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Here is the text of the exercise:

If $f:X \rightarrow Y$ is a morphism of schemes over $k$, and $P \in X$ is a point with residue field $k$, then $f(P) \in Y$ also has residue field $k$.

In the case where $X$ and $Y$ are spectra of fields, this reduces to:

If $K \subset L \subset M$ are fields and $K \cong M$, then $K \cong L$.

which I believe is false, or else we could derive, for example, $\mathbb C \cong \mathbb C (x)$.

Am I mistaken?

Al-Biruni
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2 Answers2

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The problem here is that you have forgotten the condition that $X\to Y$ be "a morphism of schemes over $k$". This condition applied in your case means that the isomorphism $K\to M$ is a morphism of $K$-algebras. Once we enforce this, the sequence of inclusions $K\subset L\subset M$ is an inclusion of $K$-vector spaces, so $K=L=M$ by dimension considerations.

Edit: This question ends up hinging on different interpretations of what exactly the statement "$P\in X$ is a point with residue field $k$" means. $k(P)$ is equipped with the structure of a $k$-algebra from the fact that $X$ is a scheme over $k$, and the intended statement here is that $k(P)$ with the $k$-algebra structure coming from $X$ being a scheme over $k$ is isomorphic to $k$ with the trivial $k$-algebra structure as a $k$-algebra, which is a more restrictive condition than just $k(P)\cong k$ as fields, as you have noticed. If one enforces this intended statement, then all opportunity for funny business is eliminated and the original answer, preserved above, is sufficient.

KReiser
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  • I'm sorry I was wasn't clear, but I did mean for the field inclusions to be field extensions. The counterexample I was thinking of involved proper subfields of $\mathbb C$ isomophic to $\mathbb C$. So, for, example, take a transcendence basis of $\mathbb C$ over $\mathbb Q$, omit one element, and then take the algebraic closure, which gives you a proper subfield of $\mathbb C$ isomoprhic to $\mathbb C$. Then adjoining the omitted transcendental you get an intermediate field isomorphic to $\mathbb C(x)$. – Al-Biruni May 11 '20 at 00:41
  • The proposed counterexample is not a map of $\Bbb C$-algebras, as would be required. I am not sure what you mean by your first sentence. – KReiser May 11 '20 at 00:44
  • Yes, you are right. I misunderstood your comment. Thank you, it is clear now. – Al-Biruni May 11 '20 at 01:03
  • Having slept on it, I no longer agree! Whenever you have a tower of field extensions $K \subset L \subset M$, the inclusion of $L$ in $M$ is a $K$-algebra map. – Al-Biruni May 11 '20 at 12:40
  • But the isomorphism of fields $K\cong M$ need not always be compatible with the $K$-algebra structure on the inclusion $K\subset M$! Your proposed example exactly shows this - $\Bbb C\cong \Bbb C(x)$ as fields, but the chosen inclusion there gives $\Bbb C(x)$ a different $\Bbb C$-algebra structure than the "standard" inclusion $\Bbb C\subset \Bbb C(x)$. – KReiser May 11 '20 at 18:54
  • $\mathbb C$ is not isomorphic to $\mathbb C(x)$, but to its algebraic closure. $\mathbb C(x)$ is the intermediate field in my example. (But I agree that this is only an isomorphism of fields, not of $\mathbb C$-algebras.) – Al-Biruni May 11 '20 at 23:54
  • You're right, this was an error - I should have written algebraic closure. The point is that the $\Bbb C$-algebra structure changes, which you agree with. – KReiser May 12 '20 at 00:03
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Here is where I believe I went wrong.

The correctness of the statement depends on what is meant by "with residue field $k$".

If "with residue field $k$" means "with a residue field isomorphic, as a field, to $k$", then the statement is incorrect. In this interpretation, the case when $X$ and $Y$ are spectra a fields reduces to:

If $K \subset L \subset M$ are fields and $K \cong M$, then $K \cong L$.

and this is false.

However, if "with residue field $k$" means "with residue field, viewed as a field over $k$, equal to $k$", then the statement is correct. In this interpretation, the case when $X$ and $Y$ are spectra of fields reduces to:

If $K \subset L \subset M$ are fields and $K = M$, then $K = L$.

which is trivially true.

But the second interpretation is the more natural in this context. The statement is about a morphism over $k$, so we should interpret the entire statement as being "over $k$".

Al-Biruni
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  • You end up at the correct interpretation here, but your process for getting there could use some clarity! "Residue field $k$" just means $\mathcal{O}{X,x}/\mathfrak{m}_x\cong k$, which is your first statement - the language "a morphism of schemes over $k$" means that all rings of sections of $\mathcal{O}_X$ and stalks $\mathcal{O}{X,x}$ are $k$-algebras in a specified way, and all relevant maps preserve this $k$-algebra structure. "Viewed as a field over $k$" exactly comes from this data - it's not a choice we make which is "more natural in this context", it's implied by our hypotheses! – KReiser May 11 '20 at 18:56
  • First, thanks for the time you are devoting to this question. That said, I am not sure I understand what you are trying to tell me. I claim that the following statement is false: "If $f:X \rightarrow Y$ is a morphism of schemes over $k$ and $P \in X$ is a point with residue field isomorphic to $k$ as a field, then $f(P)$ also has a residue field isomorphic to $k$ as a field". Do you agree that this statement is false? – Al-Biruni May 11 '20 at 23:50
  • No, this statement is true. From the map of sheaves $\mathcal{O}Y\to f*\mathcal{O}_X$ we get an injective map of residue fields $k(f(P))\to k(P)$. By the condition that $f:X\to Y$ is a $k$-morphism of $k$-schemes, this map of fields is a map of $k$-algebras. In particular, we have distinguished injective maps $k\to k(f(P))$ and $k\to k(P)=k$ so that the composite $k\to k(f(P))\to k(P)$ equals the distinguished map $k\to k(P)=k$. This gives the desired conclusion. – KReiser May 12 '20 at 00:03
  • But the assumption (which I tried to make clear) is only that that $k(P)$ is isomorphic to $k$ as a field, not that the distinguished map $k \rightarrow k(P)$ is an isomorphism. – Al-Biruni May 12 '20 at 01:17
  • Thus, my example: Let $k$ be a field. Then there is a $k$-algebra homomorphism $k(x) \rightarrow \overline{k(x)}$, and hence a morphism $f:\operatorname{Spec} \overline{k(x)} \rightarrow \operatorname{Spec} k(x)$ of schemes over $k$. If $P$ is the unique point of $\operatorname{Spec} \overline{k(x)}$, then the residue fields of $P$ and $f(P)$ are $\overline{k(x)}$ and $k(x)$, respectively. In particular, when $k = \mathbb C$, the residue field at $P$ is isomorphic to $k$ as a field (though clearly not as a $k$-algebra.) – Al-Biruni May 12 '20 at 01:18
  • The residue field of a point $x\in X$ has a $k$-algebra structure specified by the structure morphism $X\to\operatorname{Spec} k$, which is the data of the statement "$X$ is a scheme over $k$". The intended meaning "$P\in X$ is a point with residue field $k$" is that the residue field $k(P)$ is isomorphic as a $k$-algebra to $k$ with the trivial $k$-algebra structure. Hartshorne is often sloppy with things like this, and many/most people in the area understand this is what he means, to the point one can forget why anyone would think otherwise. (This has perhaps happened to me here.) – KReiser May 12 '20 at 05:54
  • Understood. Thanks for your time. I accepted the original answer, with your lucid edit. – Al-Biruni May 12 '20 at 12:18