How Do Tensor Products Relate to Supports in Algebraic Geometry?

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SUMMARY

The discussion focuses on the relationship between tensor products and supports in algebraic geometry, specifically addressing the problem of demonstrating that for any locally ringed space $(X, \mathscr{O}_X)$ and $\mathscr{O}_X$-modules $\mathscr{F}$ and $\mathscr{G}$ of finite type, the support of their tensor product satisfies the equation $\operatorname{Supp}(\mathscr{F}\otimes_{\mathscr{O}_X} \mathscr{G}) = \operatorname{Supp}(\mathscr{F}) \cap \operatorname{Supp}(\mathscr{G})$. This result is crucial for understanding the behavior of supports in the context of algebraic geometry and module theory.

PREREQUISITES
  • Understanding of locally ringed spaces
  • Knowledge of $\mathscr{O}_X$-modules
  • Familiarity with tensor products in algebra
  • Concept of support in algebraic geometry
NEXT STEPS
  • Study the properties of locally ringed spaces in depth
  • Explore the theory of $\mathscr{O}_X$-modules and their applications
  • Learn about tensor products and their implications in module theory
  • Investigate the concept of support in algebraic geometry and its significance
USEFUL FOR

Mathematicians, algebraic geometers, and students studying module theory who seek to deepen their understanding of the interplay between tensor products and supports in algebraic geometry.

Euge
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Here is this week's POTW:

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Suppose $(X, \mathscr{O}_X)$ is a locally ringed space. Show that for all $\mathscr{O}_X$-modules $\mathscr{F}$ and $\mathscr{G}$ of finite type, $\operatorname{Supp}(\mathscr{F}\otimes_{\mathscr{O}_X} \mathscr{G}) = \operatorname{Supp}(\mathscr{F})\cap \operatorname{Supp}(\mathscr{G})$.-----

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No one answered this week's problem. You can read my solution below.
Let $\mathscr{F}$ and $\mathscr{G}$ be $\mathscr{O}_X$ modules of finite type. If $x\notin\operatorname{Supp}(\mathscr{F}) \cap \operatorname{Supp}(\mathscr{G})$, then either $\mathscr{F}_x = 0$ or $\mathscr{G}_x = 0$. It follows that $\mathscr{F}_x \otimes_{\mathscr{O}_{X,x}} \mathscr{G}_x = 0$, or $(\mathscr{F}\otimes_{\mathscr{O}_X} \mathscr{G})_x = 0$. Thus $x\notin \operatorname{Supp}(\mathscr{F}\otimes_X \mathscr{G})$. Conversely, if $x\notin \operatorname{Supp}(\mathscr{F}\otimes \mathscr{G})$, then $\mathscr{F}_x \otimes_{\mathscr{O}_{X,x}} \mathscr{G}_x = 0$. As $\mathscr{F}$ and $\mathscr{G}$ are of finite type, $\mathscr{F}_x$ and $\mathscr{G}_x$ are finitely generated modules over local ring $\mathscr{O}_{X,x}$. Nakayama's lemma implies $\mathscr{F}_x = 0$ or $\mathscr{G}_x = 0$, i.e., $x\notin \operatorname{Supp}(\mathscr{F}) \cap \operatorname{Supp}(\mathscr{G})$.
 

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