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 a,b \in \mathbb Z, there exist integers c,d so that ac-bd=1.

Proof: recall from my previous blogpost “a cute application of mapping class groups” that a member of \pi_1(\mathbb T), or \mathbb Z \oplus \mathbb Z can be realized as a simple closed curve, if and only of (a,b) is such that a and b 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 (1,0) and gluing back, we see that  this can be realized as an automorphism of \mathbb Z \oplus \mathbb Z. However, this is an automorphism in SL_2(\mathbb Z) so that the determinant is one(1,0) \mapsto (a,b), which means that the matrix representation is

    \[\begin{pmatrix}a&c\\b&d \end{pmatrix}\]

with determinant one, so that ad-bc=1.

Theorem (Fundamental Theorem of Algebra): Ever nonconstant polynomial P\in \mathbb C[x] has a root.

Proof: Let p(x) be a  monic polynomial of degree n>0 polynomial. Recall that every  monic polynomial can be realized as the characteristic polynomial of an endomorphism (take \mathbb C[x]/(p(x)) and consider the linear transformation induced my multiplying by x, 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 A  \in GL(\mathbb C^n). This matrix action decends to \mathbb CP^{n-1} 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 A is homotopic to the identity map.  From this, the Lefschetz number of  A^{\prime}:\mathbb CP^{n-1} \to \mathbb CP^{n-1} is the same as the lefschetz number of the identity map, also known as \chi( \mathbb CP^{n-1})=n+1 which is not zero, implying that A^{\prime} has a fixed point, or A has an eigenvalue.

Free Corollary: every linear map \mathbb C^n \to \mathbb C^n has an eigenvalue.

Similar Corollary: Every linear map \mathbb R^n \to \mathbb R^n with n odd has an eigenvalue.

Proof: let T:\mathbb R^n \to \mathbb R^n be linear and without a loss of generality nonsingular. Define f(x)=\frac{T(x)}{\|T(x)\|}. Moreover, since f:S^{n-1} \to S^{n-1} is bijective, it induces an isomorphism on the (n-1)^{th} homology, and hence has degree \pm 1, so we can take \pm T to obtain a degree 1 map. Then its Lefschetz number is nonnegative since \chi(S^{n-1})=2 with n odd, and hence it has a fixed point.


Honorable Mention (Cayley Hamilton): Every matrix satisfies its own characteristic polynomial, as in \mathrm{det}(tI-T) annihilates A.

Proof Sketch: If T has distinct eigenvalues, then v_1, \dots v_n are an eigenbasis with eigenvalues \lambda_1 \dots \lambda_n, so C_T(T)(v_i)=0 by definition. But this was a basis, so it is the zero polynomial. Otherwise, let \delta be the discriminant of T, and note that \delta(T)=0 is the solution of a polynomial (an algebraic variety) whose complement is open, and hence dence. Since T \mapsto \C_T(V) is a polynomial map on \mathrm{End(V)} that vanishes on a Zariski dense subset, so it vanishes everywhere. See here for a full proof.