Let F be a field. Suppose that a polynomial p(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n}

Sapewa 2021-12-30 Answered
Let F be a field. Suppose that a polynomial \(\displaystyle{p}{\left({x}\right)}={a}_{{{0}}}+{a}_{{{1}}}{x}+\cdots+{a}_{{{n}}}{x}^{{{n}}}\) is reducible in \(\displaystyle{F}{\left[{x}\right]}\). Prove that the polynomial \(\displaystyle{q}{\left({x}\right)}=={x}^{{{n}}}{p}{\left({\frac{{{1}}}{{{x}}}}\right)}={a}_{{{n}}}+{a}_{{{n}-{1}}}{x}+\cdots{a}_{{{0}}}{x}^{{{n}}}\) is also reducible.

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Expert Answer

Paineow
Answered 2021-12-31 Author has 1356 answers

Step 1
Definition: Let \(\displaystyle{\left({F},\ +,\ \cdot\right)}\) be a field and let \(\displaystyle{f}\in{F}{\left[{x}\right]}\). Then f is said to be Irreducible over F if f cannot be factored into a product of polynomials all of which having lower degree than f. If f is not irreducible over F then we say that f is Reducible over F.
Step 2
We will prove with the help of principal of mathematical equation.
For \(\displaystyle{n}={1}\)
\(\displaystyle{P}{\left({x}\right)}={a}_{{{0}}}+{a}_{{{1}}}{x}\) is reducible in field \(\displaystyle\Rightarrow{a}_{{{0}}}{a}_{{{1}}}\in{F}{\left({x}\right)}\) and \(\displaystyle{q}{\left({x}\right)}={x}{\left({a}_{{{0}}}+{\frac{{{a}_{{{1}}}}}{{{x}}}}\right)}={a}_{{{0}}}{x}+{q}_{{{1}}}\) is also reducible in field as \(\displaystyle{a}_{{{0}}}{a}_{{{1}}}\in{F}{\left({x}\right)}\)
Now let for \(\displaystyle{n}={k}\) is true
\(\displaystyle{P}{\left({x}\right)}={a}_{{{0}}}+{a}_{{{1}}}{x}+\cdots{a}_{{{k}}}{x}^{{{k}}}\) is reducible in fidd
So any polynomial of degree dess than k with \(\displaystyle{a}_{{{0}}}{a}_{{{1}}}\)
\(\displaystyle{a}_{{{k}}}\in{F}{\left({x}\right)}\) is reducible
\(\displaystyle\Rightarrow{q}{\left({x}\right)}={a}_{{{0}}}{x}^{{{k}}}+{a}_{{{1}}}{x}^{{{k}-{1}}}\cdots+{a}_{{{k}}}\) is reducible in field
Now for \(\displaystyle{n}={k}+{1}\)
\(\displaystyle{P}{\left({x}\right)}={a}_{{{0}}}+{a}_{{{1}}}{x}+\cdots{a}_{{{k}}}{x}^{{{k}}}+{a}_{{{k}+{1}}}{x}^{{{k}+{1}}}\) in reducible
\(\displaystyle\Rightarrow{a}_{{{0}}}\cdot{a}_{{{k}+{1}}}\in{f{{\left({x}\right)}}}\) with polynomial of degree less or equal to k is reducible
Step 3
Now \(\displaystyle{k}={1}\)
\(\displaystyle{q}{\left({x}\right)}={a}_{{{k}+{1}}}+{q}_{{{k}}}{x}+\cdots{q}_{{{0}}}{x}^{{{k}+{1}}}\)
\(\displaystyle{q}{\left({x}\right)}={a}_{{{k}+{1}}}+{x}{\left({q}_{{{k}}}+\cdots{a}_{{{0}}}{x}^{{{k}}}\right)}\)
Since \(\displaystyle{a}_{{{k}}}+\cdots{a}_{{{0}}}{x}^{{{k}}}\) is reducible in f(x) and k+1 degree polinomial with \(q_{k+1}\in f(x)\) can also reducible
\(\displaystyle\Rightarrow{q}{\left({x}\right)}\) is reducible.

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lenkiklisg7
Answered 2022-01-01 Author has 703 answers
Step 1
Suppose \(\displaystyle{p}{\left({x}\right)}={a}_{{{0}}}+{a}_{{{1}}}{x}+\cdots+{a}_{{{n}}}{x}^{{{n}}}\in{F}{\left[{x}\right]}\) is not irreducible, i.e., there is \(\displaystyle{q},\ {r}\in{F}{\left[{x}\right]}\) such that \(\displaystyle{d}{e}{g{{\left({q}\right)}}},\ {d}{e}{g{{\left({r}\right)}}}{>}{0}\) and \(\displaystyle{p}={q}{r}\)
We know that \(\displaystyle{d}{e}{g{{\left({p}\right)}}}={d}{e}{g{{\left({q}\right)}}}+{d}{e}{g{{\left({r}\right)}}}\), so we may assume WLG that
1) \(\displaystyle{q}{\left({x}\right)}={b}_{{{0}}}+{b}_{{{1}}}{x}+\cdots+{b}_{{{m}}}{x}^{{{m}}}\)
2) \(\displaystyle{r}{\left({x}\right)}={c}_{{{0}}}+{c}_{{{1}}}{x}+\cdots+{c}_{{{n}-{m}}}{x}^{{{n}-{m}}}\)
Step 2
Then
\(\displaystyle{a}_{{{0}}}{x}^{{{n}}}+{a}_{{{1}}}{x}^{{{n}-{1}}}+\cdots+{a}_{{{n}}}=\)
\(\displaystyle{x}^{{{n}}}{p}{\left({\frac{{{1}}}{{{x}}}}\right)}=\)
\(\displaystyle{x}^{{{n}}}{q}{\left({\frac{{{1}}}{{{x}}}}\right)}{r}{\left({\frac{{{1}}}{{{x}}}}\right)}=\)
\(\displaystyle{x}^{{{m}}}{q}{\left({\frac{{{1}}}{{{x}}}}\right)}{x}^{{{n}-{m}}}{r}{\left({\frac{{{1}}}{{{x}}}}\right)}=\)
\(\displaystyle{\left({b}_{{{0}}}{x}^{{{m}}}+{b}_{{{1}}}{x}^{{{m}-{1}}}+\cdots+{b}_{{{m}}}\right)}{\left({c}_{{{0}}}{x}^{{{n}-{m}}}+{c}_{{{1}}}{x}^{{{n}-{m}}}+\cdots+{c}_{{{n}-{m}}}\right)}\)
Therefore \(\displaystyle{a}_{{{0}}}{x}^{{{n}}}+{a}_{{{1}}}{x}^{{{n}-{1}}}+\cdots+{a}_{{{n}}}\) is not irreducible.
The implication in the opposite direction is analogous.
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karton
Answered 2022-01-09 Author has 9103 answers

Substitute \(x\rightarrow\frac{1}{z}\). We get
\(a_{0}+a_{1}\frac{1}{z}+\cdots+a_{n}(\frac{1}{z})^{n}\)
\(=\frac{1}{z^{n}}(a_{0}z^{n}+a_{1}z^{n-1}+\cdots+a_{n})=\frac{1}{z^{n}}q(z)\)
\(p(z)\)is irreducible if \(q(z)\) is irreducible.

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