Algebraic Number Theory: Separability (Part 3)

January 25, 2011

Algebraic Number Theory: Separability (Part 3)

Filed under: Algebra,Number Theory — by Masoud Zargar @ 12:04 am

Starting with this post, which is a continuation of the last post, I will define separability, separable extensions, perfect fields, and discriminants, and also prove results germane to these objects. In the next post, in which norm and trace are discussed, I will apply these tools for $\mathbb{Q}(\zeta_p)/\mathbb{Q}$ to prove a beautiful result in combinatorial number theory.

To study the multiplicity of the roots of a polynomial, the following formal derivative is useful:

Definition. Given the polynomial $f\in F[X]$ , $F$ a field, such that $f=a_0+a_1X+\hdots+a_nX^n$ , we define $f'=a_1+2a_2X+\hdots+na_nX^{n-1}$ .

It is easy to see that this definition satisfies the usual properties of differentiation, e.g. linearity and the product rule.

Suppose now that we are given a polynomial $f\in F[X]$ of degree $n$ . Then we may choose the splitting field of $f$ over $F$ (or any field extension of the splitting field), say $L$ , and write $f=a\prod_{i=1}^n(X-\alpha_i)$ , $a\neq 0$ . If we have that the $\alpha_i$ s are distinct, then $f$ is said to be a separable polynomial.

Theorem 1. $f$ is separable iff $\gcd(f,f')=1$ .

Proof. First, we prove that $(f,g)=1$ in $F[X]$ iff $(f,g)=1$ in $K[X]$ , where $K/F$ is a field extension. Suppose $(f,g)_F=1$ in $F[X]$ . If $d=(f,g)_K$ over $K[X]$ , then $d|(f,g)_F=1$ . So $d~1$ , and so $(f,g)=1$ in $K[X]$ , as well. Conversely, if $(f,g)=1$ over $K[X]$ , then there are $u,v\in K[X]$ such that $uf+vg=1$ . Then if $d':=(f,g)$ in $F[X]$ , then if $d'$ is treated as an element of $K[X]$ , then $d'|uf+vg=1$ . Therefore, $(f,g)=1$ in $F[X]$ .

Having proved this, suppose $\gcd(f,f')=1$ . If $f$ is not separable, then $f=(X-\alpha)^2g$ over some splitting field of $f$ over $F$ . Then $(X-\alpha)|f$ and $f'=2(X-\alpha)g+(X-\alpha)^2g'$ , which is also divisible by $X-\alpha$ . Since $\gcd(f,f')\neq 1$ over some field extension of $F$ , $\gcd(f,f')\neq 1$ in $F[X]$ , which is a contradiction.

Conversely, if $f$ is separable, assume to the contrary that $\gcd(f,f')\neq 1$ . Then $f=(X-\alpha)g$ and $f'=(X-\alpha)h$ for some $\alpha$ . Then by the first equation, $f'=g+(X-\alpha)g'$ . Using the second equation, we see that $f'(\alpha)=0$ , so $0=f'(\alpha)=g(\alpha)\implies X-\alpha|g\implies (X-\alpha)^2|f$ , a contradiction to the separability of $f$ . $\Box$

From the above proof, we see that $\alpha$ is a simple root of $f$ iff $f(\alpha)=0$ but $f'(\alpha)\neq 0$ . Let us construct a number that will precisely determine whether $f$ is separable. Given $f\in F[X]$ of degree $n$ with roots $\alpha_1,\hdots,\alpha_n$ , $f$ is separable iff $f'(\alpha_i)\neq 0$ for all $1\leq i\leq n$ . So $f$ is separable iff $\prod_{j=1}^nf'(\alpha_j)\neq 0$ . For convenience, which will become clear soon, we define the discriminant of $f$ as $\text{Disc}(f):=(-1)^{\tbinom{n}{2}}\prod_{j=1}^nf'(\alpha_j)$ .

We will try to directly find a criterion that is more natural. $f$ is separable iff all its roots are distinct, i.e. $\alpha_i\neq \alpha_j$ for all $i\neq j$ . This is equivalent to requiring that $(-1)^{\tbinom{n}{2}}\prod_{1\leq i<j\leq n}(\alpha_i-\alpha_j)\neq 0$ , where the coefficient is chosen for convenience.

Let’s try to write $(-1)^{\tbinom{n}{2}}\prod_{1\leq i<j\leq n}(\alpha_i-\alpha_j)$ as the determinant of a matrix. It is well known that if $V(X_1,\hdots,X_n):=\det(X_j^{i-1})_{i,j}$ , then

$V(X_1,\hdots,X_n)=(-1)^{\tbinom{n}{2}}\prod_{1\leq i<j\leq n}(X_i-X_j).$

To show this, you may consider the polynomial $V(X_1,\hdots,X_n)$ as a polynomial in $X_n$ with the $X_i,\ i<n$ treated as constants, i.e. $V\in F[X_1,\hdots,X_{n-1}][X_n]$ . Then if you set $X_n=X_i,\ i<n$ , then we will have two equal rows. Therefore $V(X_1,\hdots,X_{n-1},*)$ has roots $X_i,\ i<n\implies (X_n-X_i)|V,\ \forall i<n$ . By considering $V$ as a polynomial in each of $X_j$ , we obtain that $\prod_{1\leq i<j\leq n}(X_i-X_j)|V(X_1,\hdots,X_n)$ . It remains to show $\deg V=\binom{n}{2}$ and that its leading coefficient is $(-1)^{\tbinom{n}{2}}$ (left as exercises).

As a result, we have the following theorem.

Theorem 2. $f$ is separable iff $V(\alpha_1\hdots,\alpha_n)\neq 0$ .

By using the product rule, it is easy to show that $\text{Disc}(f)=V(\alpha_1,\hdots,\alpha_n)^2$ . Of great importance to use are algebraic number fields (reminder: these are finite field extensions of $\mathbb{Q}$ ) and finite fields. So it is natural to study separability of polynomials over such fields. It would be ideal to have all irreducible polynomials to be separable. Fields over which all irreducible polynomials are separable are called perfect. In fact, we have the following theorem.

Theorem 3. All finite fields and fields of characteristic $0$ are perfect.

Proof. By theorem $1$ , this is equivalent to proving that if $F$ is finite or of characteristic $0$ with $f$ (w.l.o.g. monic) irreducible, then $\gcd(f,f')=1$ .

Case 1: characteristic $0$

Suppose $f$ is a monic irreducible polynomial over a field $F$ of characteristic $0$ . Assume to the contrary that $\deg (f,f')>0$ . Then $(f,f')|f$ , and so irreducibility of $f$ implies that $(f,f')=f$ . Since $\deg f'<\deg f$ , we have $f|f'$ iff $f'=0$ . So $f$ is a constant, a contradiction to $\deg (f,f')>0$ . So $(f,f')=1$ , as required.

Case 2: Finite Fields

In this case, we do the same as above, but when $f'=0$ , we deduce that $f(X)=g(X^p)$ for some $g$ . Thus, $f(X)=a_o+a_1X^p+\hdots+a_nX^{np}$ .

Lemma 1. The map $\varphi: F\rightarrow F$ , for finite fields $F$ , given by $\varphi(x)=x^p$ is an isomorphism.

Proof. Left as exercise [Hint: Why do we need $|F|<\infty$ ?]. $\blacksquare$

Remark. This is called the Frobenius automorphism.

Using lemma 1, we can find $b_i\in F$ such that $b_i^p=a_i$ . So then we have $f=(b_0+b_1X+\hdots+b_nX^n)^p$ , a contradiction to the irreducibility of $f$ .

And the conclusion follows. $\Box$

Exercise. If $k$ is a field of positive characteristic $p$ , prove that $f(X):=X^p-t\in k(t)[X]$ , where $k(t)$ is the field of rational functions over $k$ , is irreducible.

Now that we have discussed the separability of polynomials, it is time to think about separable extensions. Our notion of separable extensions is defined in a way that suits our needs in algebraic number theory. Suppose $K$ is an algebraically closed field extension of of the finite extension $E/F$ .

Theorem 4. If $\tau:F\hookrightarrow K$ is an injective field homomorphism, then there are at most $|E:F|$ embeddings $\sigma:E\hookrightarrow K$ .

Proof. We induct on the number $|E:F|=n$ . It is clearly true for $n=1$ . Suppose the theorem is true for all finite field extensions of degree $<n$ . Let $\alpha\in E\setminus F$ . If $E=F(\alpha)$ , then we know that $\alpha$ must be mapped to a root of $min_{F,\alpha}(X)$ , which is of degree $n$ . Therefore, there are at most $|E:F|=n$ embeddings of $E$ into $K$ . Suppose now that $1<|F(\alpha):F|<n$ . Then each embedding $E\hookrightarrow K$ is an extension of $F(\alpha)\hookrightarrow K$ . By induction, there are at most $|E:F(\alpha)|$ such embeddings that are extensions. Similarly, there are at most $|F(\alpha):F|$ embeddings that are extensions of $\tau$ . It follows that there are at most $|E:F(\alpha)||F(\alpha):F|=|E:F|$ embeddings $E\hookrightarrow K$ that are extensions of $\tau$ , as required. $\Box$

Definition. Given the above theorem, we define an extension $E/F$ separable iff for each embedding $\sigma:F\hookrightarrow K$ , there are exactly $|E:F|$ embeddings $\sigma_i:E\hookrightarrow K$ that are extensions of the map $\sigma$

By using the fact that $|K:F|=|K:E||E:F|$ if $E\subseteq F\subseteq K$ are fields, we obtain that $latex K/F$ is separable iff $E/F$ and $K/E$ are separable.

Now that we have defined separable polynomials and separable extensions, we will define elements that are separable over $F$ .

Definition. We say an $\alpha$ that is algebraic over $F$ to be separable iff $\min_{F,\alpha}$ is a separable polynomial over $F$ .

Suppose $\alpha$ is separable. Then $\min_{F,\alpha}$ is separable of degree $|F(\alpha):F|$ . Since all the roots of $\min_{F,\alpha}$ are distinct, we may map $\alpha$ to any other root of $\min_{F,\alpha}$ and obtain an extension of an embedding $F\hookrightarrow K$ , where $K$ is algebraic closed. So $F(\alpha)/F$ is a separable extension.

Conversely, if $F(\alpha)/F$ is separable, consider $\min_{F,\alpha}$ . Given the extensions $\sigma_i:F(\alpha)\hookrightarrow K$ , then this polynomial has has the roots $\sigma_i(\alpha)$ , all of which must be distinct. Since we have $|F(\alpha):F$ such extension, $\min_{F,\alpha}$ is separable.

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[...] Theory,Number Theory — by Masoud Zargar @ 8:41 pm This is a continuation of the the previous post, regarding separability, in the Algebraic Number Theory series. Now that we have worked with [...]

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A Portion of the Book

January 25, 2011