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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?

RobPratt
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While I Am
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  • Well if $a, b$ are two roots of an irreducible polynomial $f(x) $ of positive degree with rational coefficients then $\mathbb {Q} (a) $ is isomorphic to $\mathbb {Q} (b) $ and they both are isomorphic to $\mathbb {Q} [x] /(f(x)) $. In short roots of irreducible polynomials are indistinguishable via processes of algebra. – Paramanand Singh Dec 14 '21 at 02:33
  • However using processes of algebra one can sometimes show that $b\notin\mathbb {Q} (a) $ so that adjoining one root does not necessarily lead to splitting field. – Paramanand Singh Dec 14 '21 at 02:38
  • I agree with you. In the example above, we have two "conjugacy classes" of roots of irreducible $f$; the real root $\alpha$, and the complex roots ${z,\overline{z}}$. I'm confused about why the quotient by the ideal of $f$ gives the extension by $\alpha$, not $z$. That is, why is $\alpha$ "prioritized"? I can see it from the degree of the extension, but that isn't a satisfactory explanation. For example, from that I have no insight into the order of extensions by an irreducible polynomial with many distinct conjugacy classes. – While I Am Dec 14 '21 at 03:55
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    It does not give $\alpha$ or $z$ but rather it can be taken as any root. – Paramanand Singh Dec 14 '21 at 03:56
  • So you're saying $$\mathbf{Q}[x]/\langle f \rangle \cong \mathbf{Q}[z] \cong \mathbf{Q}[\alpha]?$$ I've seen this for conjugate roots, but I thought the behavior would be different since the real root is not conjugate to the complex ones. (I've assumed $f$ is not the minimal poly of $\alpha$, so maybe that's where my mistake is). – While I Am Dec 14 '21 at 04:00
  • If $f$ is irreducible then it is the minimal polynomial of each of its roots – Paramanand Singh Dec 14 '21 at 04:05
  • See this question https://math.stackexchange.com/q/2241608/72031 – Paramanand Singh Dec 14 '21 at 04:08
  • This was very helpful. Thanks! – While I Am Dec 14 '21 at 04:13

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