# Steps towards the Galois group of permutations of the roots of a polynomial

I'm having a hard time to understand Galois theory and its motivation from a beginner's perspective.

(1) What I take for granted is the fundamental theorem of algebra, thus a polynomial of degree $$n$$

$$a_nx^n + a_{n-1}x^{n-1}+\dots + a_1x + a_0$$

has $$n$$ possibly complex roots $$x_i$$.

(2) I also take Vieta' formulas for granted, i.e.

$$e_1(x_1,\dots,x_n):=x_1 + x_2 + \dots + x_n = - \frac{a_{n-1}}{a_n}$$

$$\dots$$

$$e_n(x_1,\dots,x_n):=x_1 \cdot x_2 \cdot \dots \cdot x_n = (-1)^n \frac{a_0}{a_n}$$

with $$e_i(x_1,\dots,x_n)$$ being the elementary symmetric polynomials.

(3) What is obvious: When $$a_i \in \mathbb{Q}$$ then also $$e_i(x_1,\dots,x_n) \in \mathbb{Q}$$, i.e. the complexities of the roots $$x_i$$ "cancel out" in these symmetric expressions.

(4) Next thing to be taken for granted: the fundamental theorem of symmetric polynomials: Every symmetric polynomial $$S(X_1,\dots,X_n)$$ with coefficients in $$K$$ has a unique representation

$$S(X_1,\dots,X_n) = Q(e_1(X_1,\dots,X_n),\dots,e_n(X_1,\dots,X_n))$$

for some other polynomial $$Q$$ with coefficients in $$K$$.

(5) What then is obvious again: When $$a_i \in \mathbb{Q}$$ then also $$S(x_1,\dots,x_n) \in \mathbb{Q}$$ for every symmetric polynomial $$S$$.

All one can say up to this point is:

For all polynomials $$P(X_1,\dots,X_n)$$ with rational coefficients, all symmetric polynomials $$S(X_1,\dots,X_n)$$ and all permutations $$\pi$$ of the roots $$x_i$$ of $$P$$ we have $$S(\pi(x_1),\dots,\pi(x_n)) \in \mathbb{Q}$$.

My question is:

Which observation, idea, or argument gives the next step towards the Galois group of permutations of the roots of a polynomial?

This group in general is not the full symmetric group $$S_n$$, but only a subgroup of it.

It could be defined by those permutations $$\pi$$ with

$$S(x_1,\dots,x_n) = S(\pi(x_1),\dots,\pi(x_n))$$

for every symmetric polynomial $$S$$, which would be a much stronger condition.

But why would one consider exactly this condition and these permutations? Is it possibly the only natural stronger condition than $$S(\pi(x_1),\dots,\pi(x_n)) \in \mathbb{Q}$$?

And is it the correct definition of the Galois group of a polynomial?

• It sounds like you are reading some historical sources that describe how Galois theory was originally done in the 19th century. I would recommend instead reading some more modern expositions (e.g., the exposition in Lang's Algebra is excellent) which use the language of abstract algebra to simplify many of the concepts. – Eric Wofsey Oct 16 '18 at 15:57
• @EricWofsey: Yes, you are right, but I do deliberately so. – Hans-Peter Stricker Oct 17 '18 at 7:56

The condition $$S(x_1,\dots,x_n) = S(\pi(x_1),\dots,\pi(x_n))$$ actually holds for all permutations. That's what it means for $$S$$ to be symmetric: it stays the same under any permutation of the variables at all.
The idea behind the definition of the Galois group is that it consists of permutations that preserve any "relations" that may exist between the roots. What is a "relation" between the roots? It's a polynomial $$Q(X_1,\dots,X_n)$$ with coefficients in $$\mathbb{Q}$$ such that $$Q(x_1,\dots,x_n)=0.$$ So the Galois group can be defined as the set of all permutations $$\pi$$ such that for any such $$Q$$, $$Q(\pi(x_1),\dots,\pi(x_n))=0.$$ Equivalently, the Galois group can be defined as the group of automorphisms of the field $$\mathbb{Q}(x_1,x_2,\dots,x_n)$$ generated by the roots of your polynomial. Any such automorphism must permute the roots in some way, and in order for a permutation of the roots to extend to an automorphism of the field, it needs to preserve all relations between the roots.
• This means: Because $x_1 + 1 = 0$ and $x_2 - 1 = 0$ for the two roots $x_1, x_2$ of $x^2 - 1 = 0$ the Galois group of this polynomial consists only of the identity permutation? While for $x^2 - 2 = 0$ it also contains the swap permutation? Does it hold in general that if the polynomial can be factored into polynomials with rational coefficients the Galois group is trivial? (It seems so.) – Hans-Peter Stricker Oct 17 '18 at 11:26
• The latter has to do with the fact, that the extensions $\mathbb{Q}(x_i)$ are not proper extensions, i.e. $\mathbb{Q}(x_i)= \mathbb{Q}$? – Hans-Peter Stricker Oct 17 '18 at 11:34
• To sum it up: $P(x)$ is reducible iff its Galois group is trivial. So the only "interesting" polynomials are the irreducible ones. Because a Galois group can be non-trivial in many ways. – Hans-Peter Stricker Oct 17 '18 at 14:46
• Er, that's not right. $P(x)$ splits (factors completely) over $\mathbb{Q}$ iff the Galois group is trivial. It might factor but not factor all the way to linear polynomials, in which case the Galois group is still nontrivial – Eric Wofsey Oct 17 '18 at 14:49