R is local ⟹ R[x]/(f(x)) is semilocal

Let be a commutative ring with identity. Recall that we say is local if has only one maximal ideal, and we say is semilocal if has only finitely many maximal ideals. In this note, we show that if is local and is a non-constant monic polynomial, then is semilocal. This is a generalization of a simple and well-known fact about fields. Let’s first prove it for fields.

Lemma. Let be a field, and let be a non-constant polynomial. Then the ring is semilocal.

Proof. A maximal ideal of has the form where is a maximal ideal of that contains So we just need to show that the number of maximal ideals of which contain is finite. Since is a PID, for some irreducible polynomial Since we have for some Therefore is an irreducible factor of and we are done because the number of irreducible factors of is finite.

OK, that was easy! Now, let’s make it interesting. Every field is a local ring because is the only proper ideal of a field. We now use the Lemma to extend the result given in the Lemma to any commutative local ring.

Theorem. Let be a commutative ring with identity, and let be a non-constant monic polynomial. If is local, then is semilocal.

Proof. Since a maximal ideal of has the form where is a maximal ideal of containing we just need to show that the number of maximal ideals of which contain is finite.

Let be a maximal ideal of which contains Let

Note that is both a field and an -module.

Claim 1. is a finitely generated -module.

Proof. Let Since we have

and so, by induction,

for all integers Thus

for all integers and hence as an -module, is generated by This completes the proof of Claim 1.

Now, since is local, it has a unique maximal ideal. Let be the unique maximal ideal of

Clim 2.

Proof. Since is a field, and is an ideal of we have either or If then

and so and we are done. If then, by Claim 1 and Nakayama Lemma, there exists such that Thus since is a unit of because we must have which is absurd because This completes the proof of Claim 2.

Now we can finish the proof of the Theorem. By Claim 2, is a maximal ideal of that contains But and so is isomorphic to a maximal ideal of which contains the non-constant polynomial

The result now follows from the Lemma because is a field.

Remark 1. The in the Theorem doesn’t have to be monic, what we need is for the leading coefficient of say to be a unit because if is a unit, then is monic and

Remark 2. The Theorem does not necessarily hold if the leading coefficient of is not a unit.

Proof. So we need to give an example of a local ring and a non-constant polynomial such that has infinitely many maximal ideals. Let be the ring in this post, and choose Notice that the leading coefficient of which is is not a unit of Then, by the Exercise in that post, which, as we showed in that post, has infinitely many maximal ideals.

Note. The Theorem is the second part of Lemma 16.2 in Donald Passman’s book A Course in Ring Theory.

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