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

### 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

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

$\text{Q.E.D.}\hfill\square$