# 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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(ADDITION) Late remark: you say in your question that the only application you know is Rotman's proof of the fundamental theorem of arithmetic via Jordan-Hölder, indeed that is Corollary 4.56 of his book "Advanced Modern Algebra". Just for completeness, I believe it is worth mentioning including here his discussion on the next two pages about the ...

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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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The equality $RI\otimes_R N=R\otimes_R IN$ is very subtly false: the point is that it does not hold in $I\otimes_RN$, which is the only place where it could hold. But, since tensor product is $R-$bilinear, can't we write (for example) $1\cdot i\otimes n=1\otimes i\cdot n \:$? No, we can't! Because $1\otimes i\cdot n$ does not make sense in ...

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Here's what always makes sense to me. Yoneda's lemma tells us that if we really want to understand a ring $R$ we should study the sets $\text{Hom}_{\text{Ring}}(R,S)$ for all the other rings $S$ we could possibly imagine. That said, studying ALL the rings $S$ seems a little naive--is there no way to lighten our load? Well, intuitively if we have some class ...

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