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Out of curiosity, are there any relatively quick classifications of all the Gaussian primes, the primes in $\mathbb{Z}[i]$?

I found a classification here, but the process comes off as rather tedious. No doubt the end classification is nice, but is there a quicker and cleaner process to classify them all?

If it matters at all, I have a decent grasp on elementary number theory/quadratic reciprocity and ring/field theory but not much in the way of algebraic number theory yet.

J. M. ain't a mathematician
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Koji Hamada
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  • What about primes $3 \bmod 4$? – draks ... Jul 18 '12 at 06:51
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    The paper in your link classifies primes in $\mathbb{Z}[i]$ in terms of primes in $\mathbb{Z}$, which is about as good as one can expect. To go further, you'd need a "classification" of primes in $\mathbb{Z}$. – Ted Jul 18 '12 at 06:52
  • For some proof for the uniqueness of the deomcposition, go here: $p=4n+3$ never has a Decomposition into $2$ Squares – draks ... Jul 18 '12 at 07:06
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    It is a little bit tedious to establish that the Gaussian Integers form a Euclidean domain, but what could be more elegant than the fact that rational primes $p$ such that $p \equiv 3$ (mod 4) remain prime in the Gaussian integers,while if $p = 2$ or $p \equiv 1$ mod $4,$ there are unique positive integers $a \leq b$ such that $p = a^2 +b^2$, and $a+bi$ is a prime element of the Gaussian integers. Furthermore, up to multiplication by a unit, these two cases determine all the primes in the Gaussian integers. – Geoff Robinson Jul 18 '12 at 11:52

2 Answers2

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Lattice theorem for rings: For any ring homomorphism $\psi:A\to B$ and any ideal $I$ of $B$, the set $\psi^{-1}(I)$ is an ideal of $A$. This map of ideals is inclusion-preserving, i.e. if $I\subseteq J$ are ideals of $B$, then $\psi^{-1}(I)\subseteq \psi^{-1}(J)$. Furthermore, for any prime ideal $P\subset B$, we have that $\psi^{-1}(P)$ is a prime ideal of $A$.

We have an inclusion homomorphism $j:\mathbb{Z}\to\mathbb{Z}[i]$. Our strategy is to classify the (non-zero) prime ideals $P$ of $\mathbb{Z}[i]$ according to the "value" of $j^{-1}(P)=P\cap \mathbb{Z}$.

First, we will re-express the ring $\mathbb{Z}[i]$ in a more convenient form:

Consider the ring homomorphism $\operatorname{ev}:\mathbb{Z}[x]\to\mathbb{Z}[i]$ defined by $\phi(f)=f(i)$. This function is surjective, and the kernel of $\operatorname{ev}$ is the ideal $(x^2+1)$ of $\mathbb{Z}[x]$. Therefore, by the first isomorphism theorem, $\operatorname{ev}$ descends to an isomorphism $\phi:R\to\mathbb{Z}[i]$, where $R=\mathbb{Z}[x]/(x^2+1)$.

For the remainder of this post, we will now be interested in classifying the prime ideals of $R$. The lattice theorem guarantees that $\phi$ sets up a bijective, order-preserving correspondence between the prime ideals of the two rings. I can explain more about why this is okay if you'd like.

Consider the inclusion homomorphism $k:\mathbb{Z}\to R$ (of course $k=\phi^{-1}\circ j$); we will pretend for the sake of expediency that $\mathbb{Z}$ is actually contained as a subset of $R$.

For each (non-zero) prime ideal $(q)$ of $\mathbb{Z}$, we want to classify the prime ideals $Q$ of $R$ such that $k^{-1}(Q)=(q)$, i.e. the prime ideals $Q$ of $R$ that contain $(q)$. Note that a prime ideal $Q$ of $R$ will contain the $\mathbb{Z}$-ideal $(q)$ if and only if $Q$ contains the element $q$, which is the case if and only if $Q$ contains the $R$-ideal $qR$. By the lattice theorem, the prime ideals of $R$ containing $qR$ are in order-preserving bijection with the prime ideals of $$R/qR=(\mathbb{Z}[x]/(x^2+1))/q(\mathbb{Z}[x]/(x^2+1))\cong \mathbb{Z}[x]/(q,x^2+1)\cong \mathbb{F}_q[x]/(x^2+1).$$ If $q=2$, then in $\mathbb{F}_2$ we have $x^2+1=(x+1)^2$, so that $$R/2R\cong \mathbb{F}_2[x]/(x+1)^2$$ has one prime ideal, namely $(x+1)\mathbb{F}_2[x]/(x+1)^2$. Unwinding our various isomorphisms, this corresponds to $i+1$ in $\mathbb{Z}[i]$.

If $q\equiv 1\bmod 4$, then $x^2+1$ is reducible (by quadratic reciprocity there is some $a\in \mathbb{F}_q$ such that $a^2=-1$ in $\mathbb{F}_q$), so that $$R/qR\cong \mathbb{F}_q[x]/(x-a)(x+a)$$ which has two prime ideals, $(x-a)\mathbb{F}_q[x]/(x-a)(x+a)$ and $(x+a)\mathbb{F}_q[x]/(x-a)(x+a)$. Unwinding our various isomorphisms, this corresponds to a conjugate pair of Gaussian primes $\pi$, $\overline{\pi}$ with norm $q\equiv 1\bmod 4$ in $\mathbb{Z}[i]$.

Lastly, if $q\equiv 3\bmod 4$, then $x^2+1$ is irreducible (by quadratic reciprocity there is no $a\in \mathbb{F}_q$ such that $a^2=-1$ in $\mathbb{F}_q$), so that $$R/qR\cong \mathbb{F}_q[x]/(x^2+1)\cong\mathbb{F}_{q^2}$$ is a field and therefore has one prime ideal, namely the zero ideal. Unwinding our various isomorphisms, this corresponds to a Gaussian prime $q\equiv 3\bmod 4$ that lives in $\mathbb{Z}$.

Needless to say, this method is well-known and not original to me; for example I'm pretty sure that Neukirch does precisely this at the outset of his book. I'll see if I can find some good references later. Also, I'd just like to comment that algebro-geometrically, what's happening is we're looking at the fibers of $\operatorname{Spec}\mathbb{Z}[i]$ over $\operatorname{Spec}\mathbb{Z}$; look at the curve in $\operatorname{Spec}\mathbb{Z}[x]$ corresponding to the ideal $(x^2+1)$ in Mumford's famous sketch:

enter image description here

I have to get to sleep, so I'm signing off for now, but I'll respond to any comments or questions when I can.

Zev Chonoles
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  • Thanks Zev, I appreciate this, it is very nice. Do you mind explaining how you see the isomorphism $(\mathbb{Z}[x]/(x^2+1))/q(\mathbb{Z}[x]/(x^2+1))\cong \mathbb{Z}[x]/(q,x^2+1)$? Other than that, I can follow the rest. – Koji Hamada Jul 18 '12 at 07:49
  • @Zev Chonoles Could you give a reference for this method? I looked at Neukirch, but his approach is different – bateman May 30 '13 at 09:06
  • "The lattice theorem guarantees that $\phi$ sets up a bijective, order-preserving correspondence...". Where in the lattice theorem did it say "bijective"? In fact any non-real irreducible element $z$ of $\Bbb Z[\mathbf i]$ generates a prime ideal that intersects $\Bbb Z$ in the same set as the "complex conjugate" ideal genrerated by $\overline z$, and the two ideals are distinct unless $z$ is (associated to) $1+\mathbf i$. – Marc van Leeuwen May 30 '13 at 13:26
  • @Marc: $\phi$ is a map from $\mathbb{Z}[x]/(x^2+1)$ to $\mathbb{Z}[i]$. – Zev Chonoles May 30 '13 at 13:36
  • OK, I misread that, and "the two rings" are not $\Bbb Z$ and $\Bbb Z[\mathbf i]$. But then the sentence should not have mentioned the lattice theorem, which has nothing to do with it; any isomorphism of rings sets up a bijection between their prime/maximal/whatever ideals. – Marc van Leeuwen May 30 '13 at 13:53
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The Gaussian primes are precisely the rational primes congruent to 3 modulo 4 and their associates, and the numbers $a+bi$ such that $a^2+b^2$ is a rational prime. Seems pretty clean to me, and you won't find anything better.

Gerry Myerson
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