# Galois Theory and Number of Solutions

Is the Galois group of an irreducible separable polynomial of degree $n$ isomorphic to a group on exactly $n$ letters? Is that enough to prove that a degree $n$ polynomials has $n$ roots.

What I mean by three letters is that you can write the group as permutations on three letters for example S_3 = all permutations and C_3 = just cycles are both on three letters.

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If by "group on exactly $n$ letters", you specifically mean the symmetric group $S_n$, then the answer is no, not in general. For example, see here - the Galois group of an irreducible cubic polynomial over $\mathbb{Q}$ will be $A_3\cong\mathbb{Z}/3\mathbb{Z}$ when its discriminant is the square of a rational number.

Also, the Fundamental Theorem of Algebra is really a consequence of topological properties of $\mathbb{R}$ and/or $\mathbb{C}$ - as is mentioned here, it is often remarked that it is not quite a theorem about algebra.

Reading your comments, I'm guessing perhaps you want "$G$ is a group on $n$ letters" to be synonymous with "$G$ can be viewed as a group of permutations of $n$ objects which acts transitively". If that is the case, then the Galois group of an separable irreducible polynomial of degree $n$ is always a group on $n$ letters, because the Galois group must indeed act transitively on the roots of the polynomial. If it didn't, we could break up the set of roots into the orbits of the action of the Galois group, and this would force the polynomial to factor - for example, if $f=\prod_{i=1}^n(x-\alpha_i)$, then if the Galois group permuted $\alpha_1,\ldots,\alpha_k$ amongst themselves, and permuted $\alpha_{k+1},\ldots,\alpha_n$ amongst themselves, then we could factor $f=gh$ where $g=\prod_{i=1}^k(x-\alpha_i)$ and $h=\prod_{i=k+1}^n(x-\alpha_i)$, and $g$ and $h$ are guaranteed to be polynomials over our base field because they are fixed by the action of the Galois group.

Also note that if a group $G$ acts transitively on a set of $n$ elements, then $|G|\geq n$, which also seems to coincide with what you are wanting the group to satisfy.

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The way I meant it was that both Z/3Z = {1,2,3} = {a,b,c} and S_3 are groups on three letters. –  quanta Apr 18 '11 at 19:20
"because the Galois group must indeed act transitively on the roots of the polynomial" - This is assuming that the polynomial has n roots? i.e. the fundamental theorem of algebra. (I was hoping to prove the polynomial has n roots from the fact that the galois group is transitive on n letters... perhaps that naive idea is not possible to implement?) –  quanta Apr 18 '11 at 21:01
As Qiaochu points out, the fact that a polynomial of degree $n$ has at most $n$ roots is a consequence of the division algorithm. The statement that a polynomial is separable (which is what I assumed in my answer) is precisely that its roots are distinct, and hence there are exactly $n$. –  Zev Chonoles Apr 18 '11 at 21:04
I don't understand how to conclude there are exactly n from that there are at most n and they are all distinct? –  quanta Apr 18 '11 at 21:21
A polynomial $f\in K[x]$of degree $n$ always has $n$ roots, provided you look in a large enough field (i.e. any field in which $f$ splits) and count with multiplicity. For example if $K=\mathbb{F}_p(T)$ and $f=x^p-T$, then $f$ has no roots in $K$; if we look in a splitting field and don't count with multiplicity, there is a single root $\sqrt[p]{T}$; with multiplicity, there are $p$ roots, all of which happen to be $\sqrt[p]{T}$, because $f=x^p-T=(x-\sqrt[p]{T})^p$. When I say "all the roots are distinct", I mean these $n$ roots (which a priori could be the same) are in fact distinct. –  Zev Chonoles Apr 18 '11 at 22:27

You don't need to know anything about Galois theory to prove that a polynomial of degree $n$ over a field $F$ has at most $n$ roots. This is a straightforward consequence of the division algorithm in $F[x]$.

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at most sure, but how to show that it actually has n and no less? –  quanta Apr 18 '11 at 20:56
@quanta: generally speaking, it doesn't. (Take, for example, $F = \mathbb{Q}$.) The property that all polynomials of degree $n$ over a field $F$ have exactly $n$ roots is equivalent to $F$ being algebraically closed, and proving that algebraic closures exist also does not require that you know anything about Galois theory (and is actually a great way to prove many basic theorems in Galois theory, rather than the other way around). Proving that for any $F$ and polynomial $p$ of degree $n$ there exists an extension of $F$ in which $p$ has exactly $n$ roots is even easier: this is just the... –  Qiaochu Yuan Apr 18 '11 at 21:34
...standard construction of the splitting field of $p$, which is obtained by formally adjoining all of the roots of $p$. Once again, you don't need to know any Galois theory to do this, and it is an important step in understanding Galois theory rather than the other way around. –  Qiaochu Yuan Apr 18 '11 at 21:36
In the above discussion "$n$ roots" means "$n$ roots counted with multiplicity." If you want to disregard multiplicity and you want $p$ to be irreducible, then you are talking about separability. $p$ is always separable if the characteristic of $F$ is zero, and in general $p$ is separable if and only if it is relatively prime to its formal derivative $p'$. –  Qiaochu Yuan Apr 18 '11 at 21:38

The Galois group is a permutation group on $n$ letters, i.e., a subgroup of $S_n$. See http://en.wikipedia.org/wiki/Galois_theory#The_permutation_group_approach_to_Galois_theory. BTW, you probably need to assume that the polynomial is separable, i.e., does not repeated roots.

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Good point! I've added this. Although I don't just mean subgroup (but a subgroup with "size" 4), because Z/2Z is a subgroup of S_4 and don't want that from a quartic polynomial. (but Z/4Z and Z/2Z x Z/2Z are okay) –  quanta Apr 18 '11 at 19:34
@quanta, no, the Galois group can be $S_n$. Try for instance $x^3-2$; see en.wikipedia.org/wiki/… –  lhf Apr 18 '11 at 19:40
x^3-2 has galois group S_3 which is three letters so that doesn't contradict. –  quanta Apr 18 '11 at 20:27
However, reading your comments, I'm guessing perhaps you want "$G$ is a group on $n$ letters" to be synonymous with "$G$ can be viewed as a group of permutations of $n$ objects which acts transitively". If that is the case, then the Galois group of an separable irreducible polynomial of degree $n$ is always a group on $n$ letters, because the Galois group must indeed act transitively on the roots of the polynomial. –  Zev Chonoles Apr 18 '11 at 20:47
@Zev, thank you for looking past my ignorant misuse of the words and understanding what I really meant! I will read about this group theory concept and check that is what I was trying to express. Thanks! –  quanta Apr 18 '11 at 20:57

Here is a nice definition of Galois group from Abhyankars book "Lectures on Algebra" -- Let $f$ be an irreducible polynomial of degree $n$ with coefficients in some field $K$ whose roots are $a_1,a_2,..a_n$ then the Galois group $G$ consists of those permutations which preserves the relations between them i.e. G is the set of all those permutations $\sigma$ of the symbols ${1,..,n}$ such that $\phi(a_{\sigma(1)},..,a_{\sigma(n)})=0$ for very one variable polynomial $\phi$ for which $\phi(a_1,..a_n)=0$

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We can even show a canonical isomorphism between $G$ and the galois group defined via automorphisms of splitting fields. –  Dinesh Jun 14 '11 at 17:27