Polynomial rings have infinitely many maximal ideals

Posted: February 16, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , ,

We will assume that R is a commutative ring with unity and S=R[x], the polynomial ring over R. We will denote by J(S) the Jacobson radical of S.

Problem 1. The ring S/J(S) is never Artinian.

Solution. Let I=\langle x + J(S) \rangle, the ideal of S/J(S) generated by the coset x+J(S).

Claim . I^k \neq I^{k+1}, for all integers k \geq 1.

Proof of the claim . Suppose, to the contrary, that I^{k+1}=I^k, for some k \geq 1. Then

x^k + J(S) \in I^k = I^{k+1}=\langle x^{k+1}+J(S) \rangle

and so x^k + J(S)=f(x)x^{k+1} + J(S), for some f(x) \in S. Hence x^k - f(x)x^{k+1} \in J(S) and therefore g(x)=1 -(x^k - f(x)x^{k+1})=1-x^k + f(x)x^{k+1} must be a unit in S. But the coefficient of x^k in g(x) is -1, which is obviously not nilpotent, and so g(x) cannot be a unit (see here). Contradiction!

So, by the claim, we have a strictly descending chain of ideals of S: \ I \supset I^2 \supset I^3 \supset \cdots proving that S/J(S) is not Artinian. \Box.

Problem 2. Prove that S has infinitely many maximal ideals.

Solution. Suppose, to the contrary, that the set of maximal ideals of S is finite. Let \mathfrak{m}_1, \cdots , \mathfrak{m}_n be the maximal ideals of S. Then, by the Chinese remainder theorem, S/J(S) \cong \bigoplus_{i=1}^n S/\mathfrak{m}_i. So, since each S/\mathfrak{m}_i is a field and fields have only two ideals, S/J(S) must have finitely many (2^n in fact) ideals. But then S/J(S) would obviously be Artinian, contradicting Problem 1. \Box

Suppose that R is Noetherian. Then by the Hilbert’s basis theorem, S is Noetherian too. Thus S/J(S) is Noetherian. So S/J(S) is always a non-Artinian Noetherian ring.


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