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My question: What are the most simple examples of a commutative ring R satisfying both of the following two properties: 1. R is not Noetherian. 2. R has exactly one prime ideal.

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Hint: you can get an example of this by finding a ring whose maximal ideal consists of nilpotent elements but is not nilpotent (e.g. a suitable quotient of $k[x_1, x_2, \dots]$). –  Akhil Mathew May 16 '11 at 18:07
    
discrete valuation ring is a principal ideal ring having a unique prime ideal. –  user9413 May 16 '11 at 18:08
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@Chandru : dvr's have two prime ideals -- the maximal ideal and the zero ideal. –  David Speyer May 16 '11 at 18:17
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@Chandru Any field is noetherian. –  Adrián Barquero May 16 '11 at 18:24
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@spec: Dear Spec, Actually, what you just described is precisely the example I had in mind; that ring has one prime ideal (namely, that generated by all the $x_i$), which is not nilpotent. –  Akhil Mathew May 16 '11 at 23:59

4 Answers 4

Try $k[x_1, x_2, ...]/(x_i x_j)$ for all $i, j$.

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Sadly, no. The ideal $\langle x_j \langle_{j > 1}$ is prime, since the quotient is $k[x_1]$. –  David Speyer May 16 '11 at 18:54
    
@David: I deliberately did not exclude the case $i = j$. I'm pretty sure this works. The quotient by the ideal you describe is $k[x_1]/(x_1^2)$. –  Qiaochu Yuan May 16 '11 at 19:15
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In other words, if $I = (x_1,\ldots,x_n,\ldots)$, then this is $k[x_1,x_2,\ldots]/I^2$. –  Arturo Magidin May 16 '11 at 19:19
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Oh, I misread you. You are absolutely right. –  David Speyer May 16 '11 at 19:22

$$R:= k[x_1, x_2, x_3, \cdots ]/\langle x_1^2,\ x_2^2=x_1,\ x_3^2=x_2,\ x_4^2=x_3,\ \cdots \rangle$$

Remember that prime ideals are precisely the kernels of maps to fields. If $K$ is any field, and we have a map $\phi: R \to K$, then $\phi(x_1)=0$ and hence $\phi(x_2) =0$ and hence $\phi(x_3)=0$ and so forth, so the only prime ideal is $\langle x_1, x_2, x_3, \ldots \rangle$.

To see that this is not noetherian, note that $(0) \subsetneq (x_1) \subsetneq (x_2) \subsetneq (x_3) \subsetneq \cdots$ is an ascending chain of ideals that doesn't terminate.

Of course, the specific exponent $2$ isn't important. The key is to make a sequence of variables $x_1$, $x_2$, ..., all of which are nilpotent, but so that you can kill $x_1$, $x_2$, ..., $x_k$ while not killing the later $x$'s.

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If you're willing to ask for one nonzero prime ideal, then for any prime $p$ the ring of integers of ${\mathbf C}_p$ (the $p$-adic complex numbers) is an example. The only nonzero prime ideal in that ring is its maximal ideal, which is not finitely generated.

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And if you form $\mathcal O_{\mathbf C_p}/p,$ this will be non-Noetherian with just a single prime ideal. –  Matt E May 17 '11 at 0:12

Take a localization $O:=R_M$ of the ring $R$ of all algebraic integers at a maximal ideal $M$. Then $O/xO$ has exactly one prime for every non-zero $x\in MO$.

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