Consider a commutative algebra of $w \times w$ matrices over a field $\mathbb{F}$, $A \subset M_{w}(\mathbb{F})$. Let $S = \{M_1,\ldots,M_t\}$ be a minimal generating set for $A$. That is (a) the smallest ring that contains $S$ is exactly $A$, and (b) no subset of $S$ generates all of $A$.

I want to view $A = \mathbb{F}[M_1,\ldots,M_t]$ as a quotient ring $R = \mathbb{F}[x_1,\ldots,x_t]/I$, by associating $M_i$ with $x_i$ and taking $I$ as the ideal of dependencies between $M_1,\ldots,M_t$.

What can be said about the "number of variables" in the quotient ring ($t=|S|$)? In particular, I want to know the following.

  1. Is there a tighter (than $t \leq w$) upper bound for $t$ that possibly depends on more properties of $A$? Could there be another $R' = \mathbb{F}[y_1,\ldots,y_{t'}]/J$ analogous to $A$?

    • For example, the rings $R_1 = \mathbb{F}[x_1,x_2]/\langle x_1^2 - x_2, x_2^2 \rangle$ and $R_2 = \mathbb{F}[y]/\langle y^4 \rangle$ essentially yield the same matrix algebra. Is there a way to look at a ring and determine that it is "sub-optimal" like the above $R_1$?
  2. When a set $T$ generates $A$, but is not minimal, what property does it violate? For example, a spanning set for a vector space is minimal (a basis) if and only if it is a linearly independent set.

    • I think I am looking for some sort of "algebraic dependence modulo the ideal I". Sorry, I wish I could be more precise here.

For both parts, if the answer depends on the field $\mathbb{F}$, then in what ways does it do so?

P.S.: The edits have been made after looking into a reference suggested by @Wuestenfux in a comment on an older post.

  • 1
    $\begingroup$ I'd look into Groebner bases for quotient rings. You will find this in the book ''Using Algebraic Geometry'' by Cox et al. $\endgroup$
    – Wuestenfux
    Sep 20, 2019 at 12:17
  • $\begingroup$ @Wuestenfux: Thanks a lot for the reference. I have solved the question. Although, I don't know if I should remove the question or post an answer here. $\endgroup$
    – anamay
    Sep 23, 2019 at 5:27


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