# Finding an isomorphism between polyomial quotient rings

Let $F_1 = \mathbb{Z}_5[x]/(x^2+x+1)$ and $F_2 = \mathbb{Z}_5[x]/(x^2+3)$. Note neither $x^2+x+1$ nor $x^2+3$ has a root in $\mathbb{Z}_5$, so that each of the above are fields of order 25, and hence they are isomorphic from elementary vector space theory, however I'm tasked with constructing an explicit isomorphism between the two fields, so I applied the technique I've learned here and elsewhere as follows.

First, let $\phi: F_1 \rightarrow F_2$ by $x \mapsto ax+b$ (as the base field is fixed it suffices to find the image of $x$). Then we should have that $\phi(x^2+x+1) = x^2 + 3$, and the left hand side reduces to $\phi(x)^2 + \phi(x) + 1$ as $\phi$ is a homomorphism. Simplifying, we have $(ax+b)^2 + (ax+b) + 1 = a^2x^2 + 2abx + b^2 + ax+b+1 = a^2x^2 + (2ab+a)x + (b^2+b+1) = x^2+3$.

However, this appears to be a problem. Notably, since $a^2 = 1$ mod 5, we have $a = 1,4$. Picking the former, we have $2b+1 = 0$ and thus $b = 2$, a problem as $2^2+2+1 = 7 \neq 1$ mod 5. Trying $a=4$, we have $3b+4 = 0$ so $b = 2$, and again the same problem.

Are there flaws in my technique or arithmetic that I don't see? If not, what other options do I have to attempt to construct the isomorphism?

• You don't need to get $\phi(x^2+x+1)=x^2+3$. It suffices, if $\phi(x^2+x+1)$ is some (polynomial or scalar!) multiple of $x^2+3$. Commented Jun 27, 2015 at 19:36
• Think about it this way. The element $\alpha=x+(x^2+x+1)$ is a zero of the polynomial $x^2+x+1$. Then $\phi(\alpha)=ax+b+(x^2+3)$ will also be a zero of $x^2+x+1$. This is the case if and only if $(ax+b)^2+(ax+b)+1\in (x^2+3)$. In other words, iff $(ax+b)^2+(ax+b)+1$ is divisible by $x^2+3$. Commented Jun 27, 2015 at 19:40
• I see now. Allow me to add an answer with the updated technique. Commented Jun 27, 2015 at 19:46

HINT:

You want $(\mathbb{Z}/5)[x]/(x^2+x+1) \to (\mathbb{Z}/5)[x]/(x^2+3)$, $x \mapsto a x + b$, so $x^2+x+1 \mapsto (ax+b)^2+ (ax + b) +1$, and you want the latter a multiple of $x^2+3$. Divide $(ax+b)^2+ (ax + b) +1$ by $x^2+3$ and you get the remainder $$a\,(2 b+1)\, x + (b^2 + b+1 - 3 a^2)$$

As per Jyrki Lahtonen's hint, I will instead try $\phi(x^2+x+1) = c(x^2+3)$. So then since $x \mapsto ax+b$, we have $a^2x^2 + (2ab+a)x + b^2+b+1 = cx^2 + 3c$, so $a^2 = c$, $2ab+a = 0$, and $b^2+b+1 = 3c$. Since $a^2 = c$ and the only squares mod 5 are 1 and 4, we have that $c = 1,4$. Having already gleaned that 1 will not suffice, let us try $c = 4$.

Then $a^2 = 4$ so $a = 2,3$. The former did not yield nice results, so let $a = 3$. Then we have that $b + 3 = 0$ so that $b = 2$, and from the last relation $7 = 12$ which is true mod 5.

Therefore $a = 3$ and $b = 2$, so $x \mapsto 3x+2$ is an isomorphism.

• This time you can use a constant $c$, but if the extensions were cubic, you need something else. For example to construct an isomorphism from $\Bbb{Z}_2[x]/\langle x^3+x+1\rangle$ to $\Bbb{Z}_2[x]/\langle x^3+x^2+1\rangle$ you need to map the coset of $x$ to the coset of $1/x$, which is also the coset of $x^2+x$. Then you need a cubic polynomial in place of $c$ (and consequently in that case it is better to look for a zero of the minimal polynomial). Commented Jun 27, 2015 at 20:09

Hint $\$ Complete the square $\ 4(x^2\!+\!x\!+\!1) = (2x\!+\!1)^2\!+\!3 = X^2\!+\!3,\$ for $\,X = 2x\!+\!1$