Notes on Algebraic Number Theory (I: Primitive Elements, Trace, Norm and Discriminant)

For a long time, my intention is to write a full note on basic algebraic number theory, for example, rings of integers of number fields, Minkowski's theorem, Dirichlet's unit theorem and Hilbert's theory for number fields, with some of their applications. Together with the Kummer-Dedekind's theorem, which is an important tool for factoring a prime in the ring of integers in a number fields, we will discuss another basic things, that also complement to the story to the determination of integral basis of a ring of integers. We first recall some facts about field extensions.

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.

Equations on finite fields

I copy my note from our old blog on trello.com.

PART 1

First, we begin with the circle $(C): x^2 + y^2 = 1$ over a finite field $F_q$, whose characteristic is not $2$. And the question we are interested in is that how many points lie on this circle? 

If $-1$ is a square in $F_q$, we assume that $-1=a^2$, then $x^2+y^2=x^2-a^2y^2=(x-ay)(x+ay)=1$. Hence, $(x,y)\in (C)$ is equivalent $x-ay = b, x+ay = b^{-1}$. And from this $x = (b+b^-1)/2$, and $y = (b-b^{-1})/2a$. An EXERCISE is that counting the number of points in this case.

If $-1$ is not square in $F_q$, we consider the quadratic extensions $F_q(i)/F_q$, where $i$ is the square root of $-1$ in $\overline{F_q}$. The field extension is Galois, and there exists two elements of the Galois group $\{id, x\mapsto x^q\}$, or equivalent $\{id, x+iy \mapsto x-iy\}$.

Because of the definition of norm, let $u=x+iy$, we can compute its norm by two ways: $Norm(u) = x^2 + y^2 = u^{q+1}$. And hence, points on $(C)$ are exactly the kernel of the norm map.

Because $F_q(i)^\times$ is cyclic, whose order $q^2-1$ and generator $g$, the kernel of norm is the cyclic group generated by $g^{q-1}$, and has order $q+1$. That means, $(C)$ will have $(q+1)$ points.

(NOTE: at this stage, we will also conclude that Norm is surjective, because $\#Norm(F_q(i)^\times)=|F_q(i)^\times|/|Ker(Norm)|=q-1=|F_q^\times|$)

Note that there is a 1-1 corresponding between points on $(C)$ and $Ker(Norm)$. Via this, we can define the group law for points in $(C)$ (See my earlier comment on the card named "Pairing on Linear Varieties and Curves of Genus 0").

An interesting corollary of NOTE is that: the norm map is surjective, and hence, any element in $F_q$ can be represented under the form $x^2+y^2$. And we obtain the following

Theorem 1. Let $F_q$ be a finite field with characteristic not 2, and $\alpha,\beta,\gamma$ are not zero in $F_q$. Then the equation $\alpha x^2+\beta y^2+ \gamma z^2$ always has non-trivial solutions over $F_q$.

(Or equivalent statement by another language: Any quadratic form of rank $\ge 3$ is isotropic over $F_q$)

We NOTE that the equation $\alpha x^2 + \beta y^2 = 0$ does not always has non-trivial solution over $F_q$, since in this case $\frac{-\alpha}{\beta}$ must be a square in $F_q^\times$.

The interesting story now begins. One may ask how is the case of cubic forms? That means, for what $n$, we can ensure that the equation $a_1x_1^3+a_2x_2^3+...+a_nx_n^3$ always has non-trivial solution? And more generally, $d$-forms?

PART 2

The aim of this part is to prove the theorem of Chevalley, conjectured by E. Artin.

Theorem 1. Let $f(x_1,...,x_n) \in F_q[x_1,...,x_n]$ be a multi-variables of degree $d$, where $n>d$, and $f(0,0,..,0)=0$, then $f$ has at least two solutions.

Two solutions are mentioned in the book of K. Ireland and M. Rosen "A Classical Introduction to Modern Number Theory", Chapter X, and we just explain one of them, with some further comments. 

(NOTE: that the remaining solution is very interesting too, and close to the method for supersingular elliptic curves discussed in the book of L. C. Washington about EC, Chapter IV.)

Assume that $f$ has no other solutions. That means, $f(a_1,...,a_n)\ne 0$, for all $(a_1,...,a_n)\ne (0,...,0)$, that means $1-f^{q-1}$ takes value 1 at $(0,..,0)$, and 0 otherwise. It can be seen that, in fact, it is "equivalent" (this will be defined later) to the polynomial $g(x_1,...,x_n)=(1-x_1^{q-1})...(1-x_n^{q-1})$. because $g$ takes value 1 at $(0,...,0)$, and 0 otherwise (*).

The term "equivalent" can be explained as follows.

Definition 2. The two polynomial $f,g$ over $F_q[x_1,...,x_n]$ is called equivalent if $f(a_1,...,a_n)=g(a_1,...,a_n)$ for all $a_i\in F_q$.

The properties Frobenius automorphism allows us to prove that $x^q$ and $x$ are equivalent. And hence, $x^{aq+r}$ is equivalent to $x^r$. And we can extend this observation to the following

Proposition 3. Let $f\in F_q[x_1,...,x_n]$, then $f$ is equivalent to $g$, where any monomial term of $g$ has the form $x_1^{a_1}...x_n^{a_n}$, and $a_i<q$. 

Such a polynomial $g$ is called reduced polynomial of $f$, and it is easy to see that the reduced polynomial of a given polynomial is unique.

Proposition 4. The reduced form of a polynomial is unique.

Proof. It is equivalent to prove that if $f$ and $g$ are reduced forms, i.e. their monomial term has the forms $x_1^{a_1}...x_n^{a_n}$, where $a_i<q$, and identical at all values in $F_q^n$, then they are identical. Let $h=f-g$, and fix the value $a_2,...,a_n\in F_q$, we can see $k(x_1)=h(x_1,a_2,...,a_n)$ is the polynomial of variable $x_1$, with the degree does not exceed $q-1$, and it takes $q$ zeros. Hence $k(x_1)$ is identical zero. Extend this method, it is obvious to see that $h$ is zero, and hence $f\equiv g$ (Q.E.D).

Back to (*), we can see that actually, $g$ is a reduced polynomial of $f$, and hence, $\deg(1-f^{q-1})=(q-1)\deg(f) \ge n(q-1)$. This implies $d\ge n$, a contradiction with the hypothesis of our theorem. Hence, $f$ must have another solution other than zero. And the theorem of Chevalley is now completely proved (Q.E.D).

As a consequence, we can see in any positive characteristic, the equation $\alpha x_1^2 + \beta x_2^2 + \gamma x_3^2$, where $\alpha, \beta, \gamma \ne 0$ always has non-trivial solution (NOTE that in the first part, we implied this in the case characteristic is not 2). And for $d$-form, the number of variables must be greater than $d$ to ensure that our equation always has non trivial solution.

How about in the case $n=d$? A very interesting counter example is built in the EXERCISE section of this book (ex. 17). We will conclude this part by proving the following

FACT 5. There is a $d$-form in $d$ variables such that it has no solution other than $(0,...,0)$.

Proof. We now need some Galois theory to point out this. Consider the extension $F_{q^d}/F_q$. The Galois group in this case is cyclic of order $d$, and generated by the Frobenius automorphism $x \rightarrow x^q$, and all of its elements is $\{id, x\rightarrow x^{q}, ..., x\rightarrow x^{q^{d-1}}\}$. Let $a_1,...,a_d$ is the basis for the vector space $F_{q^d}$ over $F_q$. Consider the polynomial

$$f(x_1,...,x_m)=\prod_{i=0}^{d-1}(a_1^{q^i}x_1+...+a_d^{q^i}x_d)$$

Now, $f$ is invariant under the action from the Galois group (CHECK THIS). Hence, it is defined in $F_q[x_1,...,x_d]$. And the basic facts for the basis of a vector space implies that $f=0$ iff $x_1=...=x_d=0$. And this is our counter example. (Q.E.D)

The book gave a hint for the polynomial and the solution is mine. I don't know how the idea for this counter example comes from. And I didn't think that the Galois theory helps in this case. But actually, it does!

PART 3

IMPORTANT: Fields in this note have characteristic different from 2.

For a long time (maybe 2 weeks), I did not take note on anything, there is many things here (and, of course, quite far from Math), and it makes me feel a little bit of tired. I want to write something about the orthogonal groups over finite fields today.

It should be first noted that we need to know something about quadratic forms first, Witt's theorems (Cancelation Theorem and Extension Theorem). It is not difficult, but really interesting, because it reflects the geometric intuition, and one may be curious how to define "geometry" over arbitrarily fields. 

I took a note in the past, but today I look it back, and find some stupid mistakes. I did not bring the other notes with me. But I will edit the mistakes, and then make an outline to compute the order of the orthogonal group over finite fields (associated with a given quadratic form).

But does it have anything to do with the equations over finite fields? Well, as we will see later, one of the most important steps is: compute the number of solutions for the equation $a_1x_1^2+a_2x_2^2+...+a_nx_n^2 = 1$ over $F_q$. And this can be done with the help of Jacobi and Gauss's sum over finite fields.