Minimal Polynomials

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December 2nd 2009, 10:03 AM
Haven

Minimal Polynomials

I am having a little troubles with this problem, I think there is something subtle that I am missing

a) Show that $\sqrt[4]{2} \not\in \mathbb{Q}(\sqrt{2})$ where $\mathbb{Q}(\sqrt{2})$ = $\{a+\sqrt{2}b \colon a,b\in\mathbb{Q}\}$

b) Find a minimal polynomial $m_{\sqrt[4]{2}}(x)$ in $\mathbb{Q}(\sqrt{2})[x]$ where $m_{\sqrt[4]{2}}(\sqrt[4]{2}) = 0$

c) Show if $\alpha$ is a primitive element in $GF(p^{n})$ , the degree of the minimal polynomial of $\alpha$ is n
December 2nd 2009, 10:54 AM
Bruno J.

For $a)$ . Suppose $2^{1/4}=a+b\sqrt{2}$ . Then $\sqrt{2}=(a+b\sqrt{2})^2=a^2-2b^2+2ab\sqrt{2}$ . Now we must have $a^2-2b^2=0$ and $2ab=1$ . But $a^2-2b^2=0$ is impossible (why?).

For $b)$ , I can tell you that the minimal polynomial will be $x^2-\sqrt{2}=0$ but I'll let you prove it.
December 2nd 2009, 12:17 PM
Haven

Okay, for c) i have, since $\alpha$ is a primitive element, the $ord(\alpha) = p^n - 1$

So, $\alpha^{p^n - 1} = 1 \rightarrow \alpha^{p^n} = \alpha$

So $\alpha$ is a root of the polynomial $f(x) = (x^{p})^{n} - x$

But this is of degree p*n and not n
December 2nd 2009, 06:56 PM
aliceinwonderland

Quote:

Originally Posted by Haven

Okay, for c) i have, since $\alpha$ is a primitive element, the $ord(\alpha) = p^n - 1$

So, $\alpha^{p^n - 1} = 1 \rightarrow \alpha^{p^n} = \alpha$

So $\alpha$ is a root of the polynomial $f(x) = (x^{p})^{n} - x$

But this is of degree p*n and not n

$\alpha$ is a root of the polynomial $f(x) = x^{p^n} - x$ over GF(p), but f(x) is reducible over GF(p) (check the case with p=2, n=2 s.t p^2=4).

Since $GF(p^n)$ is the splitting field of f(x) over GF(p) and it is the simple extension over GF(p) such that $[GF(p^n):GF(p)]=n$ , the degree of the minimal polynomial of $\alpha$ should be n.