The “Injectivity Implies Surjectivity” Trick

There is a standard proof technique involving the relationship between injectivity and surjectivity on “finite” structures. I rather like it—for the examples provided, it is very difficult to proceed without knowing the trick, even though the solutions are very simple.


Principle A: If S is a finite set and \phi: S\to S is a function, then \phi is injective if and only if \phi is surjective.

Proposition 1: If \mathbb{F} is a finite field and \phi:\mathbb{F}\to\mathbb{F} is a homomorphism of fields, then \phi is an isomorphism.

Proof: \ker{\phi} is an ideal of \mathbb{F}, but it cannot equal \mathbb{F}, (for \phi(1)=1) so \ker{\phi}=0 and \phi is injective. Therefore \phi is surjective.

Proposition 2: If G is a finite simple group and \phi:G\to G is a homomorphism, then \phi is an isomorphism or \phi\equiv e.

Proof: Either \ker{\phi}=G, in which case \phi\equiv e, or \ker{\phi}=\{e\}, in which case \phi is injective, so \phi is surjective.


Principle B: If V is a finite dimensional vector space and \phi:V\to V is a linear transformation, then \phi is injective if and only if \phi is surjective.

Proposition 3: If K is a field and A is an integral domain containing K with finite dimension over K, then A is a field.

Proof: It suffices to show that we can take inverses. Choose any nonzero a\in A, and consider the map \phi: A\to A defined by \phi(x) = ax. Then \phi is a linear transformation. Since A is an integral domain, \phi is injective, therefore \phi is surjective. So there exists some b\in A with \phi(b)=1, so ab=1.


Problem: Show that any finite integral domain is a field.


EDIT: As Steven points out in the comments, the following is also useful, and the proof is a good exercise:

Principle C: If M is a Notherian module then any surjective homomorphism \phi: M\to M is injective, and if M is an Artinian module, then any injective homomorphism \phi:M\to M is surjective. (hint: in fact, for any \phi:M\to M, if M is Noetherian, then for large enough n we have \ker{\phi^n} \cap \rm{image}\ \phi^n = 0. If M is Artinian, then for large enough n we have \ker{\phi^n} + \rm{image}\ \phi^n = M.)

6 Responses to The “Injectivity Implies Surjectivity” Trick

  1. Steven Sam says:

    It may be worth mentioning other instances of this:

    Let $M$ be an $A$-module, and $u \colon M \to M$ an endomorphism. Then$u$ surjective and $M$ Noetherian implies $u$ is injective. Similarly, $u$ injective and $M$ Artinian implies $u$ is surjective. ($A$ can be any commutative ring). This is Exercise 6.1 of Atiyah–MacDonald.

  2. lydianrain says:

    Yes, that’s a good idea–I’ll add an edit. I’ve seen the first one before in Dummit and Foote–the second one is new to me. (though I see that the proof is the same)

  3. David Speyer says:

    It’s also worth mentioning the amazing result of Ax:

    If F is an injective polynomial map from \mathbb{C}^n to itself, then F is surjective. The proof (1) uses model theory to replace the complex numbers by the algebraic closure of a finite field, (2) uses an elementary argument to replace the algebraic closure of a finite field by a finite extension of the finite field (3) uses the obvious observation above.

  4. theo says:

    Are there ‘tricks’ like these for morphisms of finitely generated algebras? I’m guessing fg is not finite enough, but just wanted to check if any positive results are known.

    • lydianrain says:

      I guess there are a number of theorems in algebraic geometry with this general flavor, but I don’t know of any direct analogies.

      The closest analogy I can think of is for modules over nice rings. For example, Atiyah-MacDonald has a number of similar results and exercises about modules over local rings.

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