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?

  • $\begingroup$ No, you cannot "define an isomorphism that sends a to a and b to b" because that is not an isomorphism. For a general method of constructing an isomorphism between two different representations (such as $\mathbb Z_{11}[x]/< x^2+1>$ and $\mathbb Z_{11}[x]/<x^2+x+4>$) of the same finite field $(\mathbb F_{121}$ in your case), see Arturo Magidin's answer to this question. For a different isomorphism between the representations (which works in that specific instance), see my answer to the same question. $\endgroup$ – Dilip Sarwate May 9 '16 at 15:45
  • $\begingroup$ "How would I approach a more general problem where the fields are varied or the polynomials have differing degrees?" You do know that the characteristic of the two fields must be the same, and that if the polynomials are of different degrees, then the underlying rings must be different? That is, there is no isomorphism between any quotient rings of $\mathbb Z_p[x]$ and $\mathbb Z_q[x]$ where $p$ and $q$ are distinct primes, or between polynomial quotient rings where the polynomials are of different degrees over the same ring $\mathbb Z_p[x]$? $\endgroup$ – Dilip Sarwate May 9 '16 at 16:16
  • $\begingroup$ 3.9.4 $\endgroup$ – Akash Gaur Oct 15 '17 at 13:55

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.