If $a$ is a positive integer that is not a perfect square then $[\Bbb Q(a^{\frac{1}{4}}):\Bbb Q]=4$ If $a$ were a square free number I am done by Eisenstein  criterion but if $a$ is not a perfect square how do I use Eisenstein criterion. The book by Joseph Rotman gives an argument that that there exists a prime $p$ such that $p$ divides $a$ but $p^2$ does not divide $a$, but I dont get this.
 A: May be Rotman means the following.
Consider the prime factorization 
$$
a=\prod_{i=1}^kp_i^{a_i}.
$$
Because 
$$\Bbb{Q}(a^{1/4})=\Bbb{Q}([a/p^4]^{1/4})\tag{*}$$ for any prime $p$, we can without loss of generality assume that $a_i<4$ for all $i$. As we assumed that $a$ is not a perfect square we also know that at least one of exponents, say $a_{i_0}$, is odd. It sounds like you know what to do when $a_{i_0}=1$, so the troublesome case is that of $a_{i_0}=3$.
We can deal with that case by a trick similar to $(*)$. All we need to do is to observe that
$$
\Bbb{Q}(a^{1/4})=\Bbb{Q}(a^{-1/4})=\Bbb{Q}(A^{1/4}),
$$
where
$$
A=\frac{\prod_{i=1}^kp_i^4}a=\prod_{i=1}^kp_i^{4-a_i}.
$$
Here $4-a_{i_0}=1$, so Eisenstein's criterion proves that $x^4-A$ is irreducible over $\Bbb{Q}$ and you are done.
A: It suffices to show $X^4-a$ is irreducible over $\mathbb{Q}$. If it where reducible, it would reduce in either two polynomials of degree 1 and 3 or two polynomials of degree 2. 
In the first case, the degree one polynomial should have a zero of $X^4-a$ in common, so it follows some zero of $X^4-a$ is in $\mathbb{Q}$. However, the zeroes of this polynomial are $\{a^{1/4}, -a^{1/4}, i \cdot a^{1/4}, -i \cdot a^{1/4} \}$ so the only possibilities are $\pm a^{1/4} \in \mathbb{Q}$, from which it follows that $\sqrt{a} \in \mathbb{Q}$. An elementary argument now shows $\sqrt{a} \in \mathbb{N}$, a contradiction.
Suppose now $X^4-a=pq$, with $p$ and $q$ both of degree 2. Suppose wlog that $p$ has $i \cdot a^{1/4}$ as a root. Then it must have $-i \cdot a^{1/4}$ as a root, which again shows $\sqrt{a} \in \mathbb{Q}$, a contradiction. 
A: We want to show that $X^4 - a$ is irreducible over $\mathbf Q$. Let $\alpha$ be a real 4-th root of $a$ (which exists because $a>0$) and consider the extensions $k =\mathbf Q (i), K=\mathbf Q (i, \alpha)$. Since $k$ and $\mathbf Q (\alpha)$ are linearly disjoint over  $\mathbf Q$, we have $[\mathbf Q (\alpha):\mathbf Q]=4$  iff $[K:k]=4$. But according to Kummer's theory over $\mathbf Q(i)$, $K/k$ is cyclic of exponent 4, with degree equal to the order of the subgroup $A$ of $k^{*}/(k^{*})^4$ generated by $a$. If the order of $A$ were $2$, then $\pm a \in (k^{*})^2$, or equivalently $\pm a \in (\mathbf Q^{*})^2$, or yet $\mathbf Q(\sqrt {a})= \mathbf Q$ or $\mathbf Q (i)$: the first case is impossible because $a$ is not a square, and the second is impossible because $a>0$. Hence $[\mathbf Q (\alpha):\mathbf Q]=4$ and $X^4 - a$ is irreducible.  
For information.  You can find in Lang's "Algebra", chap. VIII, §9, an inductive (induction on the degree) proof of the following irreducibility result : Let $F$ be a field and $n \geq 2$. Let $a \in F^{*}$ such that for all primes $p$ dividing $n$, $a \notin (F^{*})^p$, and if $4$ divides $n$ then $ a \notin -4(F^{*})^4$. Then $X^n - a$ is irreducible over $F$.  
