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I am getting the confusion with the definition of algebra. When we say $A$ is a finitely generated $R$- algebra then is that mean $A$ has a ring structure and finitely generated as an $R$-module. Thanks

Rajesh
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  • Perhaps language problems again...I shall check that and erase my comment. Thanks. – DonAntonio Jan 16 '13 at 18:43
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    I thin I must have been thinking of finite-dimensional algebra...oh, well. – DonAntonio Jan 16 '13 at 18:44
  • No. It means there exists finitely many $r_1,\dots,r_n \in R$ such that every $r \in R$ can be written as a polynomial in the $r_1,\dots,r_n$ with coefficients in $A$, i.e., there exists $f(x_1,\dots,x_n) \in A[x_1,\dots,x_n]$ with $r=f(r_1,\dots,r_n)$ where the action of $A$ on $R$ comes from the structure ring hom $A\to R$. – user5826 Sep 22 '21 at 18:14

1 Answers1

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No, being finitely generated as an algebra is generally not as strong as being finitely generated as a module.

Being finitely generated as an algebra means that there is some finite set of elements from the algebra, such that the subalgebra generated by those elements is the entire algebra.

This means that apart from $R$-linear combinations of the elements, we can also take all products of the elements, which may well give us a lot more elements.

As an easy example of an algebra (let's say over a field $k$) that is finitely generated as an algebra but not finitely generated as a module over $k$, we can take the polynomial ring $k[x]$ in one indeterminate. As a $k$-module (ie, a vector space), this is infinite dimensional, so not finitely generated. But as a $k$-algebra, it is generated by the element $x$.

Tobias Kildetoft
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    If I remember right, those who use "finitely generated algebra" this way like to use "module-finite algebra" for the OP's meaning. – rschwieb Jan 16 '13 at 18:37
  • @Tobias: $A$ is an algebra over a field $F$. For finitely many elements $a_1,a_2,\cdots,a_n$ in the algebra, the sub-algebra generated by $a_1,a_2,\cdots,a_n$ we mean, we take $F$-linear combinations of them as well as their products, am I right? – Groups Oct 26 '15 at 09:41
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    @Groups Yes, all possible products and $F$-linear combinations (more precisely, we take the smallest subalgebra containing those elements, which will have a slightly different description if we leave out some of the "usual" requirements on the algebra). – Tobias Kildetoft Oct 26 '15 at 09:43
  • OK. Thanks Tobias for quick reply and clarifications. – Groups Oct 26 '15 at 09:43
  • @TobiasKildetoft In the example, isn't $K[x]$ generated by $(1,x)$ instead of just $x$, as a $K$-algebra? – Takamoto Yuji Mar 11 '23 at 01:54
  • It is certainly also generated by the larger set. Whether you actually need to include $1$ depends on whether you restrict yourself to unital algebras, or if you allow non-unital ones (for non-unital algebras, you do indeed need to include the unit explicitly to get the whole thing in this case). – Tobias Kildetoft Mar 11 '23 at 20:38