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Let $F$ be a field and let $f(x)$ be a polynomial of degree $n \geq 1$ in $F[x]$. Let $g(x) = f\big(\frac{1}{x}\big)x^n$.
Assume $f(0) \neq 0$. Prove that $f(x)$ is irreducible over $F$ if and only if $g(x)$ is irreducible over $F$.

$\mathbf{My~Attempt:}$
Prove of if:
Assume that $f(x)$ is irreducible over $F$, WTP: $g(x)$ is irreducible over $F$.
First, notice that $F[x] = \{ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 ~|~ a_i \in F \text{ and for some } n \in \mathbb{N} \cup \{ 0 \} \}$.
Since, we let $f(x)$ be a polynomial of degree $n \geq 1$ in $F[x]$.
Then, $f(x) = b_n x^n + b_{n-1} x^{n-1} + \cdots + b_1 x + b_0$, for each constant coefficient $b_i \in F$.
Since, we assumed that $f(x)$ is irreducible over $F$ and we know that $f(0) \neq 0$.
Then, by defintion of irreducible, if $f(x) = h(x)j(x)$ with $h(x), j(x) \in F[x]$, then $h(x)$ or $j(x)$ is a unit in $F[x]$.

Now I don't know how to prove that $g(x) = f\big(\frac{1}{x}\big)x^n = b_n + b_{n-1} x + \cdots + b_1 x^{n-1} + b_0 x^n$ whhere $g(x) \in F[x]$ is irreducible.
Also, I have no idea only how to prove the only if which assume $g(x)$ is irreducible over $F$ and prove $f(x)$ is irreducible over $F$.

dancm
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xxxxxx
  • 603

2 Answers2

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Suppose $g$ is reducible, let $g=h_1\cdot h_2$, with $n_1=\deg h_1$, $n_2=\deg h_2$. Then $$\left(x^{n_1}h_1(1/x)\right)\cdot\left(x^{n_2}h_2(1/x)\right)=x^ng(1/x)=f(x),$$ so $f$ is reducible. Hence $f$ irreducible $\implies$ $g$ irreducible.

You can then basically run this argument in reverse.

jlammy
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If $f$ is reducible, then $f(x)=q_1(x)q_2(x)$ and so $g(x)=q_1(\frac{1}{x})q_2(\frac{1}{x})x^n$. But $n=\deg(q_1)+\deg(q_2)$, and so $$g(x)=x^{\deg(q_1)}q_1(\frac{1}{x})x^{\deg(q_2)}q_2(\frac{1}{x})$$

where $x^{\deg(q_1)}q_1(\frac{1}{x})$ is a polynomial. This tells us that $g(x)$ is reducible. This should show you the "only if" part.