Are there any commutative rings in which no nonzero prime ideal is finitely generated?
I feel like the example (or proof of impossibility) ought to be obvious, but I'm not seeing it.
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user26857
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Daniel McLaury
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1Fields, but that's just vacuously true. I suppose you want to discount those? – Casteels Sep 01 '14 at 05:27
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The standard example is $R := k[x_1, x_2, \ldots]/(x_1^2, x_2^2, \ldots)$. Then $R$ has a single prime ideal $\mathfrak{m} = (x_1, x_2, \ldots)$ which is not finitely generated ($\mathfrak{m}$ is the only prime since it is maximal, but every prime ideal of $R$ must contain $x_1^2, x_2^2, \ldots$ and hence $\mathfrak{m}$).
zcn
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2Man, I'm pretty sure I've asked questions before where that was the relevant example. I just need to get that ring tattooed on my arm or something... – Daniel McLaury Sep 01 '14 at 07:21
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I generally am not very fond of tattoos, but in your case I'll make an exception! – Georges Elencwajg Sep 02 '14 at 10:20