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

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

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turtletalk75

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 $$\displaystyle{\mathbb{{{Z}}}}_{{{3}}}$$.

###### Not exactly what youâ€™re looking for?
Hattie Schaeffer

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 $$\displaystyle{p}{\left({x}\right)}$$.

karton

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.