1

I'm having a bit of trouble with the below problem:

Consider the factor ring $R_{a,b} = \mathbb{Z}_2[x]/(x^2 + ax + b)$. For which $a,b$ is this ring a field? Can two rings $R_{a_1,b_1}$ and $R_{a_2,b_2}$ be isomorphic? If so, describe all such pairs $a,b$.

Here's my work so far:

$\mathbb{Z}_2[x]/(x^2+ax+b)$ is a field only if $p(x) = x^2 + ax + b$ is irreducible in $\mathbb{Z}_2[x]$, which occurs only if $p(0) \equiv 1 (mod \ 2)$ and $p(1) \equiv 1 (mod \ 2)$. These will yield $b \equiv 1 (mod \ 2)$ and $1 + a + b \equiv 1 (mod \ 2)$ $\Rightarrow$ $a \equiv 1 (mod \ 2)$ and $b \equiv 1 (mod \ 2)$. Thus, $\mathbb{Z}_2[x]/(x^2+ax+b)$ is a field only when both $a$ and $b$ are odd.

I'm struggling with the second question. I know that if both $R_{a_1,b_1}$ and $R_{a_2,b_2}$ are fields, then the two will be isomorphic by the classification of finite fields -- both will be finite fields with $2^2 = 4$ elements, and hence, isomorphic. But how can I consider other cases, for example, when one or both are not fields? Is there a more clever way to do this?

Thanks!

testguy807
  • 221
  • There are only four choices for $(a,b)$, so a quick census settles this one. You found the only field. With $x^2+ax+b=x^2+x=x(x+1)$ we can apply the Chinese remainder theorem to analyze the quotient ring. It should be easy for you to construct an isomorphism between the cases $x^2$ and $(x+1)^2$. Done. I think we may have handled this one on the site already. – Jyrki Lahtonen Nov 07 '19 at 04:46
  • Related but possibly a bit deeper. – Jyrki Lahtonen Nov 07 '19 at 04:50
  • @JyrkiLahtonen So, the one field can't be isomorphic to the others -- got it. There's three left to consider. I believe quotient ring involving x(x+1), by the CRT, splits as Z_2 x Z_2. I think this quotieng ring will be isomorphic to the other two quotient rings, right? Because every non-identity element is of order 2? And that isomorphism between the cases x^2 and (x+1)^2 just sends x to x + 1? – testguy807 Nov 07 '19 at 05:09
  • That's the general idea. But $\Bbb{Z}_2\times\Bbb{Z}_2$ is not isomorphic to $\Bbb{Z}_2[x]/(x^2)$. The latter ring has a non-zero element whose square is zero, the former has no such element. – Jyrki Lahtonen Nov 07 '19 at 05:19
  • The term reduced in Martin Brandenburg's answer to the linked question is about the absence/presence of nilpotent elements. – Jyrki Lahtonen Nov 07 '19 at 05:20
  • @JyrkiLahtonen by similar logic, $\mathbb{Z}_2 \times \mathbb{Z}_2$ can't be isomorphic to $\mathbb{Z}_2[x]/(x^2+1)$ either, right? In one, $x^2 = x$, but in the other, $x^2 = 1$. – testguy807 Nov 07 '19 at 18:03
  • That's one way of seeing it. You do need a bit of care. The point being that in the ring $\Bbb{Z}_2\times\Bbb{Z}_2$ $1=(1,1)$ is the only element with square equal to $1$. Because we already established that $R'=\Bbb{Z}_2[x]/(x^2+1)$ is isomorphic to $R=\Bbb{Z}_2[x]/(x^2)$ I would be tempted to use the fact that $(x+1)^2=0$ in $R'$. Again, a nilpotent element. – Jyrki Lahtonen Nov 07 '19 at 18:08

0 Answers0