# What are the obstructions of an automorphism?

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?

• 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. – hardmath Dec 29 '19 at 0:44
• 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 $R$ – reuns Dec 29 '19 at 0:56
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.