Let $K$ be a field of characteristic $0$ and $f(y)=a_0(y-\alpha_1)\cdots (y-\alpha_n)$ be a non-constant polynomial over $K$ whose roots $\alpha_i$'s are in some extension field. Let $L^*=K(\alpha_1,\ldots,\alpha_n)$. Let $\mbox{Gal}(L^*,K)$ be the group of $K$-automorphisms of the field $L^*$. For any $\sigma\in{\rm Gal}(L^*,K)$, $f(y)=a_0 (y-\sigma(\alpha_1))\cdots (y-\sigma(\alpha_n))$, hence $\sigma$ gives a permutation $\sigma^*$ of $\{\alpha_1,\ldots,\alpha_n\}$. The map $\sigma\mapsto \sigma^*$ gives an injective homomorphism from group ${\rm Gal}(L^*,K)$ to $S_n$ (permutation group on $\alpha_1,\ldots,\alpha_n$). Denote the image of it by ${\rm Gal}(f,K)$.

This is a description about Galois group of a polynomial from the book of Abhyankar: Lectures on Algebra. My question is below:

Later, author says:

Now Galois himself characterized ${\rm Gal}(f,K)$ as the set of all relations preserving permutations of the roots, i.e. the set of all $\tau\in S_n$ such that for every $P(X_1,\ldots, X_n)\in K[X_1,\ldots, X_n]$, with $P(\alpha_1,\ldots,\alpha_n)=0$ we have $P(\tau(\alpha_1),\ldots,\tau(\alpha_n))=0$

I do not understand this comment; can anyone explain it, with possibly an example?

  • $\begingroup$ The $n$ roots $\alpha_i$ determines a polynomial when they are considered as independent polynomials, which is symmetric, i.e., $P(\alpha_{\sigma(1)}, \alpha_{\sigma(2)}, \cdots,\alpha_{\sigma(n)})=P(\alpha_1, \alpha_2, \cdots,\alpha_n)$ for $\sigma \in S_n$. $\endgroup$
    – MAS
    Dec 29, 2021 at 10:09
  • $\begingroup$ The statement at the end says that there may be relationship between roots (apart from those given by the elementary symmetric functions of roots) and these also need to be preserved by the permutations of roots. $\endgroup$
    – Paramanand Singh
    Dec 30, 2021 at 0:58
  • $\begingroup$ The more the relationships between roots the smaller will be the Galois group. In case when the only relations are the ones given by elementary symmetric polynomials the Galois group is $S_n$. $\endgroup$
    – Paramanand Singh
    Dec 30, 2021 at 1:00

1 Answer 1


Let us take a (not so) simple example with $K=\mathbb {Q} $ and $$f(x) =x^4+x^3+x^2+x+1\in K[x] $$ One can prove that if $a$ is a root of this polynomial then $a^5=1$ and further none of $a, a^2,a^3$ equals $1$. Using these we can prove that $a, a^2,a^3,a^4$ are the four distinct roots of the polynomial.

Thus if we use $r_1,r_2,r_3,r_4$ to denote roots of this polynomial with $r_i=a^i$ then we have the following non-trivial relationships among roots $$r_2=r_1^2,r_3=r_1^3,r_4=r_1^4$$ Now let us try some permutations of $S_4$ and see if these are preserved.

The transposition $(1 2)$ does not work as it would lead to a relation $r_1=r_2^2$ ie $a=a^4$ which is not the case. One can check further that none of the transpositions work.

Let us however observe that $r_3=r_4^2, r_1=r_3^2$ and hence the cycle $\tau=(1243)$ does preserve the relationship among roots. And so do its powers. Thus $\tau, \tau^2,\tau^3,\tau^4=\text{id}$ form a group of permutations which preserve the relationship among roots. One can show that none of other remaining 20 permutations in $S_4$ work as desired.

So the Galois group here is essentially a cyclic group of order $4$.

The idea of preserving relationship among roots is frequently used in computation of small Galois groups.

The relationship among roots is also important for finding the splitting field explicitly. For the example above the relations between roots allow us to conclude that the splitting field is $K(a) $ which is of degree $4$ over $K$ and thus the Galois group must be of order $4$.


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