I've trouble with understanding some details about noetherian rings.

For example, I don't understand why this theorem is wrong:

A ring $R$ is noetherian if and only if it's finitely generated $\mathbb Z$-algebra.

My proof: the implication "noetherian $\Rightarrow$ finitely generated" is standard and obvious. The inverse implication also looks rather simple to me: cannot we say that finitely generated $\mathbb Z$-algebra is quotient $\mathbb Z[x_1, \ldots, x_n]$ / $\mathfrak I$ ($\mathfrak I$ is ideal), so by Hilbert basis theorem it's a quotient of the noetherian ring, so it's noetherian itself?

Can somebody help me and explain to me where is my mistake?


You should rethink the implication "noetherian $\Rightarrow$ finitely generated". Noetherianness only implies that every ideal is finitely generated as an ideal, but not necessarily as a ring. Every field is noetherian, for example, but an uncountable field cannot be finitely generated. In fact, no infinite field is finitely generated as a ring ; see If a ring is Noetherian, then every subring is finitely generated? for details.

  • $\begingroup$ Thank you very much! I've already found a mistake in this implication... $\endgroup$ – John Watson Jul 2 at 10:40
  • $\begingroup$ Could you say me, please, is it true that a submodule of a finitely generated module over finitely generated $\mathbb Z$-algebra is finitely generated? Is it true for quotients? $\endgroup$ – John Watson Jul 2 at 10:45
  • 1
    $\begingroup$ It is easy to see for quotients, and for submodules, see this question. $\endgroup$ – Arnaud D. Jul 2 at 10:51
  • $\begingroup$ Thank you for your attention and help. $\endgroup$ – John Watson Jul 2 at 10:53

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