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Let $f:X \to Y$ be an etale morphism of Noetherian schemes. Is it true that the induced morphism on the reduced schemes, i.e., $f_{\mathrm{red}}:X_{\mathrm{red}} \to Y_{\mathrm{red}}$ is etale as well?

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1 Answer 1

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Yes.

Firstly, recall that the base-change of an etale morphism is etale, so $X' := X \times_Y Y_{\text{red}} \to Y_{\text{red}}$ is etale. Also, the base-change of a closed immersion is a closed immersion, and so $X' \to X$ is a closed immersion.

Next, note that $X_{\text{red}} \to Y$ factors through each of $X$ and $Y_{\text{red}}$, and so factors through $X'$. In other words, the closed immersion $X_{\text{red}} \to X$ factors through $X'$.

Finally, recall that if $T \to S$ is etale and $S$ is reduced, then $T$ is reduced, and so $X' \to X$ factors through $X_{\text{red}}$. Thus $X' = X_{\text{red}}$, and so $X_{\text{red}} \to Y_{\text{red}}$ is etale, as claimed.

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