2
$\begingroup$

I'm trying to prove the order of a particular quotient field of $\mathbb{Z}[i]$ without using the characteristic of the field. My hesitation in using the characteristic comes from the fact that I'm working my way through Dummit and Foote's Abstract Algebra and encountered this question before the section on field theory. This question arose in the section on UFD's. There must by some way to prove my claim without using the field theoretic notion of a characteristic.

Claim: Let $q \in \mathbb{Z}$ be a prime such that $q \equiv 3$ (mod $4$). Then, $\mathbb{Z}[i]/(q)$ is a field of order $q^2$.

A few things about this claim are immediately clear. First, q is clearly an irreducible in $\mathbb{Z}[i]$ because it would only be reducible if $q=2$ or $q \equiv 1$ (mod $4$). Since $\mathbb{Z}[i]$ is a UFD, it follows that $(q)$ is prime. Moreover, it's pretty simple to show that $\mathbb{Z} \cap (q) = q\mathbb{Z}$. By the second isomorphism theorem for rings, you then recover that $\mathbb{Z} + (q)/(q) \cong \mathbb{Z}/q\mathbb{Z}$. But that seems like a dead end because I can't figure out a way to break down the rest of $\mathbb{Z}[i]$ and show that it's disjoint with $\mathbb{Z} + (q)$ and isomorphic to $\mathbb{Z}/q\mathbb{Z}$ also.

$\endgroup$
  • 2
    $\begingroup$ This is precisely an exercise in Dummit and Foote, so I tend to think that it's true. $\endgroup$ – Will Byrne Feb 15 '16 at 1:14
  • $\begingroup$ Oh yeah, I confused $\mathbb Z[x]$ and $\mathbb Z[i]$, sorry about that. $\endgroup$ – Jorge Fernández Hidalgo Feb 15 '16 at 1:21
4
$\begingroup$

It is easy to see that $a+bi\in\mathbb{Z}[i]$ is divisible by $q$ iff both $a$ and $b$ are divisible by $q$ as integers. It follows that $a+bi\equiv c+di$ mod $q$ iff $a\equiv c$ and $b\equiv d$ mod $q$. So this gives a bijection (in fact, an additive group-isomorphism) $\mathbb{Z}[i]/(q)\to \mathbb{Z}/(q)\times\mathbb{Z}/(q)$.

$\endgroup$
  • $\begingroup$ Oh wow, of course. Thanks so much! $\endgroup$ – Will Byrne Feb 15 '16 at 1:18
2
$\begingroup$

We can do it with little machinery. The elements of our ring are of the form (the equivalence class of) $a+bi$ where $0\le a\le q-1$ and $0\le b\le q-1$. So there are $q^2$ of them.

We want to show that every non-zero element $a+bi$ has an inverse. The obvious candidate is the equivalence class of $c(a-bi)$, where $c$ is the inverse of $a^2+b^2$ modulo $q$. We will know that $c$ exists once we know that $a^2+b^2$ is not divisible by $q$.

If $a^2+b^2$ were divisible by $q$, the congruence $x^2\equiv -1\pmod{q}$ would have a solution. But for primes of the form $4k+3$, it does not.

$\endgroup$
-1
$\begingroup$

Hint: All you have to prove is that $-1$ is not a square modulo $q$ if $q\equiv 3\mod 4$. This results from Little Fermat and Euler's criterion for quadratic residues.

Some details: $$\mathbf Z[i]/(q)\simeq (\mathbf Z[x]/(x^2+1))/q(\mathbf Z[x]/(x^2+1))\simeq \mathbf (Z/q\mathbf Z)[x]/(x^2+1)$$ hence $\mathbf Z[i]/(q)$ is a field if and only if $x^2+1$ is irreducible modulo $q$, i.e. if and only if $-1$ is not a square modulo $q$.

Now by Little Fermat, in $\mathbf Z/q\mathbf Z$ all non-zero elements satisfy the equation $x^{q-1}-1=0$. But since $q$ is odd, this polynomial factors as $$x^{q-1}-1=\bigl(x^{\tfrac{q-1}2}-1\bigr)\bigl(x^{\tfrac{q-1}2}+1\bigr)$$

If $-1$ were a square modulo $q$, say $-1\equiv\alpha^2\mod q$ for some $\alpha$, we would have $$(-1)^{\tfrac{q-1}2}=\alpha^{q-1}=1$$ This is impossible if $q\equiv 3\mod 4$, since $\dfrac{q-1}2$ is odd in that case. Thus, $x^2+1$ is irreducible over $\mathbf Z/q\mathbf Z$ and $\mathbf Z[i]/(q)$ is a field. As it has degree $2$ over $\mathbf Z/q\mathbf Z$, it contains $q^2$ elements

$\endgroup$

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.