Show two sets are isomorphic as fields

Question

We have two sets $A$ and $B$. $A = \{a + b \sqrt{2}: a, b \in \mathbb{Z}_{5}\}$ and $B = \{ a + b \sqrt{3}: a,b \in \mathbb{Z}_{5}\}$. We want to show that they are isomorphic as fields. I think that the structure of these two sets are a little bit similar ot $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q} [\sqrt{3}]$. However, $\mathbb{Q}[\sqrt{2}]$ is not isomorphic with $\mathbb{Q}[\sqrt{3}]$. Thus,I tried to use the definition of isomorphism. We need to find a bijection mapping $\phi$ from $A$ to $B$ such that $\phi(a + b) = \phi(a) + \phi(b)$, for all $a,b \in A$. And also we need $\phi(ab) = \phi(a)\phi(b)$, for all $a,b \in A$. But I'm confused about how to find such a mapping. Could anyone help me this please?
Thanks so much!

Answer 1

In the following we shall directly write $n$ when we mean $\bar{n}$ or $n+5\mathbb{Z}$.

More formally the question can be restated as:

Let $F_5$ be the field with $5$ elements. Then $F_5(a)\cong F_5(b)$, where ${\rm irr}(a, F_5)=x^2-2$ and ${\rm irr}(b, F_5)=x^2-3$.

Solution:

Method 1: Since $F_5(a)$, $F_5(b)$ both have dimension $2$ as vector space over $F_5$, they both have $25$ elements. Any two finite fields with same cardinality are isomorphic.

Method 2: For the second field, $b^2-3=0\implies (3b)^2-2=0$. Therefore ${\rm irr}(3b, F_5)=x^2-2$. You can check $F_5(b)=F_5(3b)$ and $3b\mapsto a$ induces an isomorphism $F_5(a)\cong F_5(b)$.