## Tag Archives: irreducible

### Normalized root tower and primitive roots of unity

$\mathbf{Problem 1:}$ Let $p$ be a prime unequal to the characteristic of the field $F$. Show that, if $a\in F$, then $x^p-a$ is either irreducible in $F[x]$ or it has a root in $F$.

$\mathbf{Proof:}$ Let $w$ be a p-th primitive root of unity, $K=F(w)$. Since $F$ has characteristic 0, $K$ contains $p$ distinct pth root of unity.

Let $U=\{\text{pth roots of unity contained in } K\}$. $U$ is a multiplicative subgroup of $K$.

Suppose $E$ is a splitting field of $x^p-a$ over $K$ and $r$ is one root of $x^p-a=0$ in $E$. All the roots of $x^p-a$ are  $zr$, $z\in U$. So $E=K(r)$. $\forall\, \xi\in \text{Gal }E/K$, $\xi(r)=zr$, $z\in U$. We can verify that this induces a monomorphism

$\text{Gal }E/K\mapsto U$

$\xi\to z$

Since $U$ is cyclic of order $p$. $G$ which is a subgroup of $U$ can only be the trivial group $\{1\}$ or the whole group $U$.

If $G=\{1\}$, $E=K=F(w)$. And $x^p-a$ splits over $F(w)$. Then it must have a root in $F$.

If $G=U$, then $[K(r):K]=[E:K]=|G|=p$, $x^p-a$ must be irreducible in $K[x]$, hence also in $F[x]$.

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

$\mathbf{Problem 2:}$ Let $E/F$ be the cyclotomic field of the pth roots of unity over the field $F$ of characteristic 0. Show that $E$ can be imbedded in a field $K$ which has a root tower over $F$ such that the integers $n_i$ are primes and $[F_{i+1}:F_i]=n_i$. Call such a root tower normalized.

$\mathbf{Proof:}$ Since  $E$ is the cyclotomic field over $F$ of characteristic 0, $G=\text{Gal }E/F$ is abelian by lemma 1 in page 252.

Then $G$ is a finite solvable group. So $G$ has a composition series

$\displaystyle G=G_1\rhd G_2\rhd \cdots\rhd G_{r+1}=\{1\}$

whose composition factor $G_i/G_{i+1}$ is cyclic of prime order $p_i$. By Galois corresponding theorem, we have an increasing chain of subfields

$F=F_1\subset F_2\subset\cdots\subset F_{r+1}=E$

such that $G_i=\text{Gal }E/F_i$ and $\displaystyle G_i/G_{i+1}=\text{Gal }F_{i+1}/F_i$. So $\displaystyle [F_{i+1}:F_i]=p_i$ is prime. This is a normalized root tower.

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

$\mathbf{Problem3:}$ Obtain normalized root tower over $\mathbb{Q}$ of the cyclotomic fields of 5th and 7th roots of unity.

$\mathbf{Proof:}$ Suppose $w$ is a primitive 5th roots of unity. $[\mathbb{Q}(w):\mathbb{Q}]=4$ and  $G=\text{Gal }\mathbb{Q}(w)/\mathbb{Q}$ is multiplicative group of $\mathbb{Z}_5$, because $\forall\,\zeta\in G$ maps $\{w,w^2,w^3,w^4,w^5=1\}$ to itself. The automorphism $\zeta:w\to w^3$ is a generator of $G=\{\zeta,\zeta^2,\zeta^3,\zeta^4=1\}$.

$G=\langle \zeta\rangle\rhd \langle \zeta^2\rangle\rhd=\{1\}$

the corresponding root tower is

$\mathbb{Q}\subset \mathbb{Q}(w+w^4)\subset \mathbb Q(w)$

It is easy to verify this is a normalized root tower.

For 7th roots of unity, suppose $w$ is a primitive 7th root of unity,

$\mathbb{Q}\subset \mathbb{Q}(w+w^2+w^4)\subset \mathbb Q(w)$

is a normalized tower where $[\mathbb{Q}(w+w^2+w^4):\mathbb{Q}]=2$.

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

$\mathbf{Problem4:}$ Prove that, if $f(x)=0$ has a solvable Galois group over a field $F$ of characteristic 0. Then its splitting field can be imbedded in an extension field which has a normalized tower over $F$.

Suppose splitting field is $E$ and $[E:F]=n$, the proof relies on adjoin nth roots of unity to $E$.

$\mathbf{Remark:}$ Jacobson p256.

### f=x^p-x-a in a field of characteritic p

$\mathbf{Problem:}$ Let $F$ be of characteritic $p$ and let $a\in F$. Show that $f(x)=x^p-x-a$ has no multiple roots and that $f(x)$ is irreducible in $F[x]$ if and only if $a\neq c^p-c$ for any $c\in F$.
$\mathbf{Proof:}$ $f'(x)=-1$, so $(f(x),f'(x))=1$ which means $f(x)$ is separable.
For the second part. If $f$ is irreducible then immediately it has no roots in $F$. Conversely, suppose $c$ is a root of $f$ in some splitting field, $c$ is not in $F$. Then by Freshman’s dream, one can verify $c+1,c+2,\cdots, c+p-1$ are also roots of $f$.

$\displaystyle f(x)=(x-c)(x-c-1)\cdots(x-p+1)$

The irreducible polynomial $g(x)$ of $c$ is a factor of $f(x)$. $g(x)$ must contain all the roots of $f(x)$, otherwise the sum of these roots will results in $c\in F$. So $f(x)=g(x)$ is irreducible.
$\mathbf{Remark:}$ Jacoboson, Algebra I, p234.