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 $E(\xi)=E$.
$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

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 $[\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5>4=[\mathbb{Q}(\omega):\mathbb{Q}]$.
$\text{Q.E.D.}\hfill\square$
$\mathbf{Problem:}$ 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.