# Do all polynomials of degree n with indeterminate coefficients have Galois groups that are isomorphic to Sn?

I just finished reading "A Book of Abstract Algebra" by Charles C. Pinter and, as someone who is studying this independently, I was having some understanding issues and many questions.

1) Does every polynomial of the form $a_0+a_1x+a_2x^2+...+a_nx^n$ have a Galois group that is isomorphic to $S_n$, and therefore there are no radical expressions for the roots of polynomials of degree $>4$ that have nonzero coefficients?? Does the polynomial being irreducible change the answer to the previous question?

2) The book only shows one example of a polynomial not being solvable by radicals, a quintic whose Galois group is isomorphic to $S_5$. Does every Galois group of a polynomial with complex roots have a transposition? Are the permutations of $\pm\sqrt2$ considered a transposition as well?

3) Finally, I've seen this question, the querstion, asked in some form but I've never seen it asked and answered fully. How, given a polynomial of degree n, can I determine the Galois group and therefore determine if it is solvable by radicals? From what I've read/seen, it seems to be a logical trial and error but I haven't seen any methods of doing so. Is there a method, besides trying to reason through it? If there isn't, are there any readings/exercises I can do to check if polynomials are solvable by radicals or if groups are solvable? I realize I must get more comfortable with the common group types. Any input would be great.

I apologize if this question is an amateur question, but I've only been studying this book for a couple months. I'm going to move on to Basic Algebra I to solidify my knowledge and understanding, and eventually go on to Abstract Algebra by Dummit and Foote (which I hear is the "holy grail" in this subject), but I'm open to any suggestions you might have/what you did that helped you learn Algebra.

1) No, even if you just consider irreducible polynomials. But if you pick a random polynomial, the chance is very very high that the group is $S_n$, so other cases can be hard to find. There are however a few cases where you can guarantee from the shape of the polynomial that the group can not be $S_n$ (for example $a+b*x^3+x^6$). There is a very hard (unsolved in general, research level) question that asks whether you can get every group over the rationals.

2) No. For example if $\zeta$ is a root of $\frac{x^{11}-1}{x-1}=1+x+\cdots+x^{10}$, and you take the minimal polynomial of $\zeta+\bar\zeta$, the Galois group is cyclic of order 5, so contains no elements of order 2.

3) One can (though most algorithms determine only the type, and not the concrete action on the roots). It is not horribly hard on a computer for small degrees, but you probably only very rarely would want to do it by hand. (The book Galois Theory by David Cox gives a nice description of what you can do.) The basic idea is to show that some expressions in the roots are rational, so must be preserved by the Galois group. For example, in degree 4, if preserves $\alpha_1\alpha_3+\alpha_2\alpha_4$ is rational, the Galois group must be a subgroup of the dihedral group. Concretely: $f(x)=x^4-3$ is such a polynomial. Its roots are $\alpha_1=\sqrt[4]{3}$, $\alpha_2=i\sqrt[4]{3}$,$\alpha_3=-\sqrt[4]{3}$, and $\alpha_4=-i\sqrt[4]{3}$. And $\alpha_1\alpha_3+\alpha_2\alpha_4=0$ is rational. So its Galois group must be in $D_8$, in fact it is this group. But if we take instead $f(x)=x^4+x-3$ the roots are (float approx. as it is horribly messy otherwise): $-.94-1.1i, -.94+1.1i, .94-.78i, .94+.78i$ and $\alpha_1\alpha_3+\alpha_2\alpha_4=-3.42165...$ is not rational, nor is it for any other labelling of the roots. So the Galois group does not lie in $D_8$. The expression $\alpha_1\alpha_3+\alpha_2\alpha_4$ comes from having the permutation groups act on polynomials by permutation of variables and searching for polynomials that are fixed.

However, while you might be able to solve by radicals as a consequence of such a test, the solutions are typically so messy to be practically useless.

• The minimal polynomial of $\zeta^2$ in part (2) is the same cyclotomic polynomial you gave. The field corresponding to the subgroup of order $5$ you mention is the maximal real subfield of $\mathbb{Q}(\zeta)$, on which complex conjugation acts trivially. Jan 14 '15 at 4:00
• @ahulpke: For 1), are you referring to the Inverse Galois Problem? Just for interest's sake. 3) I'm a little confused. What is that preserved expression actually? Like where is it coming from? And are you saying some polynomials are solvable by radicals but considered basically not solvable because of their messiness? Thanks for clearing up some things though Jan 14 '15 at 4:09
• @micky22: 1): Yes, exactly. For 3) I've added more of an example. They are solvable, but the expression you get is practically useless (unless it is so easy that you can deduce it without Galois theory as in $x^5-3$) I have never seen anyone using the solution by radicals in practice for degrees $>4$. Jan 14 '15 at 16:11
• @anomaly: Thanks -- fixed. Jan 14 '15 at 16:16
• @ahulpke: But that polynomial has only real roots, and so complex conjugation $\sigma$ acts trivially. Since $\sigma^2 = 1$, the map $\sigma$ must have order (exactly) 2 iff it's nontrivial--- that is, not all of the roots of the (real) polynomial are real. Jan 14 '15 at 16:20

1) No; for example, take the cyclotomic polynomial $X^{p-1} + \cdots + X + 1\in \mathbb{Q}[X]$ for some large prime $p$, whose splitting field is an abelian extension over $\mathbb{Q}$. The point of the unsolvability result is that isn't something like the quadratic formula (or the much less memorable versions in degree $3$ or $4$) in higher degrees; the splitting field of a polynomial $f(X) = a_n X^n + \cdots + a_1 X + a_0$ for fixed $a_i\in \mathbb{Q}$ isn't even a solvable extension of $\mathbb{Q}$ in general.

2) Complex conjugation $\sigma$ is a permutation of the roots of the polynomial $f$ (assuming the coefficients of $f$ are real), and acts nontrivially unless all the roots of $f$ are real. It has order $2$, but it's in general not a transposition. As for the map $\alpha:\sqrt{2} \to -\sqrt{2}$, you'd have to first show that operation is even well defined on the splitting field $k/\mathbb{Q}$ of $f$. For $\sigma$, you get it for free; $k$ embeds in $\overline{\mathbb{Q}} \subset \mathbb{C}$, and $\sigma$ is well-defined on $\mathbb{C}$. How do you extend $\alpha$ across $\overline{\mathbb{Q}}$, or at least $k$?

3) There are some ad hoc techniques (most notably, just exhausting the various possibilities in low degrees), but I don't think there's any convenient algorithm for doing it by hand. There's are some long, explicit calculations for the degree $3$ and $4$ cases in Hungerford's "Algebra", but they're not very illuminating.