I am seeking an example of how Noetherian approximation can be used to simplify a proof in the context of commutative algebra strictly.

To begin, I am referring to the fact that any commutative ring can be written as a direct limit of Noetherian rings, we can in fact restrict simply to subrings as every ring is the union of its finitely generated $\mathbb{Z}$-subalgebras. Noetherian approximation has uses in algebraic geometry where using this fact as a starting point we are able to prove results in the Noetherian setting, and then generalize to general schemes usally under a (locally) finite presentation hypothesis. I am seeking a (hopefully elementary) example of where we can do a similar thing however simply in the case of commutative rings. There are numerous applications whereby writing a module as a direct limit of finitely generated modules we can simply proofs, an example that comes to mind is that via the criterion for flatness involving finitely generated ideals, we can prove that in an integral domain where every finitely generated ideal is principal, a module is flat if and only if it is torsion-free. This can be used to characterize flat modules over a valuation ring. Here the crucial observation is that since direct limits are exact and commute with the tensor product, we can restrict to only finitely generated ideals where we then leverage that they are principal.

I have thought of numerous theorems which perhaps one could use Noetherian approximation to extend a result out of the Noetherian situation, however I consistently see a trend where either the result is proved for Noetherian rings, but is false for non-Noetherian rings rendering the technique useless, or the result is proven without a Noetherian hypothesis at all and the proof does not seem to be any easier in the Noetherian case anyway.

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    $\begingroup$ An example I’ve used recently is the following statement: let $R$ be a commutative ring, and $S=R[T_1,\ldots,T_n]/(f_1,\ldots,f_r)$ be such that the matrix $\frac{\partial f_i}{\partial T_j}$ has a maximal minor with invertible projection in $S$. Then $S$ is flat over $R$. In the Noetherian case, you prove it by showing that $S$ is formally smooth over $R$, and it follows that the $R$-surjection from the $(f_1,\ldots,f_r)$-adic completion of $R[T_1,\ldots,T_n]$ to $S$ splits; as the former $R$-module is flat, so is the latter. $\endgroup$
    – Aphelli
    Jan 30, 2022 at 22:44
  • $\begingroup$ What do you mean by "invertible projection in S"? That its image is invertible in S? $\endgroup$
    – user26857
    Jan 31, 2022 at 3:07
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    $\begingroup$ @user26857: Yes, that’s what I meant. $\endgroup$
    – Aphelli
    Jan 31, 2022 at 6:36

1 Answer 1


Alex, I'm not sure if I'm reading the question properly, but what about showing that finite ring maps are integral?

Let $A\to B$ be a finite ring map. If $A$ is Noetherian and $t \in B,$ then applying the ascending chain condition $A\subset A + At\subset A+At+At^2\subset \cdots$ shows that $t$ is integral over $A$. So the case for Noetherian $A$ is very easy (no Cayley-Hamilton for instance)!

I think the reduction from the general case isn't too bad. Write $A$ as a colimit of $A_i$ that are finitely generated $\mathbb{Z}$-algebras (and hence Noetherian). Then for large $i$, we can find (using just finite presentation) an $A_i$-algebra $B_i$ s.t. $B=A\otimes_{A_i}B_i$ and $A_i \to B_i$ is finite. Since $A_i$ is Noetherian, this map and its base change are integral.

  • $\begingroup$ The usual proof that finite $\implies$ integral also needs no Noetherian hypothesis, and just relies on the determinant trick $\endgroup$
    – math54321
    May 6 at 16:37
  • $\begingroup$ You're right, of course, but assuming Noetherianness makes the proof even easier. The more important issue is that the proof I gave might be circular in the reduction step, but I'm not sure. $\endgroup$
    – caagun
    May 6 at 16:55

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