## Polynomial rings have infinitely many maximal ideals

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