15
$\begingroup$

Let $A$ be a local ring over $\mathbb{C}$, which moreover is a finite dimensional $\mathbb{C}$-vector space.

When is $A$ a subring of $\mathbb{C}[t]/t^n$?

What does the minimal such $n$ have to do with $A$?

Example: the ring $\mathbb{C}[x,y]/(x^3,y^2,xy)$ is a subring of $\mathbb{C}[t]/t^5$ by $x = t^2$ and $y = t^3$.

$\endgroup$

1 Answer 1

6
$\begingroup$

OK, this is not an elegant solution, but at least something...

So, let $A$ be a local $k$-algebra where $k$ is a field. I don't think that $k=\mathbb C$ makes a difference (and $k$ is easier to type than $\mathbb C$).

Anyway, the finite dimensionality condition implies that $A$ is Noetherian (in fact Artinian as the title says), so in fact it is a finitely generated $k$-algebra, in other words a surjective image of $k[x_1,\dots,x_p]$ for some $p\in \mathbb N$.

Let $I=\ker \left( k[x_1,\dots,x_p]\to A \right)$, i.e., $A\simeq k[x_1,\dots,x_p]/I$.

Now let $S\subset \mathbb N^p$ be the set of $p$-tuples $(\alpha_1,\dots,\alpha_p)$ for which $\prod x_i^{\alpha_i}\in I$, but $\prod x_i^{\alpha_i-\delta_{ij}}\not\in I$ for any $j\in\{1,\dots,p\}$ (where $\delta_{ii}=1$ and $\delta_{ij}=0$ for $i\neq j$).

Let us assume that there exists a homomorphism $\phi: k[x_1,\dots,x_p]\to k[t]/(t^n)$ such that $I=\ker \phi$. (Clearly this is equivalent to the existence of a homomorphism $A\hookrightarrow k[t]/(t^n)$.)

Let $r_i\in\mathbb N$ be such that $\phi(x_i)=u_it^{r_i}$ for some unit $u_i\in k[t]/(t^n)$. Then it is necessary that the following inequalities are satisfied:

For any $(\alpha_1,\dots,\alpha_p)\in S$ and $j\in\{1,\dots,p\}$ , we must have

$$ n\leq \sum_i \alpha_i r_i < n + r_j $$

This certainly gives a lot of restrictions. For example this shows that the ring $$ k[x,y]/(x^my, xy^m, x^{2m}, y^{2m})$$ where $m\geq 2$ cannot be embedded into a ring of the type $k[t]/(t^n)$. (The problem is that the first two terms imply that $mr_x+r_y\geq n$ and $r_x+mr_y\geq n$ has to hold, but this implies that then $(m+1)(r_x+r_y)\geq 2n$ which implies that then either $(m+1)r_x\geq n$ or $(m+1)r_y\geq n$ but then this contradicts $(2m-1)r_x< n$ (or the same with $r_y$) as soon as $m+1\leq 2m -1$, that is as soon as $m\geq 2$.

It seems to me that it is also a sufficient condition if the above system of inequalities has an integer solution $(r_1,\dots,r_p)$. I am not saying that it is necessarily easy to check, but at least with few variables and few relations it may not be so bad. If you want a more conceptual condition, I suppose one could associate a toric variety to the set of $S$ and there might be a nice condition of that variety that gives a condition. It will probably not be easier to check, but it might be interesting. Then again, you might not want to spend too much time on this. I think I already did. :)

EDIT: Corrected argument following Hailong's comment.

Remark: This argument is secretly about valuations...

$\endgroup$
4
  • $\begingroup$ Hi Sandor, a small quibble: since this is an algebra map, I am not sure one can assume all $u_i=1$. $\endgroup$ Nov 25, 2010 at 22:35
  • 1
    $\begingroup$ @Long, I edited the answer to correct the error you point out. On the other hand: what happened to your answer? Was there something wrong with it? $\endgroup$ Nov 30, 2010 at 2:08
  • $\begingroup$ @Sandor: I made a silly mistake with the second edit. I don't have time to fix it now, so I will just delete it and come back later. $\endgroup$ Nov 30, 2010 at 2:25
  • 4
    $\begingroup$ Woo hoo! Looks like I just won the "disciplined" badge for that! $\endgroup$ Nov 30, 2010 at 2:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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