Tag Archives: primitive element

Algebraically closed and Primitive element theorem

\mathbf{Problem:} Let F be a perfect field and F\subset E an algebraic field extension, such that every non-constant polynomial f(x)\in F[x] has a root in E. Show that E is algebraically closed.

\mathbf{Proof:} Suppose f(x)\in F[x] is a non-constant polynomial. Let K be a splitting field of f.
Since F is perfect, K is a separable extension of F. Primitive element theorem implies that any finite separable extension is simple extension. That is \exists\,\alpha\in K such that K\in F(\alpha). Suppose \alpha has a minimal polynomial g(x) over F. By assumption, g(x) has a root \gamma\in E. Since there exists an isomorphism \eta:F(\alpha)/ F\to F(\gamma)/ F\subset E by \eta(\alpha)=\gamma, then E contains a splitting field of f(x) over F.
\forall\, f\in E[x], adjoin a root \xi of f(x), we get E(\xi)/E. Since E(\xi)/E and E/F are algebraic. E(\xi)/F is also algebraic. Hence, \xi is a root of some non-constant polynomial h(x)\in F[x]. By the previous proof, h(x) splits over E. We conclude that \xi\in E and E(\xi)=E.
\text{Q.E.D}\hfill \square


Simple Extension

\mathbf{Defination:} Let E/F be a finte extension, if there is u\in E such that E=F(u), then we call u is a primitive element of the extension E/F. And this extension is called simple exentsion.

It is very important to know when a extension is a simple extension.The following theorem Emil Artin

\mathbf{Thm:(Primitive\, Element\, Thm)} E/F is a simple extension if and only if there are finitely many intermediate field \displaystyle K between E and F.
We are not interested in the proof of this theorem. But in the consequence of this theorem.

\mathbf{Corollary:} Let E/F be a finite separable extension, then E/F is a simple extension. Especially when \displaystyle char(F) is 0, all finite extension is a simple one.

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.