Tag Archives: splitting field

Splitting field of f(r)=r^3+r^2-2r-1 over Q

\mathbf{Problem:} Let E=\mathbb{Q}(r) where r^3+r^2-2r-1=0. Verify that r'=r^2-2 is also a root of x^3+x^2-2x-1=0. Determined \text{Gal }E/\mathbb{Q}. Show that E is normal over \mathbb{Q}.
\mathbf{Proof:} r^3+r^2-2r-1=0 means (r+1)(r^2-2)=-1. So we only need to prove \displaystyle \frac{-1}{r+1} is another root of f(x)=x^3+x^2-2x-1=0.
Let y=r+1, f(r)=0 means y^3-2y^2-y+1=0.
Let \displaystyle z=\frac{-1}{y}, then \displaystyle\frac {1}{z^3}(z^3+z^2-2z-1)=0, obviously z\neq 0, thus \displaystyle z=\frac{-1}{r+1} is a root of f(x)=0.
f(x) is monic and has no root in \mathbb{Z}, so it is irreducible in \mathbb{Q}. f(x) is minimal polynomial of r. So [E:\mathbb{Q}]=3. \text{Gal } E/\mathbb{Q} must be the cyclic group of order 3. By theorem 4.7 chapter 4 in Jacoboson’s book, E is normal over \mathbb{Q}.
\mathbf{Remark:} Jacobson, Algebra I. p243.

Splitting field of x^5-2 over rational field and its primitive element

\mathbf{Problem:} Consturct a splitting field over \mathbb{Q} of x^5-2. Find its dimensionality over \mathbb{Q}.
\mathbf{Proof:} Let \omega is the primitive 5th root of unity in \mathbb{C}, whose minimal polynomial is x^4+x^3+x^2+x+1.
\displaystyle x^5-2 has roots \sqrt[5]{2}, \sqrt[5]{2}\omega, \sqrt[5]{2}\omega^2,\sqrt[5]{2}\omega^3,\sqrt[5]{2}\omega^4 in \mathbb{C}. Then a splitting field of x^5-2 is \mathbb{Q}(\sqrt[5]{2},\omega). And [\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}]=20.
To see this, \displaystyle [\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5, because x^5-2 is irreducible by Eisenstein criterion. Furthermore f(x)=x^4+x^3+x^2+x+1 is irreducible in \mathbb{Q}(\sqrt[5]{2}). Suppose not, since \omega\not \in \mathbb{Q}(\sqrt[5]{2}), f must split into two quadratic polynomial, namely f(x)=(x^2-(\omega+\omega^4)x+1)(x^2-(\omega^2+\omega^3)x+1). So \displaystyle \omega+\omega^4\in \mathbb{Q}(\sqrt[5]{2}). While \omega+\omega^4 satisfies x^2-5x+2=0, which has roots \displaystyle\frac{5+\sqrt{17}}{2} and \displaystyle\frac{5-\sqrt{17}}{2}. So \sqrt{17}\in \mathbb{Q}(\sqrt[5]{2}). This is impossible because [\mathbb{Q}(\sqrt{17}):\mathbb{Q}]=2 does not divide 5.
We can consider this problem in another way. [\mathbb{Q}(\omega):\mathbb{Q}]=5, x^5-2 is irreducible in \mathbb{Q}(\omega). Otherwise \mathbb{Q}(\sqrt[5]{2}) can be imbedded in \mathbb{Q}(\omega). This is impossible because [\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5>4=[\mathbb{Q}(\omega):\mathbb{Q}].
\mathbf{Problem:} Find a primitive element of a splitting field of x^5-2 over \mathbb{Q}.
\mathbf{Proof:} The Galois group of x^5-2 over \mathbb{Q} are permutations of roots. Every permutation \eta is uniquely determined by \eta(\sqrt{2}) and \eta(\omega). Let \alpha=\sqrt{2}+\omega, the only \eta fixes \alpha is \eta=id. So \mathbb{Q}(\alpha) is the splitting field of x^5-2 and \alpha is a primitive element.

Dimensionality of a splitting field of a polynomial of degree n

\mathbf{Problem:} Show that the dimensionality of a splitting field E/F of f(x) of degree n is at most n!.
\mathbf{Proof:} Assume n\geq 1(otherwise E=F). There exists an extension K of F containing a root \alpha of f(x).Consider the extension F(\alpha)\subset K of F, then the minimal polynomial of \alpha in F(\alpha) must divide f(x), which has degree n, then [F(\alpha):F]\leq n. Suppose f(x)=(x-\alpha)g(x), with g(x)\in F(\alpha)[x], deg(g)=n-1. By induction, [E:F(\alpha)]\leq(n-1)!, so

\displaystyle [E:F]=[E:F(\alpha)][F(\alpha):F]\leq n!