# What No Algebra Teacher Ever Told Me.

These are a bunch of Topological proofs for facts in  Algebra. Have you ever had the experience of  insisting to yourself that the fundamental theorem of algebra is really a topological result, only to forget the argument? Have you ever proven Cayley Hamilton by muttering something about cofactors? Showing Bezout’s Identity by stating the  Euclidian algorithm (whose proof eludes lazy undergraduates everywhere.)

I have. I am therefore writing this blogpost to once again weasel out of actually doing any math.

Theorem (Bezout’s Identity): For any coprime integers , there exist integers so that .

Proof: recall from my previous blogpost “a cute application of mapping class groups” that a member of , or can be realized as a simple closed curve, if and only of is such that and are coprime. From this, we know that they define a simple closed curve, and if we cut along it, we obtain a cylinder. Of course, there is a homeomorphism that carries this to the cylinder cut along and gluing back, we see that  this can be realized as an automorphism of . However, this is an automorphism in so that the determinant is one, which means that the matrix representation is

with determinant one, so that .

Theorem (Fundamental Theorem of Algebra): Ever nonconstant polynomial has a root.

Proof: Let be a  monic polynomial of degree polynomial. Recall that every  monic polynomial can be realized as the characteristic polynomial of an endomorphism (take and consider the linear transformation induced my multiplying by , the companion matrix.)  Note that an eigenvalue of this matrix corresponds precisely to a zero of its characteristic polynomial. Assuming that zero is not an eigenvalue (else we are done) we may take the companion matrix . This matrix action decends to by sending the image of a vector to its span. Moreover, since the general linear group is path connected (by Gauss-Jordan elimination) we also know that is homotopic to the identity map.  From this, the Lefschetz number of  is the same as the lefschetz number of the identity map, also known as which is not zero, implying that has a fixed point, or has an eigenvalue.

Free Corollary: every linear map has an eigenvalue.

Similar Corollary: Every linear map with odd has an eigenvalue.

Proof: let be linear and without a loss of generality nonsingular. Define . Moreover, since is bijective, it induces an isomorphism on the homology, and hence has degree , so we can take to obtain a degree map. Then its Lefschetz number is nonnegative since with odd, and hence it has a fixed point.

Honorable Mention (Cayley Hamilton): Every matrix satisfies its own characteristic polynomial, as in annihilates .

Proof Sketch: If has distinct eigenvalues, then are an eigenbasis with eigenvalues , so by definition. But this was a basis, so it is the zero polynomial. Otherwise, let be the discriminant of , and note that is the solution of a polynomial (an algebraic variety) whose complement is open, and hence dence. Since is a polynomial map on that vanishes on a Zariski dense subset, so it vanishes everywhere. See here for a full proof.