What, generally, is the strategy for proving if two polynomial quotient fields are isomorphic? Say from $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ to $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$? My first instinct is to see they have the same number of elements $(121)$ and that elements in both fields are of form $a + bx$. Maybe define an isomorphism which sends
$a$ to $a$ and $b$ to $b$?
How would I approach a more general problem where the fields are varied or the polynomials have differing degrees?
Best Answer
For this problem:
0) General Approach:
We first show both $x^{2}+1$ and $x^2+x+4$ are irreducible in the polynomial ring $\mathbb{Z}_{11}[x]$. Since these polynomials are of degree $2$, the irreducibility can be checked by running through $0,1,\cdots,10$ to see none of these is a root.
Now since both $x^{2}+1$ and $x^2+x+4$ are irreducible. The principal ideals generated by them are maximal in the ring $\mathbb{Z}_{11}[x]$. It follows that $\mathbb{Z}_{11}[x]/(x^{2}+1)$ and $\mathbb{Z}_{11}[x]/(x^2+x+4)$ are two finite fields of same order $11^{2}=121$. There is one and only one finite field $\mathbb{F}_{p^{n}}$ of given order prime power $p^{n}$, being the splitting field of $x^{p^{n}}-x$ over $\mathbb{F}_{p}$. We conclude $\mathbb{Z}_{11}[x]/(x^{2}+1)$ and $\mathbb{Z}_{11}[x]/(x^2+x+4)$ are isomorphic (to the finite field $\mathbb{F}_{121}$).
1) Specific Approach:
$\mathbb{Z}_{11}[x]/(x^{2}+1)=\mathbb{Z}_{11}[\alpha]$ where $\alpha$ is an abstract root of $x^{2}+1$.
$\mathbb{Z}_{11}[x]/(x^2+x+4)=\mathbb{Z_{11}}[\beta]$ where $\beta$ is an abstract root of $x^{2}+x+4$.
Suppose $i:\mathbb{Z}_{11}[\alpha]\rightarrow \mathbb{Z_{11}}[\beta]$ is an isomorphism, it must fix $1$: $i(1)=1$ since $1$ is identity in both sides. It follows that $i$ fixes $\mathbb{Z}_{11}\subset \mathbb{Z}_{11}[\alpha]$. $i$ is purely determined by $i(\alpha)$:
We know $$i(\alpha^{2}+1)=i(0)=0=i(\alpha)^{2}+1$$ $$=\beta^{2}+\beta+4$$ Hence $i(\alpha)^{2}=\beta^{2}+\beta+3.$
Now how about set $i(\alpha)=\beta+6$? It leaves for you to check this induces an explicit isomorphism that can be written down.