1. ## Help with Proof.

My book on Abstract Algebra (John Fragleigh 7th Ed.) is about to prove that every field has a extension field which has algebraic closure. Before it shows that it proves it for complex numbers. More, famously known as the Fundamental Theorem of Algebra.

Let $f(z)\in C[z]$ be a non-constant polynomial, in sake of a contradicton assume, that $f(z)$ has no zero. Then $\frac{1}{f(z)}$ is analytic everywhere. Then, $\lim_{|z| \rightarrow \infty}\left|\frac{1}{f(z)}\right|=0$. Thus, $\frac{1}{f(z)}$ is bounded (this I do not understand). Then by Liouville's Theorem it must be a constant, thus $f(z)$ must be a constant thus a contradiction.
Q.E.D.

Note it is possible that I erred because I am writing this by memory. May someone help explain this proof to me, I am dieing to know the proof to this theorem.

One more question, why is it so fundamental? There are other theorems from algebra which probably have more importance, like that every field has an extension field having algebraic closure, that is more fundamental!

2. I assume you have basic knowledge of complex analysis? Since you'll need to know what it implies for a function to be analytic and as you can see, you'll be using Liouville's theorem as well.

Let's state Liouville's theorem first.

If $f:\mathbb{C} \to \mathbb{C}$is bounded and analytic for every $z \in \mathbb{C}$, then f is a constant function.

This follows rather easy from Cauchy's inequality (one of the many ) if you write f(z) as a power series, which is possible due to its analyticity.

Now the fundamental theorem of algebra, with a bit more detail in the proof.

Proof
Let f(z) be a non-constant polynomial over C and suppose that $f\left( z \right) \ne 0\,\,\,\forall \,\,\,z \in \mathbb{C}$.
This makes 1/f(z) analytic everywhere. We then have

$\lim_{|z| \rightarrow \infty}\left|\frac{1}{f(z)}\right|=0$

But this implies that there exists an R > 0 for which we have (take epsilon = 1 in the definition):

$\left| {\frac{1}{{f\left( z \right)}}} \right| < 1$

if |z| > R. Because f(z) is continuous over the disc $\left| {z - a} \right| \leqslant R$, so is 1/f(z) and thus bounded there (since the disc is a compact set) which then implies that it is bounded on the complex plane.
But a bounded and analytic function is a constant function, according to Liouville's theorem, hence 1/f(z) is constant and so is f(z) which is a contradiction with our assumption that f(z) was non-constant.

QED.

3. Thanks for your help, what do you think of the fundamental theorem?

4. Well from this theorem it follows that the field of the complex numbers is the algebraic closure of the field of the real numbers which seems very fundamental (and historically important) theorem to me

Personally, I like this proof since it shows the power of complex analysis. You can proof this theorem without it (e.g. purely algebraically) but I find this very elegant and short for such a powerful theorem.

5. Originally Posted by TD!
Well from this theorem it follows that the field of the complex numbers is the algebraic closure of the field of the real numbers which seems very fundamental (and historically important) theorem to me

Personally, I like this proof since it shows the power of complex analysis. You can proof this theorem without it (e.g. purely algebraically) but I find this very elegant and short for such a powerful theorem.
There are many ways to prove, the quickest two proofs are this one from complex analysis and another one from topology. The algebraic proof of this I think is very long.