# Tag Info

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The Tor functors are the derived functors of the tensor product. The starting observation is that if $0 \to M' \to M \to M'' \to 0$ is a ses of modules and $N$ is any module (let's work over a fixed commutative ring $R$), then $M' \otimes N \to M \otimes N \to M'' \otimes N \to 0$ is exact, but you don't necessarily have exactness at the first step. (This is ...

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To my very regret, old-fashioned terminologies are all over the place in mathematics and still prevent us from using the universal benefit of (the deep idea of) category theory, for example the unification of various scattered notions in mathematics. So let me answer what $A \times B$ and $A \oplus B$ should denote (although most books have not adopted this ...

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You will be a lot more motivated to learn about Tor once you observe closely how horribly tensor product behaves. Let us look at the simplest example possible. Consider the ring $R=\mathbb C[x,y]$ and the ideal $I=(x,y)$. These are about the most well-understood objects, right? What is the tensor product $I\otimes_RI$? This is quite nasty, it has torsions: ...

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I can sympathize with this question because I am about to teach a first (graduate level) course in commutative algebra. No homological algebra of any sort is a prerequisite: I'll be happy if all of my students are comfortable with exact sequences. On the other hand, just a little bit of Tor is extremely helpful when studying commutative algebra (and ...

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Consider the simplest possible nontrivial (left) $R$-module: $R$ itself. It's certainly finitely generated, by $\{ 1 \}$. The submodules are exactly the (left) ideals of $R$. So you want a ring which has (left) ideals which are not finitely generated. For example, you could use a non-Noetherian commutative ring, such as $\mathbb{Z}[X_1, X_2, X_3, \ldots ]$.

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