Need help to construct a ring isomorphism from $\mathbb{Z}_3[\sqrt2]$ $\rightarrow $ $\mathbb{Z}_3[i]$

Question

I've looked at this problem for some time now and am confused as to where to start. I know that to prove these rings are isomorphic I must to construct a bijective function $\phi$ such that:

$\phi(a + b)$ = $\phi(a) + \phi(b)$

and

$\phi(ab)$ = $\phi(a)\phi(b)$.

I'm not exactly sure where to begin. If anyone could give some insight on where to begin or something to think about I would really appreciate it. Thanks!

Answer 1

Each ring has nine elements. Clearly you need to send $0$ to $0$ and $1$ to $1$, which forces you to send $2$ to $2$ to maintain addition. Now $\sqrt 2$ has to go somewhere. As $\sqrt 2 \cdot \sqrt 2=2$ you need it to go somewhere that squares to $2=-1$. Addition and multiplication will then fill out the function.

Answer 2

ADDED: It occurred to me that things might have a more pleasant look if I used separate variables, one quotient $F[x] / (x^2 + 1)$ and the other $F[t] / (t^2 -2),$ where $F = \mathbb Z / 3 \mathbb Z.$ Then the isomorphism has a concrete appearance, $$ ax+b \mapsto at+b $$

ORIGINAL:Well, $x^2 + 1 \equiv x^2 - 2 \pmod 3,$ which is promising.

Writing as $F[x] / (x^2 + n)$ for some $n$ we are writing elements as $ax+b$ with $a,b \in F = \mathbb Z / 3 \mathbb Z.$

With $x^2 + 1,$ we get multiplication $$ (ax+b)(cx+d) = (ad+bc)x + (bd-ac) $$

With $x^2 -2,$ we get multiplication $$ (ax+b)(cx+d) = (ad+bc)x + (bd+2ac) $$

These look slightly different, but in $a,b,c,d \in F,$ we always have $$ -ac = 2ac $$ because $$ -ac \equiv 2ac \pmod 3 $$

Answer 3

There is nothing to prove at all. Here $\sqrt{2}$ can only be interpreted as an element $u$ in some field extension of $\mathbb{Z}_3$ such that $u^2=2$; similarly, $i$ is an element in some field extension such that $i^2=-1=2$. Since both elements are roots of $x^2-2=0$, which is irreducible over $\mathbb{Z}_3$, the two fields are obviously isomorphic and an isomorphism is obtained by $$ a+b\sqrt{2}\mapsto a+bi $$ If you use a common extension field, such as the algebraic closure of $\mathbb{Z}_3$, then the two fields are actually equal, because $\sqrt{2}=i$ or $\sqrt{2}=-i$, both being roots of the same polynomial.

The complex numbers $\sqrt{2}$ and $i$ have nothing to do with the problem at hand.