In Hartshorne's "Algebraic Geometry" III, Definition p. 268 smoothness is defined by conditions for a morphism $f:X \to Y$ of schemes of finite type over a field $k$ from which the first two are relevant for this question:

1) $f$ is flat.

2) if $X' \subseteq X$ and $Y' \subseteq Y$ are irreducible components with $f(X') \subseteq Y'$, then $\dim X' = \dim Y' + n$.

With 1) and 2) Hartshorne defines so to say "a flat morphism of relative dimension $n$".

In the following Proposition III,10.1. b) and c) it has to be proved, that this notion behaves well under base extension and composition. But, I must say, I find the arguments he gives in the proof overly terse if not to say not convincing. In trying to rewrite the proofs I came up with a little Lemma, which seems to be able to clear up things a lot:


Let $f:X \to Y$ a flat morphism of noetherian schemes and $Y' \subseteq Y$ an irreducible component of $Y$. Furthermore let $X_1 \cup \cdots \cup X_r$ be the irreducible components of $X$. Then $f^{-1}(Y') = X \times_Y Y'$ consists of $X_{i_1} \cup \cdots \cup X_{i_s}$ with the $X_{i_\nu}$ all irreducible components of $X$ such $f(X_{i_\nu}) \subseteq Y'$.

Proof: Let $f(x) \in Y'$ so that $\eta \rightsquigarrow f(x)$ specializes to $f(x)$ for $\eta$ the generic point of $Y'$. As $f$ is flat, it has the going-down-property and therefore, there is an $x' \rightsquigarrow x$ with $f(x') = \eta$. Now $\xi \rightsquigarrow x'$ for the generic point $\xi$ of a certain irreducible component $X' \subseteq X$. It must fulfill $f(\xi) = \eta$ therefore $f(X') \subseteq Y'$. As $x \in X'$ the lemma is proven.

The point of the Lemma is, that $f^{-1}(Y')$ can not acquire irreducible components that are strictly parts of an irreducible component $X' \subseteq X$.

Now my questions: A) Is the above proof correct?

B) Is the Lemma trivial to see for reasons simpler than given in my proof?

C) Do you too think it helps understanding the argument in Hartshorne's book?

For C) take for example his proof of c) in the abovementioned proposition: Let $f:X \to Y$ and $g:Y \to Z$ be smooth of relative dimension $n$ and $m$. Then he simply says

"If $X' \subseteq X$, $Y' \subseteq Y$ and $Z' \subseteq Z$ are irreducible components such that $f(X') \subseteq Y'$ and $g(Y') \subseteq Z'$ then clearly $\dim X' = \dim Z' + m + n$ by hypothesis."

There is no argument, why such a setup is possible. I would argue as follows: $g(f(X'))$ is irreducible and lies in the union of irreducible components $Z'_i$ of $Z$ therefore in one of them $Z'$. So $f(X')$ lies in $g^{-1}(Z')$. Now with the lemma above I conclude, that $g^{-1}(Z')$ consists of a union $Y_1 \cup \cdots \cup Y_s$ of irreducible components of $Y$ with $g(Y_i) \subseteq Z'$. As $f(X')$ is irreducible and lies in $Y_1 \cup \cdots \cup Y_s$ it must lie in one $Y_i$. So we have found our $Y'$ and then can argue, as Hartshorne did.


1 Answer 1


Here we discuss only about the proof (c) of Proposition 10.1 of Chapter III of Hartshorne.

In that proof we fix two irreducible subsets $X'$ and $Z'$ of $X$ and $Z$ respectively such that $g(f(X'))\subset Z'$. Now along with the irreducible $X'$ and $Z'$, the proof also assumes the existence of an irreducible subset $Y'$ of $Y$ such that $f(X')\subset Y'$ and $g(Y')\subset Z'$. But we can always have such $Y'$ since $f(X')$ is irreducible and we can set $Y'=f(X')$. Now we can apply the results of $f$ and $g$ to conclude.


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