Let $\beta$ be a zero of $f(x) =x^5+2x+ 4$ . Show that none of $\sqrt{2}$,$ \sqrt[3]{2}$ ,$ \sqrt[4]{2}$ belongs to $Q(\beta)$. Could help me with this excercises..
Let $\beta$ be a zero of $f(x) =x^5+2x+ 4$ . Show that none of $\sqrt{2}$,$ \sqrt[3]{2}$ ,$ \sqrt[4]{2}$ belongs to $Q(\beta)$.
I try.
If $f(x)$ is irreducible over $\mathbb{Q}$, then i can use that:
$$5=[\mathbb{Q}(\beta):\mathbb{Q}]=[\mathbb{Q(\beta)}:\mathbb{Q}(\sqrt{2})]*[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$$
and how $[(\sqrt{2}):\mathbb{Q}]$=3 then $3|5$ contradiction.. Similarly $ \sqrt[3]{2}$ ,$ \sqrt[4]{2}$
But How proof that $f(x)$ is irreducible
For $p=3$, $f_1(x)=x^5+2x+1$, Then
$f_1(0)=1$,$f_1(1)=1$,$f_1(2)=1$ , How $f_1(x)$ is irreducible in $\mathbb{Z_3}$ then $f(x)$ is irreducible in $\mathbb{Q}$
But it can have quadratic factors,  
Other way for proof that $f(x)$ is irreducible over $\mathbb{Q}$
 A: When you reduce $f = x^5 + 2x + 4 \in \mathbb Q[x]$ to $\overline f = x^5 - x + 1 \in \mathbb F_3[x]$, after finding there are no linear factors, it suffices to check that there are no quadratic factors (since if $\overline f$ factors with no linear terms, it must have a term of degree 2). The only quadratic irreducibles in $\mathbb F_3[x]$ are $x^2 + 1$, $x^2 + x - 1$ and $x^2 - x - 1$. You can check by polynomial long division that none of these divides $\overline f$. 
A: Hint:
You are on the right track, and your reasoning up to this point is solid. Wolfram indicates that this polynomial is in fact irreducible over $\mathbb{Z}_3$.  
It isn't actually all that difficult to enumerate the irreducible quadratic and cubic polynomials over $\mathbb{Z}_3$.  There are a total of $9$ quadratic and $27$ cubic monic polynomials over this field, and many of those are going to have roots (recall that quadratic and cubic polynomials are irreducible $\iff$ no roots exist in the given field).  You can further reduce your workload by noting that the coefficients on the quintic are going to constrain the coefficients on the possible factors.
It might seem a bit tedious, but you can use this information to relatively quickly show that your quintic cannot factor as a product of an irreducible quadratic and an irreducible cubic.
A: A method that uses slightly less abstract algebra machinery:
Since $f$ has integer coefficients, it is irreducible over $\mathbb{Q}$ iff it is irreducible over $\mathbb{Z}$.
Suppose $f$ were not irreducible over $\mathbb{Z}$. Then it must be divisible by either a linear or a quadratic irreducible factor over $\mathbb{Z}$. We can easily check that it is not divisible by a linear factor, i.e. it has no roots in $\mathbb{Z}$, using the rational roots test.
So suppose $p$ is an irreducible quadratic dividing $f$ with integer coefficients, so that $f = pq$ for some cubic polynomial $q$ with integer coefficients, and write $p(x) = x^2+ax+b$ ($a,b\in\mathbb{Z}$). Notice that $p$ cannot have exactly 1 root in $\mathbb{R}$, since it is then the square of a linear polynomial with real coefficients, and such a polynomial must have rational coefficients for its square to have rational coefficients. So $p$ has $0$ or $2$ roots in $\mathbb{R}$. But $f$ itself only has one root over $\mathbb{R}$ since $f(x)$ is strictly increasing as a function of $x$, so it follows that $p$ has no roots in $\mathbb{R}$. Since $p$ is a monic quadratic with no real roots, we have that $p(x)>0$ for all real $x$.
Now, notice that $$p(-1)q(-1) = f(-1) = 1$$
and since $q(-1)$ is an integer and $p(-1)$ is a positive integer, it follows that $p(-1) = 1$. Thus
$$(-1)^2+a(-1)+b = 1 \implies -a+b=0\implies a=b.$$
Hence $p(x) = x^2+ax+a$. Now $$p(0)q(0) = f(0) = 4$$
so $p(0)$ is a positive integer dividing $4$, i.e. $p(0) = 1$, $2$, or $4$. Since $p(0) = a$, we know that $a = 1$, $2$, or $4$. Finally, notice that
$$p(1)q(1) = f(1) = 7$$
and hence $p(1)$ divides $7$. But
$$p(1) = 1^2+a(1)+a = 2a+1 $$
and $2a+1$ equals $3$, $5$, and $9$ for $a = 1$, $2$, and $4$, respectively. None of $3$, $5$, nor $9$ divides $7$, so we get a contradiction.
Hence, $f$ is not divisible by an irreducible quadratic over $\mathbb{Z}$ either, so it must be irreducible over $\mathbb{Z}$, and hence over $\mathbb{Q}$ as well.
