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First I will copy the definition of tensor product that I am using (I am using Grillet's Abstract Algebra):

Definition: Let $A$ be a right $R$-module and let $B$ be a left $R$-module. A tensor product of $A$ and $B$ is an abelian group $A \otimes_R B$ together with a bihomomorphism $\tau:A \times B \longrightarrow A \otimes_R B$, $(a,b) \longrightarrow a \otimes b$, the tensor map, such that, for every abelian group $C$ and bihomomorphism $\beta:A\times B \longrightarrow C$ there exists a unique homomorphism $\beta ': A \otimes_R B$ of abelian groups such that $\beta = \beta' \circ \tau$

The question I have is: If we take $C = A \times B$ and $\beta = Id$, the identity map, then $A \times_R B$ couldn't be zero because in that case, $Id = \beta = \beta' \circ \tau = 0$ but I know that indeed the tensorial product is zero in some cases.

What am I doing wrong?

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The identity map is not a bihomomorhism. In general, we don't have$$(a+a',b)=(a,b)+(a',b).$$

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  • $\begingroup$ Jesus, so sorry for not realizing that before. Thank you so much for your answer. $\endgroup$ – Cibgks Mar 2 '18 at 18:17

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