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In Atiyah Macdonald, an A-algebra is defined as a ring B together with a ring homomorphism $f:A\to B$ that induces an A-module structure. Given two A-algebras $f:A\to B$ and $g:A\to C$, an A-algebra homomorphism is defined as a map $h:B\to C$ that is a both a ring homomorphism and an A-linear map.

The book then goes on to say that the map $h$ is an A-algebra homomorphism iff $g=h\circ f$, and this is where I'm running into trouble.

Suppose $g=h\circ f$. I would need to show in particular that $h(b_1 + b_2) = h(b_1) + h(b_2), \forall b_1,b_2 \in B$. If $b_1,b_2 \in Im(f)$, then by letting $b_1 = f(a_1)$ and $b_2 = f(a_2)$ for $a_1, a_2 \in A$ I can proceed as $h(b_1 + b_2) = h(f(a_1) + f(a_2)) = h(f(a_1 + a_2)) = g(a_1 + a_2) = g(a_1) + g(a_2) = h(f(a_1)) + h(f(a_2)) = h(b_1) + h(b_2)$. This would be sufficient if $f$ was surjective, but I don't understand how to proceed in general. Any help would be much appreciated - thanks in advance!

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  • $\begingroup$ I think it's meant that $h$ is always a ring homomorphism. $\endgroup$
    – user920957
    Commented Jan 26 at 9:19

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I think you are misreading their statement (which I agree could be phrased more precisely). The statement should be that if $(B,f)$ and $(C,g)$ are $A$-algebras, and if $h:B\to C$ is a ring homomorphism, then $h$ is an $A$-algebra homomorphism if and only if $g=hf$.

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