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I'm reading a proof of the FTA (taken in the book Algebra : Chapter 0 by Paolo Aluffi) and I'm having a hard time understanding the beginning :

"Let $f(x) \in \mathbb{C}[x]$ be a nonconstant polynomial; we have to prove that $f(x)$ has roots in $\mathbb{C}$. Let $F$ be a splitting field for $f(x)$ over $\mathbb{R}$; embed $F$ in an algebraic closure of $\mathbb{R}$, and consider the extension

$$\mathbb{R} \subseteq F(i)$$

This extension is Galois: it is the splitting field of the square-free part of $f(x)(x^2 + 1)$."

Why is this true ? Also, did the author meant that $f(x)$ is in $\mathbb{R}[x]$ instead of $\mathbb{C}[x]$ ? Is there something I'm missing ?

Thank you for reading

Ceru
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  • You should cite your sources. Is this a textbook? A handout? Course notes? Who is "the author"? – Arturo Magidin May 26 '24 at 00:03
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    Maybe they intended to write $f(x)\overline{f}(x)(x^2+1)$ – ameg May 26 '24 at 00:05
  • @ArturoMagidin Corrected – Ceru May 26 '24 at 00:08
  • Thank you. You do not want just a real polynomial, because that would not establish the FTA. But the reference ot a "square part" suggests that the author meant to consider the splitting field of $f(x)\overline{f(x)}$ over $\mathbb{R}$. – Arturo Magidin May 26 '24 at 00:08
  • @ArturoMagidin Well he does say I cite "Note that if $f(x)$ has no roots in $\mathbb{C}$, then neither does $f(x) \overline{f(x)} \in \mathbb{R}[x]$; that is, we may assume that $f(x)$ has real coefficients." I think that you are right – Ceru May 26 '24 at 00:13
  • I checked in the book. A few lines before, they had replaced $f(x)$ with $f(x)\overline{f(x)}$ – ameg May 26 '24 at 00:13
  • @ameg That seems to be the case. Thank you – Ceru May 26 '24 at 00:17
  • Yes, you want the conjugate polynomial as a factor to force it into $\Bbb{R}[x]$. Using the factor $x^2+1$ also simplifies the argument a bit. Can do without, (discussed here) but the argument is a lot cleaner with that extra factor in there. – Jyrki Lahtonen May 26 '24 at 03:28

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