Splitting field of irreducible polynomial

by joeblow
Tags: field, irreducible, polynomial, splitting
joeblow is offline
Feb23-12, 02:52 PM
P: 71
I need to find the splitting field in [itex] \mathbb {C} [/itex] of [itex] x^3+3x^2+3x-4 [/itex] (over [itex] \mathbb{Q} [/itex]).

Now, I plugged this into a CAS and found that it is (probably) not solvable by radicals. I know that if I can find a map from this irreducible polynomial to another irreducible polynomial of the same degree, I can find an expression for the roots of the original polynomial by using the roots of the second (if they are solvable, that is). I cannot find such a polynomial, though.

By the conjugate root theorem in conjunction with FTA, we can have (1) one real root and two complex roots which must be conjugates of each other or (2) three real roots. In the case of (2), if one root is a radical expression, then its conjugate must also be a root. Thus, the third root cannot be a radical expression. Similarly for (1), the real root cannot be a radical expression.

Thus, I have no idea what a splitting field for this polynomial would be. Any ideas of a useful way of describing this splitting field? Thanks.
Phys.Org News Partner Science news on Phys.org
Going nuts? Turkey looks to pistachios to heat new eco-city
Space-tested fluid flow concept advances infectious disease diagnoses
SpaceX launches supplies to space station (Update)
micromass is offline
Feb23-12, 02:59 PM
micromass's Avatar
P: 16,565
Every third degree polynomial is solvable by radicals. The roots can be found by a general formula: http://en.wikipedia.org/wiki/Cubic_function
joeblow is offline
Feb23-12, 03:09 PM
P: 71
You're darn right it is. Why'd I forget that?

Norwegian is offline
Feb23-12, 04:10 PM
P: 144

Splitting field of irreducible polynomial

The roots of this polynomial, you can find immediately, by just a minor rewriting, then it becomes obvious what the spliting field is. It has degree 6 over Q btw, and can be described in the form Q(ω,[itex]\sqrt[3]{b}[/itex]).
Deveno is offline
Feb23-12, 08:15 PM
Sci Advisor
P: 906
to amplify Norwegien's answer, try "completing the cube".

if we can obtain all roots of f by adjoining a real cube root and a primitive complex cube root of unity, then to find the degree of the splitting field of f, it suffices to find the degree of the minimal polynomials for the (real) cube root, and the primitive cube root of unity.

that is:

[tex][\mathbb{Q}(\sqrt[3]{b},\omega):\mathbb{Q}] = [\mathbb{Q}(\sqrt[3]{b},\omega):\mathbb{Q}(\sqrt[3]{b})][\mathbb{Q}(\sqrt[3]{b}):\mathbb{Q}][/tex]

Register to reply

Related Discussions
Irreducible polynomial over finite field Calculus & Beyond Homework 12
Splitting field of a polynomial over a finite field Calculus & Beyond Homework 2
Irreducible polynomial in extension field Calculus & Beyond Homework 6
Determining the irreducible polynomial for field sqrt{10} Calculus & Beyond Homework 0
Irreducible polynomial on polynomial ring Calculus & Beyond Homework 1