# Help determining if a field is finite?

I am currently working on a homework assignment, and I am stuck. The problem is to show that $\mathbb{Z}(\sqrt{2})$ / (a prime in $\mathbb{Z}(\sqrt{2})$) is a finite field. I have shown that $\mathbb{Z}(\sqrt{2})$ is PID, so that the quotient must be a field, but I am struggling with showing that it is finite. I am leaving some information out because I really just ant help with getting started. I have tried naively writing out elements, but it seems like there are infinitely many. I can give more information if you'd like it!

I can't tell if it is that I have been looking at this set for too long, or if I haven't touched algebra in a while, or what. Any and all help is appreciated! The course has not touched field extensions, Galois groups, etc yet.

• Do you mean the ring $\Bbb Z [\sqrt 2]$? – Fabio Lucchini Feb 19 '18 at 16:35
• Yes! $\mathbb{Z}$ adjoined with $\sqrt{2}$. I will edit this. – paranomasia Feb 19 '18 at 16:41

You can do the following. Let $\mathfrak{p}$ be the prime ideal of $\Bbb{Z}[\sqrt2]$ in question. We need to assume that $\mathfrak{p}$ is not the trivial ideal containing zero alone (that would qualify as a prime ideal according to many definitions) - otherwise your claim is false :-)

Consider the intersetion $I:=\mathfrak{p}\cap\Bbb{Z}$. Prove the following

1. If $a+b\sqrt2$ is a non-zero element of $\mathfrak{p}$, then $(a+b\sqrt2)(a-b\sqrt2)=a^2-2b^2$ is a non-zero element of $I$.
2. If $n$ is the smallest positive integer in $I$, then $n$ and $n\sqrt2$ are both elements of $\mathfrak{p}$.
3. Every coset of $\mathfrak{p}$ in $\Bbb{Z}[\sqrt2]$ contains an element of the form $a+b\sqrt2$ with $0\le a<n, 0\le b<n$.
4. There are at most $n^2$ elements in your quotient ring.

For extra credit you can prove that $n$ must actually be a prime number. Or, equivalently, that $I$ is a prime ideal of $\Bbb{Z}$.

• I like this a lot! I actually ended up defining a homomorphism $\phi:\mathbb{Z}\rightarrow\mathbb{Z}(\sqrt{2})/(3+\sqrt2)$ by $\phi(z)=a+b\sqrt{2}+(3+\sqrt{2})$, then finding its kernel, showing it was surjective, and then using first isomorphism theorem to show that $\mathbb{Z}(\sqrt{2})\cong\mathbb{Z}/7\mathbb{Z}$. ($3+\sqrt{2}$ was the prime we were using). – paranomasia Feb 21 '18 at 3:15

$\mathbb{Z}[\sqrt{2}]/(\pi)$ it is finite because is a finitely generated $\mathbb{Z}$-module with finite exponent since $N(\pi)=\pi \bar \pi \in \mathbb{Z}$ kills every element in it.