## Fundamental Theorem of Algebra

The reader is likely familiar with the result that there exist real polynomials without any roots in the ordered field $\mathbb{R}$. Indeed, consider the quadratic polynomial $p \in \mathbb{R}[x]$, defined by $p(x) := x^{2} + 1$. We know that there cannot exist an element $a \in \mathbb{R}$ such that $a^{2} = -1$ since every square in an ordered field is positive. So, we cannot hope to factor an arbitrary polynomial $p \in \mathbb{R}[x]$ over $\mathbb{R}$; we must look to the complex numbers, instead. In fact, it is easy to see that the algebraic closure of $\mathbb{R}$ must contain $\mathbb{C}$ by explicitly constructing a nonzero real polynomial $p \in \mathbb{R}[x]$ with $p(z) = 0$, for a given $z \in \mathbb{C}$. Indeed, fix $z \in \mathbb{C}$, and define

$\displaystyle p(x) := (x-z)(x-\overline{z}) = x^{2} - \overline{z}x - zx + \left|z\right|^{2} = x^{2} - 2\text{Re}(z)x + \left|z\right|^{2}$

As we will show below by proving a stronger statement, every polynomial $p \in \mathbb{R}[x]$ has a root in $\mathbb{C}$. We now if every polynomial $p \in \mathbb{C}[z]$ has a root in $\mathbb{C}$. The answer is yes. I learned to prove the Fundamental Theorem of Algebra using Liouville’s theorem of complex analysis, so this proof is new to me. I found it thumbing through Baby Rudin.

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