Let $A\subseteq B\subseteq C$ be commutative unital rings. Recall that the extension $A \subseteq B$ is finite / of finite type / integral, when $B$ is a finitely generated $R$-module / when $B$ is a finitely generated $A$-algebra / when $\forall b \in B$ $\exists$ monic polynomial $f \in A[x]$ with $f(b)=0$. Notation $_AB$ means "$A$-module $B$".

We know that $A\subseteq C$ is integral iff $A\subseteq B$ and $B\subseteq C$ are integral (Grillet, Abstract Algebra, 7.3.3).

Do we also have the following:

  1. $A\subseteq C$ is finite $\Leftrightarrow$ $A\subseteq B$ and $B\subseteq C$ are finite.
  2. $A\subseteq C$ is of finite type $\Leftrightarrow$ $A\subseteq B$ and $B\subseteq C$ are of finite type.

I'm having problems with ($A\subseteq C$ finite $\Rightarrow$ $A\subseteq B$ finite) and with ($A\subseteq C$ of finite type $\Rightarrow$ $A\subseteq B$ of finite type). If $_AB$ is a direct summand of $_AC$ (for example when $A$ is a field), i.e. $_AC= _A B\oplus _A B'$ for some submodule $B'$ of $C$, then $C=Ac_1+\cdots+Ac_n$ implies $c_i=b_i+b'_i$ for some $b_i\in B$ and $b'_i \in B$, hence $B = Ab_1 + \cdots + Ab_n$. But what if $_AB$ is not a direct summand of $_AC$? And what about the 'finite type' case?

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    $\begingroup$ There is very little reason to insist in fighting the spacing rules of TeX, really... $\endgroup$ Commented Apr 1, 2012 at 18:01

2 Answers 2


Neither of the implications that you are having trouble with hold in general. As an example, consider $A=k[X_1,X_2,\ldots]$, $C=A[Y]/(Y^2)$ and $B\subseteq C$ be the (unital) algebra over $A$ generated by terms of the form $X_iY$. You should be able to show that $C$ is finitely generated both as a module and as an algebra over $A$, but $B$ is not finitely generated over $A$ in either sense.

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    $\begingroup$ For a proof assuming $A$ is Noetherian, see this (open access) paper of Artin-Tate: projecteuclid.org/… $\endgroup$
    – B R
    Commented Apr 1, 2012 at 18:24
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    $\begingroup$ Careful with Noetherian assumptions, BR! See the warning in my answer. And thanks for the link. $\endgroup$ Commented Apr 1, 2012 at 19:21
  • $\begingroup$ Serves me right for not thinking about what I'm writing! :) $\endgroup$
    – B R
    Commented Apr 1, 2012 at 19:24
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    $\begingroup$ No wonder I didn't find this in the literature, since it isn't true. Regarding the counterexample: we see that $A\leq C=A[y]/\langle y^2\rangle$ is finite, since $C=A+Ay$, but $A\leq B$ is not of finite type, because for any finite subset $\{f_1,\ldots,f_k\}\subseteq B$, we have $x_ny\notin A[f_1,\ldots,f_k]$ for sufficiently large $n$ (because $f_1,\ldots,f_k$ contain only finitely many $x_n$). Thank you George Lowther & @GeorgesElencwajg & B R for your help. $\endgroup$
    – Leo
    Commented Apr 1, 2012 at 21:21
  • $\begingroup$ Dear George, is my understanding that $B$ is is a hyperplane in $C$ with complement the line $k\cdot Y$ correct? $\endgroup$ Commented Apr 1, 2012 at 21:45

You are right to have problems because the implications you cannot prove are false!

1) Let $A$ be a ring and $M$ be an $A$-module.
Define an $A$-algebra structure $A*M$ on $A\oplus M$ by decreeing that that $m\cdot m'=0$ for all $m,m'\in M$.
Now if $A$ is non-noetherian and if $M$ is a finite $A$-module with a non finitely generated sub-module $N\subset M$, you can take $B=A*N\subset C=A*M$ as an example of a non-finite subalgebra of a finite algebra .

2) Take $A = k$, a field, $C=k[x,y]$, the polynomial ring .
Then $B=k[x,xy, xy^2,...,xy^n,...] $ is a subalgebra of the finitely generated $k$-algebra $C$ which is not finitely generated over $k$.

Noetherian alert
If $A$ is noetherian, then a subalgebra of a module-finite $A$-algebra is module-finite so that no counterexample like 1) is possible.
However, since a field $k$ is a highly noetherian ring, counterexample 2) above proves that a subalgebra $B$ of a finitely generated $A$-algebra need not be finitely generated even if the base ring $A$ is noetherian.
If however you add to noetherianity of $A$ the hypothesis that $C$ is module-finite over $B$, then indeed $B$ will be finitely generated over $A$: this is the Artin-Tate theorem that BR links to in a comment to George (Lowther!) 's answer.

  • $\begingroup$ thank you for the clarification of the Noetherian case, and for interesting examples $\endgroup$
    – Leo
    Commented Apr 1, 2012 at 21:25

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