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Let $f(x)\in\mathbb{Z}[x]$ denote a monic irreducible polynomial. Denote by $K$ its splitting field. My question is how can one tell by simply looking at the polynomial $f(x)$, that it lacks a specific symmetry. I.e., if one can enumerate the roots of a degree $n$ irreducible polynomial $(\alpha_i)_{i=1}^n$, then is there a way of checking that for a specific $\sigma\in S^n$, the mapping $\tau_{\sigma}(\alpha_i) = \alpha_{\sigma(i)}$ is not an automorphism of the field $K$?

EDIT: (TYPO, $\tau$ was clearly meant to be a transposition as one of the commenters noticed)

For example, consider the polynomial $x^3 - 3x + 1\in\mathbb{Z}[x]$. It is known that the Galois group of its splitting field is $A_3$, implying that it doesn't have any transpositions. How can I see that a mapping $\tau: K\rightarrow K$ defined by $\tau(\alpha_1) = \alpha_1$, $\tau(\alpha_2) = \alpha_3$ and $\tau(\alpha_3) = \alpha_2$ does not define an automorphism of K?

Thanks in advance!

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    $\begingroup$ You seem to be familiar with the fact that an automorphism of field $K$ must permute the roots of a polynomial over some base field (presumably $\mathbb Q$ when integer polynomials are stipulated). But the example in the last paragraph calls for $\tau$ not to permute those roots, and I suspect $\tau(\alpha_2) = \alpha_3$ was intended. $\endgroup$ – hardmath Dec 29 '19 at 0:44
  • $\begingroup$ Do you know how to construct the splitting field as a tower of simple extensions, the automorphisms are given by the isomorphisms $F(u)\to F(v)$. The obstructions are in those successive factorization of $f$ adding the roots one by one. Even if $f$ is reducible it has a root $y$ in $\Bbb{Q}[y]/(f(y))$ so $f(x)=(x-y)g(x,y)$, then you can repeat with $g$ having a root in $\Bbb{Q}[y,z]/(f(y),g(z,y))$, you'll get a splitting ring $R$ where $S_n$ is a group of automorphism permuting the variables and you'll see the obstructions from $R$'s prime ideals equivalently from the other roots of $f$ in $R$ $\endgroup$ – reuns Dec 29 '19 at 0:56
  • $\begingroup$ I vaguely understand from your comment that you are saying that when there exists such an obstruction, it should appear by one of the polynomials in the chain being reducible all of a sudden. So you mean to say that this happens when fixing several roots causes a fixing of another, supposedly independent root. Can this information be studied from the coefficients somehow? $\endgroup$ – kindasorta Dec 29 '19 at 11:27
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In the splitting field of a cubic with three real roots and square discriminant, all roots can be expressed as polynomials in one root. Therefore, you cannot choose $\tau(\alpha_i)$ independently.

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