2
$\begingroup$

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.

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

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 $\Bbb{Z}$.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – paranomasia Feb 21 '18 at 3:15
1
$\begingroup$

$\mathbb{Z}[\sqrt{2}]/(\pi)$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.