2
$\begingroup$

Let $k$ be a field and $A\subset \mathbb{N}^d$ a vector configuration. Let $R,S$ be commutative $k$-algebras, both graded by the affine semigroup $\mathbb{N}A$. Is the 'multidgraded Segre product' $R \otimes_{\mathbb{N} A} S := \bigoplus_{\mathbb{N} A} R_a \otimes_k S_a$ known in the literature?

$\endgroup$
1
  • $\begingroup$ A very similar construction appears in Hochster's "Some applications of the Frobenius in characteristic 0" in the examples in Section 3. $\endgroup$ Commented Oct 22, 2011 at 9:02

2 Answers 2

1
$\begingroup$

I believe that Segre product, at least in the particular case of grading over $\mathbb Z$ or $\mathbb N$ (I don't know what a vector configuration is), appears in literature in many places. One such place I am aware of is the book by Polishchuk and Positselski "Quadratic Algebras", AMS, 2005, where it is used to construct non-Koszul algebras with mutually inverse Hilbert series (p. 59 onwards). It is used also in the operadic environment (see, for example, a draft of the book by Loday and Vallette at http://math.unice.fr/~brunov/Operades.html ) what encompasses the above-mentioned algebra case and, probably, many other cases, including yours.

$\endgroup$
1
  • $\begingroup$ Ordinary Segre products are long known. According to Wikipedia Segre's 1883 thesis was on quadrics in projective space. I assume that is how the name came about. I'm specifically looking for the multigraded analogue where the grading is by an affine semigroup. (A vector configuration $A = \{a_1,\dots,a_n\}$ is a finite subset of $\mathbb{N}^d$, defining the semigroup $\mathbb{N}A := \{\sum_i n_i a_i : n_i \in \mathbb{N}\}$) $\endgroup$ Commented Oct 22, 2011 at 8:59
0
$\begingroup$

I found a precursor of the notion in the paper "On unmixedness theorem" by Chow (American Journal of Mathematics, Vol. 86, No. 4, Oct., 1964). He considers a Segre-type product of the form $\bigoplus_i R_{id} \otimes S_{ie}$ for given bi-degree $(d,e)$. This captures the idea of giving linear relations on the generating degrees (of the sub-algebra of the tensor product).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .