In class, we were discussing splitting fields, and I was wondering why the quotient of a polynomial ring by an irreducible polynomial "prioritizes" certain roots in the case that all of a polynomial's roots are not conjugate.
Here is the concrete example which led to my question. By Eisenstein's criterion, $f(x) = x^3 - 3x - 15$ is irreducible in $\mathbf{Q}[x]$. We claim that if $K$ is the splitting field of $f$ over $\mathbf{Q}$, then $[K : \mathbf{Q}] = 6$.
Let $L=\frac{\mathbf{Q}[x]}{\langle f \rangle}$. Then since $f$ is degree three, we know that $[L : \mathbf{Q}] = 3$
Suppose $\alpha$ is the real root of $f$. Then $L \cong \mathbf{Q}[\alpha]$, and since $f/(x-\alpha)$ is irreducble in $\mathbf{Q}[\alpha,x]$, it follows that for a complex root $z$ of $x$, $[\mathbf{Q}[\alpha,z] : \mathbf{Q}] = 6$, as claimed.
Here is the question: why is $L \cong \mathbf{Q}[\alpha]$, and not $\mathbf{Q}[z]$?
I can see that if $L \cong \mathbf{Q}[z]$, the degree of $L$ over $\mathbf{Q}$ would not be $3$, but what if there was a polynomial $g$ with multiple roots $\alpha_1,\dots,\alpha_n$ of the "correct" type of degree? Would they have to be conjugate: $\mathbf{Q}[\alpha_i] \cong \mathbf{Q}[\alpha_j]$, or does something else occur? Is this situation even possible; i.e., is it always the case that the correct order of field extensions by quotients of polynomials is completely determined, and it's never the case that one could choose nonconjugate roots of the same "right" degree?