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I have to factor the polynomial at the field of 9 elements. For it I view the field $GF(3)[x]/(x^2+1)$. But if I view the field $GF(3)[x]/(x^2+x+2)$ I get another decomposition of this polynomial. So, why is it? And is it right to factor this polynomial over only one field?

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closed as unclear what you're asking by user223391, GNUSupporter 8964民主女神 地下教會, B. Mehta, José Carlos Santos, Chris Godsil May 13 '18 at 0:01

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  • $\begingroup$ Neither of those things you call fields are. $\endgroup$ – Lord Shark the Unknown May 12 '18 at 15:26
  • $\begingroup$ @LordSharktheUnknown, I correct to GF(3). The question is actual $\endgroup$ – alexhak May 12 '18 at 15:27
  • $\begingroup$ What's your polynomial, and what are your decompositions? $\endgroup$ – Billy May 12 '18 at 15:28
  • $\begingroup$ @Billy, For example, f(x)=$x^5+2x^4+x^2+2x+1$. If $x^2+1$: $f(x) = ((x+2\alpha+2)(x+\alpha+2)g(x)$. If $x^2+x+2$: $f(x) = ((x+2\alpha)(x+\alpha+1)g(x)$, where g(x) is irreducible polynomial of degree 3. $\endgroup$ – alexhak May 12 '18 at 15:33
  • $\begingroup$ You have too many $x$'s. $\endgroup$ – ancientmathematician May 12 '18 at 15:43
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There should only be one decomposition up to isomorphism, and I'm going to guess that your problem is a notational one.

Firstly, $GF(3)[x]/(x^2 + 1)$ and $GF(3)[x]/(x^2 + x + 2)$ are both fields of 9 elements, and they are isomorphic. But the variable $x$ doesn't play the same role in each. To keep the notation clean, I'm going to define some "generic" field of 9 elements, say $F$, and assume we have isomorphisms $GF(3)[x]/(x^2 + 1)\to F$ and $GF(3)[x]/(x^2 + x + 2)\to F$. Now:

  • Let $\alpha$ be the image of $x$ in $F$ under the map $GF(3)[x] \to GF(3)[x]/(x^2 + 1) \to F$.
  • Let $\beta$ be the image of $x$ in $F$ under the map $GF(3)[x] \to GF(3)[x]/(x^2 + x + 2) \to F$.

We can use this to write $F$ explicitly in two ways:

  • $F \cong GF(3)[\alpha]$, i.e. $F = \{0, 1, 2, \alpha, \alpha+1, \alpha+2, 2\alpha, 2\alpha+1, 2\alpha+2\}$ as a set,
  • $F \cong GF(3)[\beta]$, i.e. $F = \{0, 1, 2, \beta, \beta+1, \beta+2, 2\beta, 2\beta+1, 2\beta+2\}$ as a set,

but $\alpha$ and $\beta$ are still not the same thing, because the arithmetic works differently. After all, inside $F$, the following things are true:

  • $\alpha^2 + 1 = 0$ (but $\beta^2 + 1 \neq 0$)
  • $\beta^2 + \beta + 2 = 0$ (but $\alpha^2 + \alpha + 2 \neq 0$).

The issue is that we have two ways of representing $F$ - as $GF(3)[\alpha]$ and $GF(3)[\beta]$ - and they're isomorphic, but the isomorphism is not just the map $\alpha\mapsto \beta$. Can you work out what it is?

Now, work out the decomposition of your polynomial in terms of $\alpha$, and separately in terms of $\beta$. What happens when you take the $\alpha$-decomposition and apply the isomorphism $GF(3)[\alpha]\to GF(3)[\beta]$ to it?

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    $\begingroup$ Thank you very much! Your explanations are very detailed and informative!! $\endgroup$ – alexhak May 12 '18 at 15:50

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