Galois group of a cubic polynomial
Proof. , so any automorphism of fixing is determined by the image of each . But must take each to some (where possibly ), since is a homomorphism of and thus . Thus permutes the roots of and is determined by the resulting permutation.
We now restrict our attention to the case . If is a cubic, its Galois group is a subgroup of . We can then use the knowledge of the group structure of to anticipate the possible Galois groups of a cubic polynomial. There are six subgroups of , and the three subgroups of order are conjugate. This leaves four essentially different subgroups of : the trivial group, the group that consists of a single transposition, the group , and the full group . All four of these groups can in fact appear as the Galois group of a cubic.
Let be a splitting field of over .
If factors into a linear and an irreducible quadratic term, then , where is the discriminant of the quadratic. Hence and the order of is , so ; the nontrivial element of the Galois group takes each root of the quadratic to its complex conjugate (i.e. it maps ). Thus any cubic that has exactly one rational root will have Galois group isomorphic to .
The Galois groups and arise when considering irreducible cubics. Let be irreducible with roots . Since is irreducible, the roots are distinct. Thus has at least elements, since the image of may be any of the three roots. Since , it follows that or and thus by the fundamental theorem of Galois theory that or .
Now, the discriminant of is
Thus can be written as a polynomial in the coefficients of , so . since is irreducible and therefore has distinct roots; also clearly and thus . If , then its discriminant is (see the article on the discriminant for a longer discussion).
If , it follows that has degree over , so that . Hence , so we can derive the splitting field for by adjoining any root of and the square root of the discriminant. This can happen for either positive or negative , clearly. Note in particular that if , then is imaginary and thus is not a real field, so that has one real and two imaginary roots. So any cubic with only one real root has Galois group .
If , then any element of must fix . But a transposition of two roots does not fix - for example, the map
Then does not include transpositions and so it must in this case be isomorphic to . Thus , so since . This proves:
Let be an irreducible cubic and its splitting field. Then if is any root of ,
where is the discriminant of . Thus if is rational, and the Galois group is isomorphic to , otherwise and the Galois group is isomorphic to .
Thus if is an irreducible cubic with Galois group , it must have three real roots. However, the converse does not hold (see Example 4 below).
of looking at the above analysis is that for a “general” polynomial of degree , the Galois group is . If the Galois group of some polynomial is not , there must be algebraic relations among the roots that restrict the available set of permutations. In the case of a cubic whose discriminant is a rational square, this relation is that , which is a polynomial in the roots, must be preserved.
. By the rational root test, this polynomial has the three rational roots , so it factors as over . Its Galois group is therefore trivial.
. Again by the rational root test, this polynomial factors as , so its Galois group has two elements, and a splitting field for is derived by adjoining the square root of the discriminant of the quadratic: . The nontrivial element of the Galois group maps .
- Example 3
. This is irreducible since it is Eisenstein at (or by the rational root test), and its discriminant is , which is not a rational square. Thus the Galois group for this polynomial is also ; note, however, that has three real roots (since but ).
. This is also irreducible by the rational root test. Its discriminant is , which is a rational square, so the Galois group for this polynomial is . Explicitly, the roots of are
and we see that
Let’s consider an automorphism of sending to , i.e. sending . Given the relations above, it is clear that this mapping uniquely determines the image of as well, since
|Title||Galois group of a cubic polynomial|
|Date of creation||2013-03-22 17:40:33|
|Last modified on||2013-03-22 17:40:33|
|Last modified by||rm50 (10146)|