*This post assumes familiarity with some basic concepts in algebraic topology, specifically what a group is and the definition of the fundamental group of a topological space.*

The fundamental theorem of algebra has quite a few number of proofs (enough to fill a book!). In fact, it seems a new tool in mathematics can prove its worth by being able to prove the fundamental theorem in a different way. This series of proofs of the fundamental theorem also highlights how in mathematics there are many many ways to prove a single theorem, and in re-proving an established theorem we introduce new concepts and strategies. And perhaps most of all, we accentuate the unifying beauty of mathematics.

**Problem:** Let be a non-constant polynomial with coefficients in . Prove has a root in .

**Solution:** We may assume without loss of generality that is monic. So let

.

Supposing has no roots in , we will show is constant. First, consider for a fixed the loop

Indeed, by assumption the denominators are never zero, so this function is continuous for all . Further, each value is on the unit circle in by virtue of the scaling denominator ( for all ). Finally, and yields the same value, so this is a closed path based at 1.

We note this function is continuous in both and (indeed, they are simply rational functions defined for all ), so that is a homotopy of loops as varies. If then the function is constant for all , and so for any fixed the loop is homotopic to the constant loop.

Now fix a value of which is larger than both and . For , we have

And hence . It follows that the polynomial has no roots when both and . Fixing this , and replacing with in the formula for , we have a homotopy from (when , nothing is changed) to the loop which winds around the unit circle times, where is the degree of the polynomial. Indeed, plug in to get , which is the loop .

In other words, we have shown that the homotopy classes of and are equal, but is homotopic to the constant map. Translating this into fundamental groups, as , we note that , but if then it must be the case that , as is the free group generated by . Hence, the degree of to begin with must have been 0, and so must be constant.

### Like this:

Like Loading...

*Related*

Thank you! The details you provided really helped me to understand the proof.

LikeLike