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Let $k$ be a field and let $\phi: S\to R=S/I$ be $k$-algebras. Then given a homomorphism $\varphi: I\to R$, it's said that it induces a homomorphism $\psi:I/I^2\to R$ since $\varphi$ kills $I^2$.

I am not sure about this: if $\varphi(x)$ is nontrivial in $R$, then $\varphi(x^2)=\varphi(x)^2$ can be nontrivial in $R=S/I$, right? Hope someone can help. Thanks!

Yuyi Zhang
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1 Answers1

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$\varphi(x^2) = x\varphi(x)$. One possible point of confusion: why does $\varphi(x) = x\varphi(1)$ not hold?

RghtHndSd
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  • So, it's not a ring homomorphsim, but more like a morphism of modules? – Yuyi Zhang Oct 08 '20 at 07:04
  • If you require rings to have a $1$, then $I$ isn't even a ring. Even if you don't require this, I believe it is meant to be a module homomorphism. – RghtHndSd Oct 09 '20 at 01:18