Specialization of PBW-algebras over rational function field

Question

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 $A_0$?

Answer 1

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.

Answer 2

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.