Determine field degree extension of certain number fields. Let $p(x)\in \mathbb{Q}(x)$ be an irreducible monic polynomial of degree 3 and suppose all of its roots are real; call the roots $r_1,r_2,r_3$. Let $K=\mathbb{Q}(r_1,r_2,r_3)$. Is it the case that $K=\mathbb{Q}(r_1)$ (Or equivalently, if the extension $[K:\mathbb{Q}]=3$)?
I am being told this is true, but have no idea how to justify this. I know that showing $\text{Disc}(p)$ is a perfect square in $\mathbb{Q}$ is sufficient, but have no idea showing that either.
Also, if this is true, does this result generalize for degree prime irreducible polynomials with all real roots?
 A: I don't know about your background, but I'm afraid that an answer to your question requires a minimal amount of Galois theory. Your field $K=\mathbf Q (r_1, r_2, r_3)$ is the splitting field of an irreducible cubic $f$, whose roots $r_i$ are necessarily distinct (because we are in characteristic $0$). The Galois group $G$ of $K/\mathbf Q$ acts on the roots $r_i$, hence can be viewed as a subgroup os the symmetric group $S_3$, whose only non trivial subgroups are $S_3$ itself, of order 6, and the alternate group $A_3$ (consisting of the even permutations), of order 3. The distinction between the two cases is given by the following (see any textbook on Galois theory): $G=A_3$ iff $Disc (f)$ (which belongs to $\mathbf Q$ because it is fixed by $G$) is a square in $\mathbf Q$. NB : This result is valid over any base field of characteristic $\neq 2$.
Being a cubic with real coefficients, $f$ admits at least one real root. If only one root is real, the other two are conjugate complex, hence $G$ contains a transposition, and $G=S_3$. Suppose that all the roots $r_i$ are real. By definition, $Disc (f)$ is the square of the product of the factors $(r_i - r_j)$ for $i<j$, so is a priori a square in $\mathbf R$, but not necessarily a square in $\mathbf Q$, and it is easy to give counter-examples, using e.g. the reduced form $f=aX^2+bX+c$, with discriminant $-4a^3-27 b^2$.   
