Question regarding the proof of the Fundamental Theorem of Algebra

Question

Every non-constant polynomial in $\mathbb C[x]$ has a complex root

To prove this we first consider a polynomial with complex coefficients $f(x)=a_{0}+a_{1}x+.......a_{n}x^{n}$ and its conjugate polynomial $\bar f(x)=\bar a_{0}+ \bar a_{1}x+\bar a_{2}x^{2}+....+\bar a_{n}x^{n}$

Then we find $$f(x)\bar f(x) \in \mathbb R[x]$$ Then it says

$f(x)$ has a complex root iff $f(x)\bar f(x)$ has a complex root Now if $f(x)$ has a complex root then definitely $f(x)\bar f(x)$ has that complex number as its root but I am stuck with the reverse .

Then there is another problem. Proving the above ensures that It's sufficient to prove the statement for real polynomials.

And then it takes the polynomial $$f(x)=p(x)(x^{2}+1)$$ where $p(x)$ is irreducible over $\mathbb R$ and considers the Galois group of $f(x)$ .

I cannot understand why they needed to take a product of two irreducible polynomials instead of one polynomial irreducible in $\mathbb R[x]$ or at what step of this proof it was put to use.

Please don't mind anybody, I am uploading an image of the proof. Please help me understand these two things properly.

Thanks for any help.

Answer 1

For the first question, if $z_0$ is a root of $f(x)\bar{f}(x)$, $f(z_0)\bar{f}(z_0) = 0$ which means $f(z_0)\overline{f(\bar{z_0})} = 0$ .

thus either $f(z_0) = 0$ or $f(\bar{z_0}) = 0$ (in both case $f$ admits a complex root).

So we can conclude that if $f(x)\bar{f}(x)$ has a complex root, then $f(x)$ has a complex root.

For the second question. $\mathbb{C}$ is the splitting field of $x^2 + 1$ over $\mathbb{R}$ so in the answer $x^2 + 1$ is multiplied to make sure that $E$ "contains" $\mathbb{C}$($\mathbb{C}$ is isomorphic to a subfield of $E$).

Answer 2

Consider the homomorphism $\varphi\colon\mathbb{C}[x]\to\mathbb{C}[x]$ defined by $\varphi(a)=\bar{a}$ for $a\in\mathbb{C}$ and $\varphi(x)=x$. It is clear that $\varphi\circ\varphi$ is the identity, so $\varphi$ is an automorphism. In particular, since having a root is equivalent to being divisible by a degree one polynomial, $f\in\mathbb{C}[x]$ has a root if and only if $\varphi(f)$ has a root; note that $\varphi(f)=\bar{f}$ according to your notation.

If $f$ has a root, then clearly $f(x)\bar{f}(x)$ has a root. If $f(x)\bar{f}(x)$ has a root, then either $f$ has a root or $\bar{f}$ has a root; in the latter case also $f$ has a root.

Clearly $\overline{f(x)\bar{f}(x)}=f(x)\bar{f}(x)$, so every coefficient of $f(x)\bar{f}(x)$ is real.

Finally, considering the splitting field $E$ of $(x^2+1)p(x)$ ensures that $\mathbb{C}$ is (isomorphic to a) subfield of $E$.