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A few questions relevant formally, but quite different in nature:

From now on, let R denote a ring.

  1. If R is a UFD , is R[x] also a UFD?

  2. If R is Noetherian, is R[x] also Noetherian?

  3. If R is a PID, is R[x] also a PID?

4. If R is an Artin ring, is R[x] also an Artin ring?

For 1, we all know it's Gauss's lemma.

For 2, we all know it's Hilbert's basis theorem.

For 3, we all know that in Z[x], the ideal (2,x) is not a principal ideal, so the answer is negative.

But what about 4?

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2 Answers 2

The answer to 4 is "no." If $R$ is an Artin ring, then it is Noetherian of Krull dimension zero. It follows from dimension theory that $R[X]$ is Noetherian of dimension one, i.e., not every prime ideal in $R[X]$ is maximal, so $R[X]$ can't be Artin.

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8  
Another really easy way to see that $R[X]$ fails to be Artin is the descending chain $(X)\supseteq (X^2)\supseteq\cdots$. –  Keenan Kidwell Apr 2 '10 at 1:19
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...assuming $0 \neq 1$ :) –  S. Carnahan Apr 2 '10 at 3:20

Good argument. But let's give a down-to-earth counterexample:

Let $R$ be a field. Then consider $$(x)\supset (x^2)\supset(x^3)\supset\ldots.$$ This is a descending chain of ideals that doesn't become stationary so $R[X]$ is not Artin.

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Yes, I realized this and posted a comment. –  Keenan Kidwell Apr 2 '10 at 1:32
    
Hm, sorry. I didn't see it early enough. g –  Tilemachos Vassias Apr 2 '10 at 1:36
    
Don't worry about it. –  Keenan Kidwell Apr 2 '10 at 2:45
    
Note that one can give a very similar proof of fact that if integral domain is an artinian ring then it is a field (this also follows from consideration involving dimension). –  ifk Apr 2 '10 at 10:12

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