are the $n$ embeddings from $L$ to $\overline{K}$, then $$tr(x_ix_j)=\sigma_1(x_ix_j)+...+\sigma_n(x_ix_j)=\sigma_1(x_i)\sigma_1(x_j)+...+\sigma_n(x_i)\sigma_n(x_j)$$
This yields, in fact, $tr(x_ix_j)_{ij}=(\sigma_i(x_j))_{ij}(\sigma_i(x_j))_{ij}^T$. In particular, $D(x_1,...,x_n)=\det(\sigma_i(x_j))_{ij}^2$. And from Proposition 2.2, we get
Proposition 2.3. Let $x_1,...,x_n$ be elements in $L$, then $(x_1,...,x_n)$ is the basis for $L$ as $K$-vector space iff $\det(\sigma_i(x_j))_{ij}\ne 0$.
Proof. It is easy to get from what we have discussed so far.
(Q.E.D)
The non-degenrate of the trace form has another consequence, i.e. the dual basis. If $(x_1,...,x_n)$ is a basis of $L$ as $K$-vector space, then there exists another basis of $L$ as $K$-vector space, $(y_1,...,y_n)$ such that $tr(x_iy_j)=\delta_{ij}$, where $\delta_{ij}$ is the Kronecker's delta. Due to this, one gets the important
Proposition 2.4. Let $A$ be a integrally closed domain of characteristic 0 in its quotient field $K$. Let $L/K$ is field extension of degree $n$, and $A'$ is the integral closure of $A$ in $L$. Then $A'$ is a submodule of a free $A$-module of rank $n$. In particular, If $A$ is a P.I.D, then $A'$ is a free $A$ module of rank $n$.
Proof. Let $(x_1,...,x_n)$ be a basis for $L$ as $K$-vector space. Because $L/K$ is finite extension, $x_i$ is algebraic over $K$. And hence, there exists $f_1(X)=a_0+...+a_nx^n\in A[x]$ the minimum polynomial of $x_1$, i.e. $a_0+...+a_nx_1^n=0$. This implies $a_0a_n^{n-1}+...+(a_nx_1)^n=0$, i.e. $x_1':=a_nx_1$ is integral over $A$, i.e. $x_1'\in A'$. Similarly, we obtain $x_i'\in A'$ is integral over $A$, and $(x_1',...,x_n')$ is a basis for $L$ as $K$-vector space.
By the non-degenate of the trace form, there exists $(y_1,...,y_n)$ a basis of $L$ as $K$-vector space such that $tr(x_i'y_j)=\delta_{ij}$. Now, take any $z\in A'$, we can represent $z = b_1y_1+...+b_ny_n$, for $b_i\in L$. Because $x_i'\in A'$, we have $x_i'z\in A'$. And hence, by Proposition 1.2, $tr(x_i'z)\in A$. But $tr(x_i'z) = tr(x_i'(b_1y_1+...+b_ny_n))=b_i\in A$. This yields $A'$ is a submodule of $Ay_1\oplus ...\oplus Ay_n=:\mathscr{A}$.
Now, if $A$ is a P.I.D, $A'$ is also a free submodule of $\mathscr{A}$, and $A'$ contains a basis of $L$ (which is $(x_1',...,x_n')$). And hence, $A'$ is a free $A$-module of rank $n$.
(Q.E.D)
3. Some examples of computing the discriminant. In this section, we will compute the $D(1,\zeta,...,\zeta^{p-1})$, where $\zeta$ is the primitive $p^\text{th}$ root of unity, where $p$ is an odd prime.
We first need the following
Proposition 3.1. Let $L,K$ be fields defined in Section 2, $x\in L$, and $f_x(X)\in K[X]$ is the minimum polynomial of $x$. Then $D(1,x,...,x^{n-1})=cN(F'(x))$, where $c=(-1)^{n(n-1)/2}$.
Proof. Let $\sigma_1,...,\sigma_n$ be $n$ embeddings of $L$ into $\overline{K}$. We denote $x_i=\sigma_i(x)$. Then by our earlier remarks, $D(1,x,...,x^{n-1})=\det (\sigma_i(x^j))_{ij}^2=\det (x_i^j)_{ij}^2$. The last matrix is exactly the Vandermonde's matrix, and hence, $\det(x_i^j)_{ij}=\prod_{1\le i<j\le n}(x_i-x_j)$. This yields
$$D(1,x,...,x^{n-1})=\prod_{1\le i<j\le n}(x_i-x_j)^2=c\prod_{i\ne j}(x_i-x_j)=$$
$$=c\prod_{i}(\prod_{j\ne i}(x_i-x_j))=c\prod_i(f_x'(x_i))=cN(f_x'(x))$$.
(Q.E.D)
Now, let $\zeta$ be the primitive $p^\text{th}$ root of unity, the minimal polynomial of $\zeta$ is $f(X) := \frac{X^p-1}{X-1}=X^{p-1}+...+1$. And hence, $f'(X)=\frac{pX^p-pX^{p-1}-X^p+1}{(X-1)^2}$. In particular, because $\zeta^p=1$, we have $f'(\zeta)=p\frac{1-\zeta{p-1}}{(1-zeta)^2}$. And hence
$$D(1,\zeta,...,\zeta^{p-1})=cp^{p-1}\frac{N(1-\zeta^{p-1})}{N(1-\zeta)^2}$$
Now, one can see that $f(X)$ is the minimum polynomial of $\zeta$ implies $f(X+1)$ is the minimum polynomial of $\zeta-1$, and $f(X+1)=(X+1)^{p-1}+...+1= ... + p$. This yields, $N(\zeta-1)=p$, due to the proof of Proposition 1.1. And hence, $N((1-\zeta)^2)=p^2$. Also, $N(1-\zeta^{p-1})=N(1-1/\zeta)=N(\zeta-1)/N(\zeta) = p$, since $N(\zeta)=1.\zeta...\zeta^{p-1}=1$. It follows from our previous calculation that $D(1,\zeta,...,\zeta^{p-1})=cp^{p-2}$.
If we develop some more theory, and let $K$ be any number field of degree $n$ over $\mathbb{Q}$, with $\mathscr{O}_K$ the ring of integers in $K$, i.e. they are integral closure of $\mathbb{Z}$ in $K$, we then have, by Proposition 2.4 that $\mathscr{O}_K$ is free $\mathbb{Z}$-module of rank $n$. This yields, there exists a basis $(x_1,...,x_n)$ of $\mathscr{O}_K$ as $\mathbb{Z}$-module. And $D(x_1,...,x_n)$ is called the discriminant of $K$, and denoted by $\Delta_K$. It is an important invariant, because we will look at the discriminant of $K$ and determine which primes are ramified. It is stated that a prime $p$ ramified in $\mathscr{O}_K$ iff $p|\Delta_K$. But in general, it is difficult to compute $\Delta_K$, and we should develop a method to detect possibly ramified primes.
Let $\alpha\in \mathscr{O}_K$ be the primitive element of $K$ over $\mathbb{Q}$, then $\mathbb{Z}[\alpha]$ is a full rank $\mathbb{Z}$-submodule of $\mathscr{O}_K$. If we denote $D(\mathbb{Z}[\alpha]):=D(1,\alpha,...,\alpha^{n-1})$, then we have
$$D(\mathbb{Z}[\alpha])=[\mathscr{O}_K:\mathbb{Z}[\alpha]]^2\Delta_K$$
Also, we have another important criterion, that $p$ is singular over $\mathbb{Z}[\alpha]$ iff $p$ divides the index $[\mathscr{O}_K:\mathbb{Z}[\alpha]$. And from the formula above, a ramified prime $p$ possibly divides $D(\mathbb{Z}[\alpha])$, which can be computed by Proposition 3.1. In particular, if $K=\mathbb{Q}[\zeta]$, and $\alpha=\zeta$, then one can see the possible ramified prime (and singular prime over $\mathbb{Z}[\alpha]$) is only $p$. By Kummer-Dedekind's theorem, $p$ is not singular above $\mathbb{Z}[\alpha]$, and hence, the index is just 1, and $\mathscr{O}_K=\mathbb{Z}[\zeta]$.
As we will see, this formula gives us a convenient way to determine the ring of integers.
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