chezmarylou1i

2022-01-01

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

turtletalk75

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

Hattie Schaeffer

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

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 ${\mathbb{F}}_{3}\left[x\right]$ the polynomial p factors as
$p={x}^{6}+1=\left({x}^{2}+1{\right)}^{3}$
where ${x}^{2}+1\in {\mathbb{F}}_{3}\left[x\right]$ is irreducible. It follows that every irreducible factor of p in $\mathbb{Q}\left[x\right]$ has even degree. Now note that $p=h\left({x}^{2}\right)$ where $h:={x}^{3}-6{x}^{2}+12x-11\in \mathbb{Q}\left[x\right]$
A quick check shows that h has no roots in ${\mathbb{F}}_{7}$, and therefore it is irreducible in ${\mathbb{F}}_{7}\left[x\right]$. This means the subring of the quotient ring $\frac{\mathbb{F}\left[x\right]}{\left(p\right)}$ 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 ${\mathbb{F}}_{7}\left[x\right]$, and hence in $\mathbb{Q}\left[x\right]$ and we are done.
If p has an irreducible cubic factor in ${\mathbb{F}}_{7}\left[x\right]$, then this is the reduction of an irreducible factor of p in $\mathbb{Q}\left[x\right]$.
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 $\mathbb{Q}\left[x\right]$ and we are done. If it is quartic then its reduction in ${\mathbb{F}}_{7}\left[x\right]$ is the product of a cubic and a linear factor. But p has no roots in ${\mathbb{F}}_{7}$ because $p=h\left({x}^{2}\right)$ and h has no roots in ${\mathbb{F}}_{7}$, a contradiction.

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