# Isomorphisms between finite fields

I know that all finite fields of the same size are isomorphic to one another. I also know that if a polynomial $$f(x)$$ is irreducible over $$\mathbb{Z}[x]$$ and of degree $$n$$ then

$$\frac{\mathbb{Z}_k[x]}{(f(x))} \cong \mathbb{Z}_{k^n}$$

For instance, we should have that $$\frac{\mathbb{Z}_5[x]}{(x^2+2)} \cong \mathbb{Z}_{25}$$

However, I'm unsure of how to construct an explicit isomorphism between the two. I want an isomorphism so I can identify things like every generator of the field on the LHS (which will correspond to the inverse under the isomorphism of $$2, 3, 4, 6, 7$$ etc.). What is an isomorphism for this specific case and how do I then generalise this?

• $\mathbb Z_{25}$ is not a field – J. W. Tanner Apr 30 '19 at 13:28
• Okay, should it be $\mathbb{Z}_k^n$ (so in the specific example $\mathbb{Z}_5^2$)? – Adam Apr 30 '19 at 13:32
• Keep in mind: a field has no non-zero zero divisors – J. W. Tanner Apr 30 '19 at 13:49
• Perhaps you mean $\mathbb F_{25}$ instead of $\mathbb Z_{25}$ – J. W. Tanner Apr 30 '19 at 16:37

Your method of constructing a field of $$25$$ elements by the quotient ring of a quadratic polynomial is fine,

but it is a misconception that the ring of integers modulo $$25$$ is a field.

The ring $$\mathbb Z_{25}$$ has non-zero zero divisors -- for example, in it $$5\times5=0$$ even though $$5\ne0$$ --

so it is not a field. Integers modulo $$n$$ are a field when $$n$$ is prime.

The elements of the field of $$25$$ elements are of the form $$a+b\alpha,$$

where $$a,b\in\mathbb F_5=\mathbb Z_5$$ and $$\alpha$$ is a root of a quadratic polynomial that is irreducible in $$\mathbb Z_5$$.

• In other words, $\mathbb Z_{25}$ is not isomorphic to $\mathbb F_{25}$ – J. W. Tanner Apr 30 '19 at 19:29