A Non-Commutative Nullstellensatz

2010-05-18T21:46:45Z

In studying presentations of pro-$p$-groups via generators and relations, one is led (via the so-called Magnus embedding) to questions involving power series in non-commuting variables. Results from local algebraic geometry occasionally shed some insight on how to make progress, but more often that not, I find myself lacking appropriate analogs of major theorems from the commutative case. I haven't had much luck in books on non-commutative ring theory or non-commutative algebraic geometry -- the focus seems to be on completely different ideas (though I'll happily stand corrected). In any case, here's an important and seemingly basic question that I don't know how to answer.

Let $\mathbb{F}_p\langle\langle x,y\rangle\rangle$ be the ring of formal power series over $\mathbb{F}_p$ in two non-commuting variables $x$ and $y$. This ring has a unique two-sided maximal ideal $I=(x,y)$. Suppose $f,g\in I$. Can anything be said about the smallest $n$, if one exists, such that $I^n\subset (f,g)$? Namely, when does this quantity exist? Is this quantity computable? Boundable?

It's trivial to come up with examples for which there is no $n$, e.g., $(xy,yx)$, since no $x^n$ is contained in this ideal. I'm not sure how exactly to quantify this observation. Is there some kind of non-commutative resultant at play here?

Edit: I think it might be helpful for me to update with some examples as we go along. Here's one that I thing captures at least some of the interesting parts of this question.

Take $p=3$, $f=x+y$, and $g=x^3$. Then the inclusion $I^3\subset (f,g)$ can be seen by taking each of the 8 monomials in $I^3$ verifying that they are in $(f,g)$, e.g., $yxy=yfy-f^3+g\in (f,g)$. The same argument applies with the same $f$ and taking $g=x+y+x^3$. This seems to me evidence that this question can't be answered only by looking at the leading monomials (though admittedly it might be easy enough to exclude these trivial counter-examples).

Answer by Victor Protsak for A Non-Commutative Nullstellensatz

2010-05-18T23:09:51Z

Suppose that $f$ and $g$ are monomials and that $(f,g)$ contains a power of $I$. Then every word of sufficient length must contain $f$ or $g$ as a subword (and conversely). Thus your argument with $x^n$ shows that either $f$ or $g$ is a power of $x$, and likewise, one of them is a power of $y$. If $f=x^k, g=y^m, k,m\geq 2$ then the word $(x^{k-1}y)^N$ can be arbitrarily long and doesn't contain $f$ or $g$ — contradiction. Thus up to relabelling, $f=x^k, g=y$ and $n=k$ is minimal with the property that $I^n\subset (f,g)$.

I recommend Algebraic Combinatorics on Words by M. Lothaire (google it) for related ideas.

A Non-Commutative Nullstellensatz - MathOverflow

A Non-Commutative Nullstellensatz

Answer by Victor Protsak for A Non-Commutative Nullstellensatz