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.

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. 
