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I want to know how to calculate the minimum polynom of an element $\alpha$ in $K=\mathbb{F}_5[X]/(X^2-2)$ where $\alpha$ is the image on $K$ of $X+2$

I'm already verficated that $K$ is a field. As I know, the minimum polynom is $g$ with the smallest degree satisfying $g(\alpha)=0$.

In my notes there is an indication: If $f(X)=X^2-2$, calculate $f(X-2)$.

But I don't understand at all. I would very appreciate any help. Thanks.

MaríaCC
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1 Answers1

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Let's write $\beta$ for the residue class of $X$ in ${\mathbb F}_5[X]/(X^2-2)$, so we're looking at the field ${\mathbb F}_5[\beta]$. What is the minimum polynomial of $\beta$ over ${\mathbb F}_5$? (Trivial!)

Now $\alpha = \beta + 2$. So how do you now turn the minimum polynomial of $\beta$ into the minimum polynomial of $\alpha$? (This is what the hint tells you to do.)

Magdiragdag
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  • So, the minimum polynom is simply $f(x-2)$? – MaríaCC Mar 09 '14 at 12:45
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    Yes; if this is a exercise, you're probably expected to work this out, but that's it. – Magdiragdag Mar 09 '14 at 12:48
  • And do you know how to prove that $\alpha$ is primitive? – MaríaCC Mar 09 '14 at 12:51
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    Well, that's a totally different question. I'd just compute $\alpha^{12}$ and show that that's $-1$. – Magdiragdag Mar 09 '14 at 12:53
  • Sorry for my question but... why 12? My notes say I have to prove $\alpha^{12}$ and $\alpha^{8}$ are different from $1$... – MaríaCC Mar 09 '14 at 12:56
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    Well, you have to show that $\alpha$ generates ${\mathbb F}_{25}^*$, which is cyclic of order 24. So you could check it does not have order 2,3,4,6,8,12 (which is true if $\alpha^8$ and $\alpha^{12}$ are both not 1). But for a primitive element $\gamma$, you'd reach $-1$ halfway down the list $\gamma, \gamma^2, \dots, \gamma^{24}$: because $\gamma^{12}$ and $-1$ both have order 2 and there is only one element of order 2. – Magdiragdag Mar 09 '14 at 13:03