Prove x^{6}-6x^{4}+12x^{2}-11 is irreducible over \mathbb{Q}

chezmarylou1i 2022-01-01 Answered
Prove \(\displaystyle{x}^{{{6}}}-{6}{x}^{{{4}}}+{12}{x}^{{{2}}}-{11}\) is irreducible over \(\displaystyle{\mathbb{{{Q}}}}\)

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Expert Answer

turtletalk75
Answered 2022-01-02 Author has 643 answers

Step 1
Let \(\displaystyle{p}{\left({x}\right)}={x}^{{{6}}}-{6}{x}^{{{4}}}+{12}{x}^{{{2}}}-{11}\), whith we can transform into a polynomial in \(\displaystyle{\mathbb{{{Z}}}}_{{{3}}}{\left[{x}\right]}\):
\(\displaystyle{x}^{{{6}}}+{1}\)
Since none of the three elements 0,1,2 in \(\displaystyle{\mathbb{{{Z}}}}_{{{3}}}\) is a root of the polynomial, the polynomial has no factor of degree 1 in \(\displaystyle{\mathbb{{{Z}}}}_{{{3}}}{\left[{x}\right]}\). So the only possible factorings into non constant polynomials are
\(\displaystyle{x}^{{{6}}}+{1}={\left({x}^{{{3}}}+{a}{x}^{{{2}}}+{b}{x}+{c}\right)}{\left({x}^{{{3}}}+{\left.{d}{x}\right.}^{{{2}}}+{e}{x}+{f}\right)}\)
or
\(\displaystyle{x}^{{{6}}}+{1}={\left({x}^{{{4}}}+{a}{x}^{{{3}}}+{b}{x}^{{{2}}}+{c}{x}+{d}\right)}{\left({x}^{{{2}}}+{e}{x}+{f}\right)}\)
From the first equation, since corresponding coefficients are equal, we have
1) \(\displaystyle{x}^{{{0}}}:\quad{c}{f}={1}\)
2) \(\displaystyle{x}^{{{1}}}:\quad{b}{f}+{c}{e}={0}\)
3) \(\displaystyle{x}^{{{2}}}:\quad{a}{f}+{b}{e}+{c}{d}={0}\)
4) \(\displaystyle{x}^{{{3}}}:\quad{c}+{f}+{b}{d}+{a}{e}={0}\)
5) \(\displaystyle{x}^{{{5}}}:\quad{a}+{d}={0}\)
From (1), \(\displaystyle{x}={f}=\pm{1}\), and from (5), \(\displaystyle{a}+{d}={0}\).
Consequently, \(\displaystyle{a}{f}+{c}{d}={c}{\left({a}+{d}\right)}={0}\), and by (3), \(\displaystyle{e}{b}={0}\). But from (2) (since \(\displaystyle{c}={f}\)), \(\displaystyle{b}+{e}={0}\), and therefore \(\displaystyle{b}={e}={0}\). It follows from (4) that \(\displaystyle{c}+{f}={0}\) which is impossible since \(\displaystyle{c}={f}=\pm{1}\).
We have just shown that \(\displaystyle{x}^{{{6}}}+{1}\) cannot be factored into two polynomials each of degree 3.
For the second equation, however, \(\displaystyle{x}^{{{6}}}+{1}={\left({x}^{{{2}}}+{1}\right)}^{{{3}}}\) in \(\mathbb{Z}_{3}[x]\). So we cnnot say \(\displaystyle{p}{\left({x}\right)}\) is irreducible over \(\displaystyle{\mathbb{{{Q}}}}\) because \(\displaystyle{x}^{{{6}}}+{1}\) is irreducible over \(\displaystyle{\mathbb{{{Z}}}}_{{{3}}}\).

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Hattie Schaeffer
Answered 2022-01-03 Author has 3318 answers

Let \(\displaystyle{p}{\left({x}\right)}={x}^{{{6}}}-{6}{x}^{{{4}}}+{12}{x}^{{{2}}}-{11}\)
Substitute \(\displaystyle{x}^{{{2}}}={y}\)
\(\displaystyle{h}{\left({y}\right)}={y}^{{{3}}}-{6}{y}^{{{2}}}+{12}{y}-{11}\)
Letteing \(\displaystyle{g{{\left({y}\right)}}}={h}{\left({y}+{2}\right)}\)
\(g(y)=(y+2)^{3}-6(y+2)^{2}+12(y+2)-11\)
\(\displaystyle{g{{\left({y}\right)}}}={y}^{{{3}}}-{3}\)
\(\displaystyle{g{{\left({y}\right)}}}\) is obviously irreducible, thus so is \(\displaystyle{h}{\left({y}\right)}\) and \(\displaystyle{p}{\left({x}\right)}\).

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karton
Answered 2022-01-09 Author has 9103 answers

Step 1
If you are comfortable with a little more abstract algebra, here is an approach that does not require such ad hoc calculations. First it is easy to see that in \(\mathbb{F}_{3}[x]\) the polynomial p factors as
\(p=x^{6}+1=(x^{2}+1)^{3}\)
where \(x^{2}+1\in\mathbb{F}_{3}[x]\) is irreducible. It follows that every irreducible factor of p in \(\mathbb{Q}[x]\) has even degree. Now note that \(p=h(x^{2})\) where \(h:=x^{3}-6x^{2}+12x-11\in\mathbb{Q}[x]\)
A quick check shows that h has no roots in \(\mathbb{F}_{7}\), and therefore it is irreducible in \(\mathbb{F}_{7}[x]\). This means the subring of the quotient ring \(\frac{\mathbb{F}[x]}{(p)}\) generated by \(x^{2}\) is a cubic field extension of \(\mathbb{F}_{7}\), and therefore p has an irreducible cubic or sextic factor. In the latter case p is irreducible in \(\mathbb{F}_{7}[x]\), and hence in \(\mathbb{Q}[x]\) and we are done.
If p has an irreducible cubic factor in \(\mathbb{F}_{7}[x]\), then this is the reduction of an irreducible factor of p in \(\mathbb{Q}[x]\).
As we saw before, the degree of this factor is even, so it is either quartic or sextic. Again, if it is sextic then p is irreducible in \(\mathbb{Q}[x]\) and we are done. If it is quartic then its reduction in \(\mathbb{F}_{7}[x]\) is the product of a cubic and a linear factor. But p has no roots in \(\mathbb{F}_{7}\) because \(p=h(x^{2})\) and h has no roots in \(\mathbb{F}_{7}\), a contradiction.

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