Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S$ be some base ring (a commutative ring or even just a field), and $R$ a commutative ring containing $S$ which is finitely generated (as an algebra) over $S$. What conditions guarantee that any two minimal systems of generators of $R$ over $S$ have the same size?

I'm especially interested in a geometric picture to explain the situation, and whether it links to other geometrical ideas such as height or the Cohen-Macaulay property.

I'd also like to know what happens for graded rings - for instance, when are the degrees of two minimal systems of homogeneous generators the same (up to permutation)?

What's the geometry behind this?

share|cite|improve this question

I think this almost never happens without grading. Consider the simple example with $S=\mathbb Q$ and $R=\mathbb Q[X]$. Then $\{ X\}$ and $\{ X^2+X, X^2\}$ are minimal systems of generators of different sizes. Similar constructions should work for any algebra over a field with a transcendental element.

Of course if $R/S$ is a finite extension of prime order, then any minimal system of generators is a singleton. But this fails as soon as we remove the condition of prime order: let $S=\mathbb Q$ and $R=\mathbb Q[\sqrt{2}, \sqrt{3}]$. Then $\{\sqrt{2}, \sqrt{3}\}$ and $\{ \sqrt{2}+\sqrt{3}\}$ are minimal systems of generators of different sizes.


Graded case.

For a homogeneous algebra $R$ over a field $S$, a set of homogeneous elements of $R$ generates $R$ if and only if it contains a set of generators of $R_1$ has vector space. So the minimal systems are exactly the basis of $R_1$ as $S$-vector space.

More generally, if $R$ is a positive graded algebra over a field $S$, we can describe the minimal systems of homogeneous generators as follows. For any $d\ge 1$, denote by $R'_d$ the subvector space of $R_d$ generated by products of homogeneous elements of lower degrees ($R'_1=0$). Then $F\subset R$ is a minimal system of homogeneous generators if and only if for all $d\ge 1$, $F\cap R_d$ is a lifting of a basis of $R_d/R'_d$. In particular, for all $d\ge 1$, any two such systems share the same number of elements of degree $d$.

share|cite|improve this answer
    
Isn't the example $S=\mathbb{Q}$ and $R = \mathbb{Q}[X]$ an example of a homogeneous algebra over a field? I believe if one restricts generators to being inside of $S_1$ your last comment is true, but as your first example shows, this need not be the case. – RghtHndSd Aug 7 '13 at 15:37
    
@rghthndsd: in the first example, we don't ask the generators to be homogeneous. – user18119 Aug 7 '13 at 15:38
    
In the sentence "of course if $R/S$ is a finite extension of prime order..." do you mean a finite field extension of prime order? B/c take $S=k$, $R=S[x,y]/(x,y)^2$ for an order 3 extension (i.e. $R$ is a free $S$-module of rank $3$) such that there is no singleton generating (set since the algebra generated by any single nonconstant element is just $2$-dimensional over $k$). – Ben Blum-Smith Dec 11 '15 at 20:51
    
@user18119 - In my last comment I was assuming that by "finite extension of prime order" you meant a finite extension that is free and of prime rank as a module over the ground ring, but perhaps I misunderstood? – Ben Blum-Smith Dec 11 '15 at 23:07

For a connected, non-negatively graded algebra $A$ over a field, we can compute $Tor_1^A(k,k)$, which is a graded vector space: this is is isomorphic to every minimal space of homogeneous generators of $A$.

share|cite|improve this answer
    
What do you mean by "minimal space of homogeneous generator"? – user26857 Mar 2 '14 at 23:15
    
What mean is, every graded vector subspace of $A$ which generates $A$ and which is minimal is isomorphic to $Tor_1^A(k,k)$ as a graded vector space. – Mariano Suárez-Alvarez Mar 3 '14 at 8:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.