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?


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

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?

  • 2
    $\begingroup$ Thank you very much! Your explanations are very detailed and informative!! $\endgroup$ – alexhak May 12 '18 at 15:50

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