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I am reading Qing Liu’s book and the whole proof on pp 30 goes like this:

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I do not understand the conclusion in the yellow-highlighted line. Let $f$ be the isomorphism highlighted in red. Why $f$ maps $\alpha_i$ to $\alpha_i$?

user150248
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1 Answers1

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These corollaries (explicitly) talk about $k$-algebras, so every morphism is $k$-linear. This implies that $f(\alpha_i)=\alpha_i+\mathfrak{m}$.

In the category of $k$-algebras $k$ is an initial object, so for any $k$-algebra $A$ there is exactly one $k$-algebra homomorphism $g\colon k\to A$. This must be a ring morphism, so $g(1)=1_A$. And it must be $k$-linear, so $g(\alpha)=\alpha \cdot 1_A$.

Pedro
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  • It may not be $k$-linear in the desired way. Perhaps $f^{-1}$ maps all the scalar elements to a proper subfield of $k$. – user150248 Mar 10 '19 at 10:15
  • Oh I think I get it. Qing has been always talking about k-algebra homomorphisms since the proof of the normalization lemma, although he did not give any definition of “finite homomorphism”. – user150248 Mar 10 '19 at 10:23