Splitting field of irreducible polynomialby joeblow Tags: field, irreducible, polynomial, splitting 

#1
Feb2312, 02:52 PM

P: 71

I need to find the splitting field in [itex] \mathbb {C} [/itex] of [itex] x^3+3x^2+3x4 [/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. 



#2
Feb2312, 02:59 PM

Mentor
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




#3
Feb2312, 03:09 PM

P: 71

You're darn right it is. Why'd I forget that?




#4
Feb2312, 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]).




#5
Feb2312, 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] 


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