# Specialization of PBW-algebras over rational function field

I have been thinking about the concept of specialization of algebras defined over the field of rational functions $k = \mathbb{C}(t)$ (I'm using $\mathbb{C}$ for my work, but the question can be asked over any field). In my examples the algebras are quadratic, but I don't think that this is essential. The main question I want to ask is this: suppose I have a PBW-algebra over $k = \mathbb{C}(t)$ that is specializable in an appropriate sense (described below). When can I conclude that the specialized algebra has the same size as the original?

## Setup

The setup uses the language of reduction systems as in George Bergman's (excellent) paper The Diamond Lemma for Ring Theory. Start with a finite set $X = \{ x_1, \dots, x_n \}$ of generators, let $\langle X \rangle$ denote the free semigroup with unit on $X$ (i.e. the words on the alphabet $X$), and let $k \langle X \rangle$ be the free algebra on $X$.

A reduction system is a set $S$ of pairs $\sigma = (W_\sigma, f_\sigma)$, where $W_\sigma \in \langle X \rangle$ and $f_\sigma \in k \langle X \rangle$. Then the associated $k$-algebra is $$A = k \langle X \rangle / (W_\sigma - f_\sigma \mid \sigma \in S).$$ Now we assume that we have a semigroup partial order $\leq$ on $\langle X \rangle$ (with descending chain condition), compatible with multiplication in the sense that $A \leq A'$ implies that $BAC \leq BA'C$ for all $B,C \in \langle X \rangle$. Furthermore we assume that for each $\sigma = (W_\sigma, f_\sigma)\in S$, the element $f_\sigma \in k \langle X \rangle$ is a linear combination of monomials that are less than $W_\sigma$.

The idea is that you replace (reduce) certain monomials (the $W_\sigma$) with lower-order terms (the $f_\sigma$). A monomial $w \in \langle X \rangle$ is defined to be irreducible with respect to $S$ if $w$ does not contain any $W_\sigma$ as a subword. The classes of the irreducible monomials always span the algebra $A$; the Diamond Lemma gives a criterion (all ambiguities must be resolvable) in order for the classes of irreducible monomials to be independent. In that case we say that $A$ is a PBW-algebra and that $X$ is a set of PBW-generators.

## Specialization

Say that the algebra $A$ is specializable (at $0$) if each $f_\sigma$ is of the form $$f_\sigma = \sum_{w \in \langle X \rangle} c_\sigma^w w,$$ where each coefficient $c_\sigma^w$ lies in the local subring $k_0 \subseteq k$ consisting of rational functions that do not have a pole at $0$.

Then we define a reduction system $S_0$ for the specialized algebra $A_0$ by simply evaluating all of the relations for $A$ at $t = 0$. Formally, for each $\sigma = (W_\sigma, f_\sigma) \in S$ we define $$\sigma_0 = (W_\sigma, f_\sigma(0)) \in S_0,$$ where $$f_\sigma(0) = \sum_{w \in \langle X \rangle} c_\sigma^w(0) w.$$ This is well-defined because we assumed that none of the rational functions $c_\sigma^w$ had a pole at zero. Finally, define the specialization of $A$ at $0$ to be the $\mathbb{C}$-algebra $$A_0 = \mathbb{C} \langle X \rangle / (W_\sigma - f_\sigma(0)).$$ Note that the specialized reduction system $S_0$ has the same set of irreducible words as $S$ did.

## Question

If the irreducible words are independent in $A$, are they independent in $A_0$? Another way of putting it is this: if $X$ was a set of PBW-generators for the original $\mathbb{C}(t)$-algebra $A$, is it also a set of PBW-generators for the $\mathbb{C}$-algebra $A_0$?

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I fail to figure out where exactly you see potential problems.

Diamond lemma indeed says that each ambiguity is resolvable, that is for each "common multiple" of some $W_\sigma$ and $W_\tau$, WLOG $aW_\sigma=W_\tau b$, the element $af_\sigma-f_\tau b$ can be reduced to zero using your rewriting rules. Reducing to zero means that you find in that element a term $c\cdot m$, where $c$ is in the ground ring, and $m$ is a monomial divisible by some $W_\lambda$, $m=m_1W_\lambda m_2$ and replace that term by $c m_1f_\lambda m_2$. Clearly, this does not create poles at 0 if there were no poles in the first place. Hence, you can specialise the whole reduction procedure at 0, and Diamond lemma applies.

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Vladimir, thanks for your answer. The potential problem I saw was in your "clearly". Saying that $a f_\sigma - f_\tau b$ can be reduced to zero means it is in the span of terms of the form $d(W_\mu - f_\mu)e$, where $d W_\mu e \leq a W_\sigma$. I had thought that perhaps coefficients could arise in that sum that had poles at $0$. It feels like this shouldn't be possible but I don't have a rigorous argument why not. –  MTS Jul 4 '13 at 12:47
(The elements $d,e$ in my previous comment should be monomials, of course.) –  MTS Jul 4 '13 at 14:36
Matt, I am even more puzzled. The coefficients that arise are explicitly extracted from the reduction procedure I described (iteratively replacing $cm_1W_\lambda m_2$ by $cm_1f_\lambda m_2$). If $c$ has no pole at zero (which is true in the beginning before we started reducing), then, since coefficients of $f_\lambda$ do not have poles at zero, arising coefficients won't have poles at zero either. Really, part of the beauty here that besides being a theoretical result, Diamond lemma is most practical: it amounts to the fact that one can do long division. –  Vladimir Dotsenko Jul 4 '13 at 16:54
I see, thanks very much for the explanation. That makes total sense. –  MTS Jul 4 '13 at 17:24

Suppose all your rewriting rules have right members in the free algebra over $R=\mathbb C[t]_{(0)}$ generated by $X$. In the process of showing that each ambiguity is resolvable, no coefficients will ever arise which are not in that ring. It follows —since Bergman's lemma works over a commutative ring, which need not be a field— that the set of irreducible words are a basis of the $R$-algebra $A=R\langle X\rangle/(W_\sigma-f_\sigma, \sigma\in S)$ as an $R$-module.

Let now $T$ be the quotient of $R$ at its maximal ideal; this is of course just $\mathbb C$. It follows from the above that the images of the irreducible words in $A\otimes_RT$ are the elements of a basis of this $T$-algebra as a $T$-module.

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Thanks for your answer, Mariano. What do you mean by "right members"? And how can we be sure that no coefficients arise outside of $R$? That was precisely the question that I was asking myself when thinking about this. –  MTS Jul 4 '13 at 1:14
Your rewriting rules are of the form $W_\sigma\leadsto f_\sigma$, and the right hand side of such a thing is the $f_\sigma$. When you resolve ambiguities, you never divide by scalars (this is implicit in the act that, for example, Bergmann works over a ring) so you simply cannot get a polt at zero at any point of the process. –  Mariano Suárez-Alvarez Jul 4 '13 at 19:10