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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