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I am reading an algebra book now.

Let $p(x) \in \mathbb{R}[x]$ and $\deg p = n$.
Let $\alpha \in \mathbb{C} - \mathbb{R}$ and $p(\alpha) = 0$.

Then, $p(\bar{\alpha}) = \overline{p(\alpha)} = 0$.

So, $p(x)$ is divisible by $(x-\alpha)(x-\bar \alpha)$.

$(x-\alpha)(x-\bar \alpha) = x^2 - 2 \Re(\alpha) x + |\alpha|^2 \in \mathbb{R}[x]$.

So, there is a polynomial $q(x) \in \mathbb{R}[x]$ such that $p(x) = (x^2 - 2 \Re(\alpha) x + |\alpha|^2) q(x)$.

Why $q(x) \in \mathbb{R}[x]$?

Is it necessary to prove that $q(x) \in \mathbb{R}[x]$?
Or,
is it obvious that $q(x) \in \mathbb{R}[x]$?

My proof is here:

There is $q_1(x) \in \mathbb{C}[x]$ such that $p(x) = (x-\alpha)(x-\bar \alpha) q_1(x)$ in $\mathbb{C}[x]$.
There is $q_2(x), r_2(x) \in \mathbb{R}[x]$ such that $p(x) = (x^2 - 2 \Re(\alpha) x + |\alpha|^2) q_2(x) + r_2(x)$ in $\mathbb{R}[x]$ and $2 > \deg r_2$.
$p(x) = (x-\alpha)(x-\bar \alpha) q_1(x) = (x^2 - 2 \Re(\alpha) x + |\alpha|^2) q_2(x) + r_2(x)$ in $\mathbb{C}[x]$.
By a theorem about long division, $q_1(x) = q_2(x), r_2(x) = 0$.
So, $p(x) = (x^2 - 2 \Re(\alpha) x + |\alpha|^2) q_2(x)$, and $q_2(x) \in \mathbb{R}[x]$.

tchappy ha
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  • Is this proposition obvious or not? This was my original question. – tchappy ha Dec 24 '18 at 02:20
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    As you can see from the link, there are various methods to prove it. Whether or not any of them are "obvious" depends on the knowledge level of the person making that judgement. At undergraduate level this typically would require proof. – Bill Dubuque Dec 24 '18 at 02:39
  • Thank you very much, Bill Dubuque. – tchappy ha Dec 24 '18 at 02:56
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    You can carry out a polynomial division in $\mathbb R$ and since you are dividing by a quadratic the remainder is at most linear. The same division is valid in $\mathbb C$ (step by step), but in this context the remainder is zero. It must therefore be zero in $\mathbb R$ too and you have a factorisation in $\mathbb R$. – Mark Bennet Dec 24 '18 at 03:23
  • Thank you very much, Mark Bennet. – tchappy ha Dec 24 '18 at 07:57

2 Answers2

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The key point is that if $f,g$ are nonzero polynomials, all of $f$s coefficients are real, and $g$ has some non-real coefficient, then their product $f\cdot g$ also has some non-real coefficient. This is easy to prove - think about the lowest-degree term of $g$ with non-real coefficient ...

Now consider $f(x)=x^2-2\mathfrak{R}(\alpha)x+\vert\alpha\vert^2$ and $g(x)=q(x)$, and note that $p=f\cdot g$ has all-real coefficients ...

Noah Schweber
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r(x) = (x-a)(x-$\bar{a}$) $\in$ $\mathbb{R}[x]$. Since p(x) $\in$ $\mathbb{R}[x]$ and r(x) | p(x), there exists a polynomial q(x) such that p(x) = q(x)r(x). q(x) could not have imaginary coefficients because after multiplication p would have to have complex coefficients as well.

Joel Pereira
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