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Let $B, C$ be two $A$-algebras, $f:A \to B, g: A\to C$ the corresponding ring homomorphisms. From this we can construct an $A$-algebra $B \otimes _A C$ and the mapping $ a \mapsto f(a) \otimes g(a)$ is the corresponding ring homomorphism $A \to B \otimes _A C$.

The above is a summary of a definition on pages 30,31 of Atiyah's Commutative Algebra. I wonder why the mapping $ a \mapsto f(a) \otimes g(a)$ is a ring homomorphism.
$$ a+b \mapsto f(a+b) \otimes g(a+b) $$ $$= f(a) \otimes g(a) + f(a) \otimes g(b) + f(b) \otimes g(a)+ f(b) \otimes g(b)$$ To be a ring homomorphism $f(a) \otimes g(b) + f(b) \otimes g(a) $ must be zero.
How can I know this?

darij grinberg
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Jeong
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1 Answers1

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This is an error in Atiyah-Macdonald. As Georges Elencwajg says in this MathOverflow thread,

This is false since that map is not a ring morphism. The correct structural map $A\to D$ is actually $a\mapsto 1_B\otimes g(a) =f(a)\otimes 1_C$.

Community
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Zev Chonoles
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