I want to prove that $p(x)=a_0+a_1x+\ldots+a_nx^n$ is irreducible in $K[x]$ (where $K$ is a field) if and only if $a_0x^n+a_1x^{n-1}+\ldots+a_n$ is irreducible, knowing that $a_0\cdot a_n \neq 0$. How can I do this?
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It suffices to note that the map $(a_0 + \cdots + a_nx^n) \mapsto a_0 x^n + \cdots + a_n$ is a homomorphism – Ben Grossmann Oct 08 '20 at 14:32
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2the last polynomial is $x^{n}p(1/x)$ – Mittens Oct 08 '20 at 14:33
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Note that in English $K$ is a field, not a "body". – David C. Ullrich Oct 08 '20 at 15:29
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1Does this answer your question? A transformation of F[x]. See also Irreducibility of the polynomial $a_{n}+a_{n-1}x+\cdots+a_{0}x^{n}$ when irreducibility of $a_{0}+a_{1}x+\cdots+a_{n}x^{n}$ is given and Prove that the polynomial $x^nf(1/x)$ with reverted coefficients is also irreducible polynomial over $\mathbb{Q}$, – Sil Oct 08 '20 at 18:31
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Irreducibility of reciprocal polynomials, Prove that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible. – Sil Oct 08 '20 at 18:34
2 Answers
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Substitute $x\to \frac{1}{z}$. We get
$$a_0+a_1\frac{1}{z}+\ldots+a_n\left(\frac{1}{z}\right)^n=\frac{1}{z^n}\left(a_0z^n+a_1z^{n-1}+\ldots+a_n\right)=\frac{1}{z^n}q(z)$$ $p(z)$ is irreducible iff $q(z)$ is irreducible.
Raffaele
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1I don't think your first statement if true. $p(x)=(x^2+1)\cdot(x^2+2)$ is reducible in $\mathbb{R}[x]$ and it doesn't have any roots in $\mathbb{R}$ – kubo Oct 08 '20 at 14:42
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Suppose $p(x) = a_0 + a_1 x + \dots + a_n x^n\in \Bbb{K}[x]$ is not irreducible, i.e., there is $q, r\in \Bbb{K}[x]$ such that $\deg(q), \deg(r) > 0$ and $p = qr$.
We know that $\deg(p) = \deg(q) + \deg(r)$, so we may assume WLG that
- $q(x) = b_0 + b_1 x + \dots +b_m x^m$
- $r(x) = c_0 + c_1 x + \dots + c_{n-m} x^{n-m}$
Then
$\begin{aligned} a_0 x^n + a_1 x^{n-1} + \dots + a_n &=\\ x^n p\left(\frac{1}{x}\right) &=\\ x^n q\left(\frac{1}{x}\right)r\left(\frac{1}{x}\right) &=\\ x^m q\left(\frac{1}{x}\right) x^{n-m} r\left(\frac{1}{x}\right) &=\\ (b_0 x^m + b_1 x^{m-1} + \dots + b_m)(c_0 x^{n-m} + c_1 x^{n-m-1} + \dots + c_{n-m}) \end{aligned}$
Therefore $a_0 x^n + a_1 x^{n-1} + \dots + a_n$ is not irreducible.
The implication in the opposite direction is analogous.
Alma Arjuna
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