Finite fields isomorphisms and ideals

Question

What map is an isomorphism between the two fields $\mathbb{F}_3[x]/(x^2+1)$ and $\mathbb{F}_3[x]/(x^2+2x+2)$?

Now, the elements of $\mathbb{F}_3[x]/(f_1)$ where $f_1=x^2+1$ are $1+(f),2+(f),0+(f),x+(f),2x+(f_1),2x+1+(f_1),2x+2+(f_1)$.

I know that both of the above fields have 9 elements, however, I'm not particularly sure what isomorphism would work.

Answer 1

Working in $ \mathbb F_3[x]/(x^2 + 2x + 2) $, you have

$$ (ax + b)^2 = a^2 x^2 + 2ab x + b^2 = a^2 (x+1) + 2ab x + b^2 = (2ab + a^2) x + (a^2 + b^2) $$

You want an element whose square is equal to $ 2 $ (which would be a root of $ x^2 + 1 $), so you want $ 2ab + a^2 = 0 $ and $ a^2 + b^2 = 2 $. It's trivial to see the choice $ a = b = 1 $ works, so the desired isomorphism is given by $ [x] \to [x+1] $ as a map $ \mathbb F_3[x]/(x^2 + 2x + 2) \to \mathbb F_3[x]/(x^2 + 1) $.

Answer 2

Observe that: $x^2+2x+2=(x+1)^2+1$

Hint:

Look at the $\mathbb F_3$ algebra homomorphism

$f:\mathbb F_3 [x] \to \frac{\mathbb F_3 [x]}{(x+1)^2+1)}$

Such that $f(x)=x+1$

Here is an exercise for you : $Kerf=(x^2+1)$

Then apply first isomorphism theorem.

Answer 3

Note that we have the following abstract result:

If $k$ is a field and $f \in k[X]$ with deg($f$)=d, then $k[X]/(f)$ is a $k$-vector space of dimension d.

Since any two finite dimensional $k$-vector spaces are (non-naturally) isomorphic if they have the same dimension, so it holds for any two polynomials $f,g \in k[X]$ with deg($f$)=deg($g$), that $k[X]/(f) \cong k[X]/(g)$(as vector spaces!).

I know this doesn't answer your question in particular, but maybe it helps for your intuition.

Edit:

As @EthanBolker noted correctly, an isomorphism of vector spaces obviously doesn't give us an isomorphism of fields.

However we can use the structure of finite fields that have, up to isomorphism, a number of elements $p^d$, with a prime $p$.

We use the fact that every finite field with $p^d$ elements is isomorphic to $\mathbb{F}_{p^d}$.

Since $\mathbb{F}_p[X]/(f)$ ($f$ must be irreducible for this to be a field) is isomorphic to a d-dimensional $\mathbb{F}$-vector space, we know by a little combinatorics that $\mathbb{F}_p[X]/(f)$ has $p^{deg(f)}$ elements and is (as a field!) thus isomorphic to $\mathbb{F}_{p^{deg(f)}}$