## Cubic polynomial

There was a question about a classic exercise: Let be a cubic irreducible polynomial, which we take to be monic, and let be the roots. Let and . Clearly, , but the exercise is about the degree of .

Fact: is 1 or 2.

Proof: We have , so that we can divide by in . Therefore, and are roots of the quadratic polynomial . Since , we have . Thus, if , then . If , then .

By the tower theorem corresponding to these two possibilities, we have that is 3 or 6.

Fact: if and only if .

Proof: Note that , since it is a symmetric polynomial in . Furthermore, . If , then and this degree must divide . So we have . On the other hand, suppose . The polynomial has the roots , and . So has roots and . Therefore,

. But we are assuming that . So . Since as well, we have and , so that .

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