Today I’m going to describe a simple, yet powerful, trick to deduce specific problems from a “universal” case. The construction is common in algebra; given an object and a property that we would like to deduce about this object, we construct a certain “universal” object such that

- has property ;
- there exists a surjective homomorphism that “preserves” .

In order to justify this abstraction, it should be easier to show that satisfies than it is to show that satisfies . However, should not be a “trivial” object that vacuously satisfies . Instead, should have several “nice” properties that does not have, but it should not have properties that you feel are irrelevant to the current problem. Terry tao discusses a similar trick in analysis (trick 10) in his “problem solving strategies,” post.

One of my favorite examples of this strategy is the following standard qual problem:

Problem 1Let be a commutative ring with unity. Let and be matrices with entries in . Then and have the same characteristic polynomial.

The first thing we notice is that if or is invertible, the problem is easy. Indeed, if is invertible, then so

This observation shows that a possible candidate for a universal space , should contain a lot of invertible matrices. This may not be the case for because it may have many zero divisors and few units. Even if has a non-zero determinant, it might not be invertible. Another possible reduction could involve working over a field. Indeed, in this case, a proof involving row and column operations exists, but it is tedious.

A clever way of proceeding, is to consider the following ring:

where and are indeterminates. Define the matrices and . We consider the the quotient field

Observe that over , and are non-zero. Hence, is invertible. Our remark before shows that the characteristic polynomial of is the characteristic polynomial of . Moreover, the characteristic polynomial of has integer coefficients by the definition of the determinant. Thus, the remark is almost trivially true over . Further, we can define a map

where and . Because the coefficients of , the characteristic polynomial of , has coefficients that are polynomials in the entries of , with a bit of thought, we see that the characteristic polynomial of is equal to . The same statement holds for and . Thus, and have the same characteristic polynomial.

Because is the initial object in the category of commutative rings with unity, we can always define such a map. If does not contain an identity element, is it possible to apply this proof to that case as well? The map is not so well defined in this case. However, this technique exposes this problem for what it really is, (symbol manipulation), so it appears that the result should continue to hold.

It’s not necessary to actually pass to the quotient field; it’s only necessary to observe that U is an integral domain, so we can freely cancel terms in the identity det(N) det(Ix – MN) = det(Nx – NMN) = det(Ix – NM) det(N).

Yes! That shortens the proof quite a bit.

Thank you for pointing that out.

This idea also gives a quick proof of Cayley-Hamilton over an arbitrary commutative ring. The theorem is obvious for diagonalizable matrices, and the matrix (X_ij) is diagonalizable inside the algebraic closure of the fraction field of Z[X_ij] (its characteristic polynomial has no repeated roots).

Wow! I was just thinking that I didn’t like the proof of Cayley-Hamilton by Cramer’s rule. This is a very simple proof.