This is a soft question. I have encountered two very different proofs of what seems like "basically the same theorem," and I want to understand how they relate and "what the real explanation is." Please forgive my imprecision here; but I am extremely interested in your thoughts.

In Shafarevich's Basic Algebraic Geometry I, Ch. 1 Sec. 5.3, it is proven that a finite map has finite fibers, i.e.:

Let $X,Y$ be quasiprojective varieties over an algebraically closed ground field $k$, and let $f:X\rightarrow Y$ be a regular map that is finite in the sense that $Y$ admits a cover by affine open subsets $V$ such that $U=f^{-1}(V)\subset X$ is affine for each $V$, and we have $f^*:k[V]\rightarrow k[U]$ is an integral ring map. Then $f^{-1}(p)$ is a finite set for $p\in V$.

The proof is that since the ring maps are integral, the coordinate functions on $X$ satisfy polynomials over the pullbacks of the coordinate functions on $Y$, which become actual polynomials over $k$ when an image point in $y$ in a specified $V$ is chosen. These polynomials limit the coordinates on $X$ to a finite set of values.

Meanwhile, in Atiyah and MaDonald's Introduction to Commutative Algebra, Ch. 5 Exercises 12-15, the "same result" is established in a different setting, namely:

If $A$ is an integrally closed domain, $K=\operatorname{Frac}A$, $L$ is a finite field extension of $K$, and $B$ is $A$'s integral closure in $L$, then the map $\operatorname{Spec}B\rightarrow \operatorname{Spec}A$ coming from the inclusion $A\hookrightarrow B$ has finite fibers.

The proof is completely different. We look at $L$ as a purely inseparable extension of a separable extension of $K$ and handle each extension separately. For the separable extension, we embed it in its Galois closure and prove that the Galois group (which is finite) acts transitively on the fibers of the map. The argument uses a form of the Going-Up Theorem and the lemma that if an ideal is contained in a finite union of prime ideals, it is fully contained in one of them. For the inseparable part, we show using a calculation that the map on $\operatorname{Spec}$s is actually injective.

Obviously the scope of the two statements is a little different. For the purposes of this question I don't care about the differences coming from the fact that Shafarevich is dealing with quasiprojective rather than affine varieties, so let's just imagine he's in the affine case. Then he's dealing exclusively with finitely generated algebras over an algebraically closed field, but I don't think he needs to make any assumption of integral closure or domainhood. And his argument is only dealing with the closed points of the $\operatorname{Spec}$. Meanwhile, Atiyah and MacDonald insist on starting with an integrally closed domain, but there's no assumption that there's some underlying algebraically closed field over which everything is a f.g. algebra.

Nonetheless, the two books seem to be trying to tell me, in some sense, "the same story." Yet in one case we get the result from the basic fact that the degree of a polynomial bounds the number of its roots over a field (although the Nullstellensatz is sitting in the background to identify the closed points of the $\operatorname{Spec}$ with tuples from the field), while in the other case we have to bring a Galois group and a case handling of separable vs. purely inseparable extensions into it!

What's "the real story" here? I.e.:

How general is the result? Is it true for an arbitrary map of schemes? And what's the general proof? How does it relate to these two proofs? Is there a "moral" argument that subsumes both of these arguments? How do you sort this out for yourself?

Again, please forgive the softness. I hope the question will be of interest in any case.

  • 3
    $\begingroup$ Finite maps of schemes have finite fibers (the ring of the fiber will be finite dimensional over a field, so it's Artinian and this implies a lot of interesting stuff). The difficulty in the A-M setting is that taking the integral closure of a domain (even a Noetherian one; I think the keyword here is Nagata ring) in a finite extension of its field of fractions doesn't always result in a finite extension of rings. In many cases it does, though — notably when the ring is f.g. over a field or if it's a Dedekind domain. $\endgroup$
    – Hoot
    Commented Jan 17, 2015 at 18:38
  • 2
    $\begingroup$ You would expect that the real reason for something would be a rational reason, even a natural reason. But more often than not, those end up being quite complex, on the border of imaginary! :-) $\endgroup$
    – Asaf Karagila
    Commented Jan 17, 2015 at 22:44

1 Answer 1


It is general. Let $f : X \rightarrow Y$ be a finite morphism of schemes. Then the fibers of $f$ are finite. This is implied by fact that any finite algebra over a field is an artinian ring.

Proof : let $f : X \rightarrow Y$ be a finite morphism of schemes. By definition there is an open affine covering $(V_i)_{i\in I}$ of $Y$ such that the $f^{-1} (V_i)$ are affine and the corestrictions $f : f^{-1} (V_i) \rightarrow V_i$ are finite, and the inverse is true (that is, if there such a covering then the morphism is finite.) We can therefore assume that $X$ and $Y$ are affine. Then any fiber of $f$ is the spectrum of a finite algebra over a field (the residue field of the point of $Y$ we're taking the fibre of). Such an algebra is artinian and has only finitely many primes ideals, showing that the spectrum is finite, that is, that the fiber is finite, and also that it is discrete.

Remark : Being a finitely generated module is stable by scalar extension.

Proof : This is standard linear algebra. Let $A$ be a commutative ring with unit, $M$ a finitely generated $A$-module (by let's say elements $m_1, \ldots, m_n$) and $\varphi : A \to B$ a morphism a commutative rings with unit. Then through $\varphi$ the $B$-module $M\otimes_A B$ is finitely generated. Indeed, elements of $M\otimes_A B$ are finite sums of elements of the form $m\otimes b$ for $m\in M$ and $b\in B$, now $m$ could be written $\sum_{i=1}^n a_i m_i$ for $a_i$'s in $A$. So that $$m \otimes b = \left( \sum_{i=1}^n a_i m_i \right) \otimes b = \sum_{i=1}^n (a_i m_i) \otimes b = \sum_{i=1}^n m_i \otimes (a_i b) = \sum_{i=1}^n m_i \otimes (\varphi(a_i) b) = \sum_{i=1}^n (\varphi(a_i) b) (m_i \otimes 1)$$ by definition of the $B$-module structure of $M\otimes_A B$, which show that any $m\otimes b$ is a $B$-linear combination of the $m_i \otimes 1$'s, which in turn show that any element in $M\otimes_A B$ is a $B$-linear combination of the $m_i \otimes 1$'s, hence that $M\otimes_A B$ is a finitely generated $B$ -module.

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    $\begingroup$ Of course, here we are using the fact that the class of finite morphisms is closed under base change. $\endgroup$
    – Zhen Lin
    Commented Jan 17, 2015 at 18:45
  • $\begingroup$ Note that $f$ is then also projective. For all of this, see EGA II 6.1.11. "Conversely", if you are proper, locally of finite presentation and with finite fibres, you're finite EGA IV 3, 8.11.1. Finally, when $Y$ is noetherian, being finite is equivalent to be proper with finite fibres. $\endgroup$
    – Olórin
    Commented Jan 17, 2015 at 18:50
  • $\begingroup$ @ZhenLin ;-) ;-) $\endgroup$
    – Olórin
    Commented Jan 17, 2015 at 18:51
  • $\begingroup$ If $X = \text{Spec }A$ and $Y= \text{Spec }B$ and you know that $A$ is a finite $B-$module, how do you know that $X_y = \text{Spec }(A \otimes_B k(y))$ is a finite $k(y)$-algebra? $\endgroup$ Commented May 7, 2023 at 15:48
  • $\begingroup$ Meaning, how do you know $A \otimes_B k(y)$ is a finite $k(y)$-algebra? $\endgroup$ Commented May 7, 2023 at 19:51

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