1
$\begingroup$

Consider the "nice" ring $(\mathbb{Z}/20\mathbb{Z})[x]$ and I am trying to list all the prime and maximal ideals of this.

The reason I call this a nice (or manageable) ring is because we immediately use the Chinese remainder theorem to deduce that $\mathbb{Z}/20\mathbb{Z}\cong\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$, and since ring homomorphisms (in this case ring isomorphisms) preserve "ring structure" (namely, preserve ideals), it suffices for us to look at the prime and maximal ideals on $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$.

Furthermore, to make our "searching" efficient, since we know that every maximal ideal is prime (this follows from using lattice isomorphism theorem and realizing that every field is an integral domain, but the proof is not essential at the moment...), it is most efficient to search first for prime ideals and then look for maximal ideals within that set.

A nice characterization of prime ideals is that they contain nilpotent ideal. But how do I proceed from here?

$\endgroup$
1
$\begingroup$

We have an isomorphism of rings $$\mathbb{Z}/20 [x] \cong (\mathbb{Z}/4 \times \mathbb{Z}/5)[x] \cong \mathbb{Z}/4[x] \times \mathbb{Z}/5[x]$$ It is well-known how to describe (prime) ideals in products in terms of the factors, so that we only have to look at $\mathbb{Z}/4[x]$ and $\mathbb{Z}/5[x]$. Well, $\mathbb{Z}/5$ is a field, hence $\mathbb{Z}/5[x]$ is a principal ideal domain. The non-zero prime ideals are those of the form $(f)$, where $f \in \mathbb{Z}/5[x]$ is irreducible and monic.

Although $\mathbb{Z}/4$ is not a field, we have that $(\mathbb{Z}/4)_{red}=\mathbb{Z}/2$ is a field. Hence, $(\mathbb{Z}/4 [x])_{red} \cong \mathbb{Z}/2[x]$ and again the prime ideals are known. Now use that for any commutative ring $A$, the projection $A \twoheadrightarrow A_{red}$ induces a homeomorphism $\mathrm{Spec}(A_{red}) \cong \mathrm{Spec}(A)$.

Alternatively, if you already know the prime ideals of $\mathbb{Z}[x]$, just look at those which contain $20$. Geometrically, you look at the fiber of $\mathbb{A}^1_{\mathbb{Z}} \to \mathrm{Spec}(\mathbb{Z})$ of the two points $(2)$ and $(5)$.

$\endgroup$
  • $\begingroup$ What does $(\mathbb Z /4)_red$ mean? $\endgroup$ – anonymous489 Aug 17 '14 at 0:09
  • $\begingroup$ $A_{red}:=A/\sqrt{0}$, the smallest quotient of $A$ which is reduced. $\endgroup$ – Martin Brandenburg Aug 17 '14 at 5:52

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.