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)) \ $.