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 is a finite set and is a function, then is injective if and only if is surjective.

Proposition 1: If is a finite field and is a homomorphism of fields, then is an isomorphism.

Proof: is an ideal of , but it cannot equal , (for ) so and is injective. Therefore is surjective.

Proposition 2: If is a finite simple group and is a homomorphism, then is an isomorphism or .

Proof: Either , in which case , or , in which case is injective, so is surjective.

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

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

Proof: It suffices to show that we can take inverses. Choose any nonzero , and consider the map defined by . Then is a linear transformation. Since is an integral domain, is injective, therefore is surjective. So there exists some with , so .

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 is a Notherian module then any surjective homomorphism is injective, and if is an Artinian module, then any injective homomorphism is surjective. (hint: in fact, for any , if is Noetherian, then for large enough we have . If is Artinian, then for large enough we have .)

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6 Responses to The “Injectivity Implies Surjectivity” Trick

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.

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)

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

If is an injective polynomial map from to itself, then 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.

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.

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.

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.

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)

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

If is an injective polynomial map from to itself, then 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.

Thanks–I learned this result in model theory but forgot all about it for this post!

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.

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.