User:J D Bowen/Math735 HW11

Section 13.4, Problems 5, 6

Section 13.5, Problems 5, 6, 10

13.4.5 Let K be a finite extension of F. Prove that K is a splitting field over F iff every irreducible polynomial in F[x] that has a root in K splits completely in K[x].

($$ \Rightarrow $$) Suppose K is a splitting field over F. Let $$ f(x) \in F[x] $$ be irreducible with a root $$ \alpha \in K $$. Let $$ \beta $$ be an arbitrary root of f. By theorem 13.8, we can extend the identity isomorphism from F to itself to an isomorphism $$ \sigma : F(\alpha )\to F(\beta ) $$ such that $$ \alpha \to \beta $$. By theorem 13.27, we can extend this isomorphism to $$ \phi : K(\alpha ) \to K(\beta ) $$. Since $$ \alpha \in K $$, $$ [K(\beta ):K] = [K(\alpha ):K] = 1 $$. Therefore, $$ \beta \in K $$. Thus, for any irreducible polynomial $$ f(x) \in F[x] $$, all its roots are in K, which implies that f splits completely in K[x].

($$ \Leftarrow $$) Let $$ K=F(\alpha_1, ..., \alpha_n), \alpha_i\in K $$. Let $$ f_i(x) $$ be the irreducible minimal polynomial of $$ \alpha_i $$. Then if every polynomial with one root in K splits completely in K[x], all the $$ f_i $$'s split in K. Thus K is the splitting field of the product $$ f_1f_2...f_n(x) $$ over F.

13.4.6 Let $$ K_1, K_2 $$ be finite field extensions of F contained in the field K, and assume both are splitting fields.

(a) Prove that their composite $$ K_1K_2 $$ is a splitting field over F.

Let $$ K_1 $$ be the splitting field of $$ f_1 $$ and $$ K_2 $$ be the splitting field of $$ f_2 $$. And let $$ K $$ be the splitting field of $$ f = f_1f_2 $$. We aim to show that $$ K=K_1K_1 $$. Since all the roots of f lie in $$ K_1K_2 $$, $$ K\subseteq K_1K_2 $$. And if K splits f, then it splits $$ f_1 $$ and $$ f_2 $$. Thus $$ K_1\subseteq K $$ and $$ K_2\subseteq K $$. Therefore, $$ K_1K_2\subseteq K $$ and thus, $$ K=K_1K_1 $$. This means that $$ K_1K_2 $$ is also a splitting field over F.

(b) Prove that $$ K_1\cap K_2 $$ is a splitting field over F.

Let $$ f(x)\in F[x] $$ with a root $$ \alpha \in K_1\cap K_2 $$. Then $$ \alpha \in K_1 $$ and $$ \alpha \in K_2 $$ and since both $$ K_1 $$ and $$ K_2 $$ are splitting fields, all the roots of f(x) are in both $$ K_1 $$ and $$ K_2 $$. Therefore, f splits completely in $$ K_1\cap K_2 $$, and by 13.4.5, this implies that $$ K_1\cap K_2 $$ is a splitting field over F.

13.5.5 For any prime $$ p \ $$ and any nonzero $$ a \in F_p \ $$ prove that $$ x^p - x + a \ $$ is irreducible and separable over $$ F_p \ $$.

Note that by Proposition 37 from the text, it suffices to show that f(x) = $$ x^p - x + a \ $$ is irreducible over $$ F_p \ $$. Then let $$ \alpha \ $$ be a root of f(x). Then

f($$ \alpha \ $$ + 1) = $$( \alpha + 1)^p - ( \alpha + 1) + a \ $$

= $$ \alpha^2 + 1 - \alpha -1 + a \ $$

= $$ \alpha^2 - \alpha + a \ $$

Then, for any $$ \alpha \ $$ the root of $$ f \ $$, $$ \alpha + 1 \ $$ is also a root. Thus by induction, each $$ \alpha' \in F_p \ $$ is a root of $$ f \ $$.

Now consider, f(0) = $$ 0^p + 0 + a \ $$ = 0 $$ \implies a \ $$ = 0, which is a contradiction. And therefore $$ f \ $$ has no roots.

So suppose that $$ f \ $$ is reducible, where $$ f = g_1, g_2, \dots, g_n \ $$. Then $$ \exists \ $$ some extension of $$ F_p \ $$, which contains the root $$ \beta \ $$ of $$ f \ $$. But, we proved that each $$ \beta + k \ $$ is also a factor for $$ k \in F_p \ $$, and thus our extension field is a splitting field.

Now since our $$ \beta \ $$ is arbitrary, the deg ($$ g_i \ $$) = [$$ F_p(\beta) : F_p \ $$] for any i. Since $$ f \ $$ has no roots

$$ \prod_{1 \le i \le n}^{} \ $$ deg ($$ g_i \ $$) = $$ p \ $$

for $$ p \ $$ prime $$ f \ $$ must be irreducible.

13.5.6) From the example on page 549 of Dummit and Foote, we know that $$\alpha^{p^n-1}=1 \ $$ for all $$\alpha\in\mathbb{F}^\times_{p^n} \ $$.  Observe then that $$\mathbb{F}_{p^n}^\times \subset \left\{{x:x^{p^n}-1=0 }\right\} \ $$.  Since there are $$p^n-1 \ $$ elements of $$\mathbb{F}_{p^n}^\times \ $$, and since $$x^{p^n}-1 \ $$ is of degree $$p^n-1 \ $$, we must have equality: $$\mathbb{F}_{p^n}^\times = \left\{{x:x^{p^n}-1=0 }\right\} \ $$.

Hence, $$x^{p^n-1} -1 \ $$ is the unique monic polynomial whose roots are precisely $$\mathbb{F}_{p^n}^\times \ $$. Since $$\Pi (x-\alpha) \ $$ is a monic polynomial whose roots are precisely $$\mathbb{F}_{p^n}^\times \ $$, we must have $$\Pi (x-\alpha)=x^{p^n-1} -1 \ $$.

Observe that the only way to get a term without an $$x \ $$ from the product $$\Pi (x-\alpha) \ $$ is to multiply all the $$-\alpha \ $$ together. This immediately gives

$$-1=\Pi (-\alpha) = (-1)^{p^n-1} \Pi \alpha \ $$

or

$$\Pi \alpha = (-1)^{p^n} \ $$.

If we take $$p \ $$ odd, $$n=1 \ $$, observe this implies

$$(p-1)!=\prod_{a=1}^{p-1}a = (-1)^{\text{odd number}} \ \text{mod}(p)=-1 \ \text{mod}(p) \ $$.

13.5.10) Suppose $$f(x_1, \dots, x_n)\in\mathbb{Z}[x_1, \dots, x_n] \ $$.

We aim to show that $$p|(f(x_1,\dots,x_n)^p - f(x_1^p, \dots, x_n^p)) \ $$.

If we define $$g=f \ \text{mod}(p) \ $$, this is equivalent to showing $$g(x_1,\dots,x_n)^p -g(x_1^p,\dots,x_n^p) = 0 \ $$ in $$\mathbb{F}_p \ $$.

Observe

$$g(x_1,\dots, x_n)=\sum_{a_1, \dots, a_n=0}^m A_{a_1 a_2 \dots a_n}x_1^{a_1}x_2^{a_2}\dots x_n^{a_n} \ $$,

where the $$A \ $$ are the coefficients of $$f \ $$ modulo $$p \ $$ and $$m \ $$ is the highest power appearing in $$f \ $$.

We have

$$g(x_1,\dots, x_n)^p = \left({\sum_{a_1, \dots, a_n=0}^m A_{a_1 a_2 \dots a_n}x_1^{a_1}x_2^{a_2}\dots x_n^{a_n}}\right)^p = \sum_{a_1, \dots, a_n=0}^m \left({ A_{a_1 a_2 \dots a_n}x_1^{a_1}x_2^{a_2}\dots x_n^{a_n} }\right)^p \ $$,

since $$\mathbb{F}_p \ $$ has characteristic $$p \ $$.

But this is just

$$=\sum_{a_1, \dots, a_n=0}^m A^p_{a_1 a_2 \dots a_n}x_1^{pa_1}x_2^{pa_2}\dots x_n^{pa_n} \ $$.

Since $$A^p = A \ $$, this is simply $$g(x_1^p, x_2^p, \dots, x_n^p) \ $$.

Hence $$g(x_1,\dots,x_n)^p-g(x_1^p,\dots,x_n^p)=0 \ $$ and so $$p|(f(x_1,\dots,x_n)^p - f(x_1^p, \dots, x_n^p)) \ $$.