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 , defined by . We know that there cannot exist an element such that since every square in an ordered field is positive. So, we cannot hope to factor an arbitrary polynomial over ; we must look to the complex numbers, instead. In fact, it is easy to see that the algebraic closure of must contain by explicitly constructing a nonzero real polynomial with , for a given . Indeed, fix , and define

As we will show below by proving a stronger statement, every polynomial has a root in . We now if every polynomial has a root in . 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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