Talk:Galois theory

The Galois group of a polynomial - a basic example

By reasoning similar to the above, it can be shown that the Galois group of the fourth-degree polynomial $x^{4}-5$ - again with the coefficients viewed as elements of Q - is isomorphic to the dihedral group of the square. This group has 8 elements as illustrated in fig. 1, and a normal subgroup structure as illustrated in fig. 2.

Also, there are exactly 6 intermediate fields between the smallest field L containing all the roots and Q itself containing some of the roots, making 8 all together, as illustrated in fig. 3.

The Galois correspondence is illustrated in fig. 4.

The Galois group of a polynomial - a trivial example

As an example, let us look at the second-degree polynomial $x^{2}-5$, with the coefficients {-5,0,1} viewed as elements of Q.

This polynomial has no roots in Q. However, from the fundamental theorem of algebra we know that it has exacly two roots in C, and can be written as the product of two first-degree polynomials there - i.e. $x^{2}-5=(x-r_{0})(x-r_{1}),r_{0},r_{1}\in C$. From direct inspection of the polynomial we also realize that $r_{0}=-r_{1}$.

L = $\lbrace a+br_{0},a,b\in Q\rbrace$ is the smallest subfield of C that contains Q and both $r_{0}$ and $-r_{0}$.

The are exactly 2 automorphisms of L that leave every element of Q alone: the do-nothing automorphism $\phi _{0}:a+br_{0}\rightarrow a+br_{0}$ and the map $\phi _{1}:a+br_{0}\rightarrow a-br_{0}$.

Under composition of automorphisms, these two automorphisms together form a group isomorphic to $S_{2}$, the group of permutations of two objects.

The sought for Galois group is therefore $S_{2}$, which has no nontrivial subgroups.

It can be shown that in this case the Galois correspondence exists, so we may conclude from the subgroup structure of $S_{2}$ that there is no intermediate field extension containing Q and also roots of the polynomial.

Basic summary of Galois theory

The core idea behind Galois theory is that given a polynomial $\alpha$ with coefficients in a field K (typically the rational numbers), there exists

• a "splitting field" for $\alpha$ over K. This is just a field containing K (or a field isomorphic to K) as a subfield and also all the roots of $\alpha$.
• a smallest possible field L that contains K (or a field isomorphic to K) as a subfield and also all the roots of $\alpha$. This field is known as the extension of K by the roots of $\alpha$. It is a basic theorem in Galois theory that for any field K and any polynomial with coefficients in K, such a field exists.

• fields containing K (or a field isomorphic to K) as a subfield and

• a group containing all automorphisms in L that leave the elements in K untouched - the Galois group of the polynomial $\alpha$.

Providing certain technicalities are fullfilled, the structure of this group contains information about the nature of the roots, and whether the equation $\alpha =0$ has solutions expressible as a finite formula involving only ordinary arithmetical operations (addition, subtraction, multiplication, division and rational powers) on the coefficients.

The following is just a scratch to work out the 1st non-stub version of the article

Text from below here is not in the article yet

However, we may create an extension field L containing two elements $r_{0},r_{1}$ such that $x^{2}-5=(x-r_{0})(x-r_{1})$. By the fundamental theorem of algebra this is always possible - there exists a subfield L of C such that $K\subset L\subseteq C$

As an example, the second-degree polynomial $x^{2}-5$ - when the coefficients {0,1,5} are viewed as elements of Q - turns out to have the Galois group $S_{2}$.

From the subgroup structure of $S_{2}$ - the only proper subgroup is the trivial group $S_{1}$ - we may conclude that the chain of extension fields from Q to the smallet extension field of Q such that the polynomial splits is trivial - no intermediate extension fields exist.

Finding the Galois group of a polynomial is in general a tedious process, in this example it was easy, since the group had to be contained in $S_{2}$.

Looking again at the polynomial $x^{2}-5$, one may wonder exactly what it's "Galois group" is, and how to find it.

... Mention something about the Fundamental theorem of algebra, which implies that there is a subfield in C such that $x^{2}-5$ can be split into linear factors $(x-r_{1})(x-r_{2}),r_{1},r_{2}\subset C$ ...

...Mention $S_{n}$, where n is the degree of the polynomial ...

Basic concepts/glossary

• Polynomial over a field K: An expression of the form $a_{n-1}x^{n-1}+...+a_{1}x^{1}+a_{0}$, with $a_{0},a_{1},...a_{n-1}\in K$.
• Root of a polynomial $\alpha$: a number r such that $\alpha (r)=0$
• A splitting field for a polynomial $\alpha$: A field which contains the original field K as a subfield, and also contains all the roots of $\alpha$.

Summary of the theory

Given a polynomial $\alpha$ with coefficients in some field K, it may be the case that the equation $\alpha =0$ has no solutions in K. In that case, $\alpha$ is said to be irreducible in K.

Anyway, if K is a subfield of C, we are guaranteed by the fundamental theorem of algebra that there exists a subfield of C containing K and all the roots.

...blabber about field of characteristic <> 0 ...

Field extensions

Any field K can be "extended" by including one or more "foreign" elements, f.i. the field Q can be extended by including sqr(2). The resulting field is the subset of R described by a+b sqrt(2), a,b in Q.

Similarly, if r1, r2, ... rn are roots of a polynomial α , a lattice of extension fields may be constructed. ...

Algebraic extension vs transcendental...

The order of an extension ...

Normal extensions and splitting fields ...

Given a polynomial $\alpha$ with coefficients in a field K, there exists a field M ⊇ K - known as a splitting field of $\alpha$ - which contains all the roots of $\alpha$.

The Galois correspondence

The correspondence between the Galois group subgroup structure and the field extension lattice ...

Caveat - separability - only relevant with non-zero characteristic fields.

Soluble groups ... Why neither the quintic nor its friend S5 are "soluble". Why 60 degree angles won't let themselves be "trisected". Why this was a triumph for Galois theory, 2000+ year old riddles solved.

How much to rely on an extra "Field extensions" article?

Ragnar Schroder 05:38, 12 December 2007 (CST)