Cubic and Quartic Equations

Cubic Equations

Galois Theory of Cubic Equations

Let be a splitting field of an irreducible cubic polynomial over a field , let be the discriminant of , and be the Galois group of . Then the following statements hold:

• If is a square in , then and is the alternating group .
• If is not a square in , then and is the symmetric group .

Proof Let be an irreducible cubic polynomial over with splitting field . In , we have . Since , we can conclude that , and there is a chain of field extensions: Let , then in , the polynomial can be written as for some quadratic polynomial , which factors as in . There are only two cases: if is irreducible in , then , otherwise, and . As is irreducible over , acts transitively on the roots , so must be either or . Consider the square root of the discriminant, say . If is square, , it is then fixed by . Note that can only be fixed by even permutations, so , and . Otherwise, if is not square, , then must be , and .

Indeed, the above statement can be generalized:

Theorem

Let be a splitting field over of an irreducible polynomial of degree in , and let be the discriminant of . Then the Galois group is a subgroup of the alternating group if and only if is a square in .

Quartic Equations

For quartic equations, we immediately have the following:

Corollary

Let be the Galois group of an irreducible quartic polynomial . The discriminant of is a square in if and only if contains no odd permutation. Therefore,

• If is a square in , then is either the alternating group or the Klein four-group .
• Otherwise, is , , or .

Lagrange found another useful expression in the roots, one that is special to quartic polynomials:

Resolvent Cubic

Let be an irreducible quartic polynomial over . are four roots of . Define then the polynomial is called the resolvent cubic of .

Remark

Given a quartic polynomial , a more useful expression of the resolvent polynomial is . Notice that the resolvent polynomial is independent of the cubic coefficient . This is because our construction of , and cannot produce .

Proposition

Let be an irreducible quartic polynomial over . Then the discriminant of equals to the discriminant of its solvent polynomial

Proof Note that , so by symmetry, the discriminant of the solvent polynomial is exactly the discriminant of .

Proposition

Let be the Galois group of an irreducible quartic polynomial over , and let be the resolvent cubic of . Then is irreducible if and only if the order of is divisible by . Specifically,

• If splits completely in , then .
• If has one root in , then or .
• If is irreducible, then or .

Proof First note that the three elements , and are distinct. In fact, since is irreducible quartic polynomial, its four roots cannot be multiple roots (otherwise the it shares a root with its derivative, contradicting the irreducibility). Thus and similarly for and . Let . The operation in on the roots of induces an action on . If splits completely in , then is fixed by , so . If has one root in , say . Let then is a normal subgroup of by the theorem. So has to be or . If is irreducible, then , we can conclude that is either or . The converse is also true because can only be irreducible if is either or , as we’ve already showed all other cases above.

To summarize, we can use the following table to classify the Galois group of a quartic polynomial based on the the resolvent cubic and the discriminant of the resolvent cubic:

	is square	is not square
is reducible		or
is irreducible
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