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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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2 Answers 2

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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

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