Decomposition of the polynomial in isomorphic fields

Question

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?

Answer 1

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?