# Application of "Tower Theorem"

• Apr 14th 2009, 08:14 AM
Amanda1990
Application of "Tower Theorem"
I can deduce from the "Tower theorem" (for extensions of fields) that if [K:F] is prime then there is no field E such that $F \subset E \subset K$. I can't see how this relates to the last part of the question - how do we go about it?

"Assuming that fields of order 32 exist, show that there are exactly 6 irreducible polynomials of degree 5 in the ring $Z_2[x]$"
• Apr 14th 2009, 10:36 AM
ThePerfectHacker
Quote:

Originally Posted by Amanda1990
I can deduce from the "Tower theorem" (for extensions of fields) that if [K:F] is prime then there is no field E such that $F \subset E \subset K$. I can't see how this relates to the last part of the question - how do we go about it?

Say that $F\subseteq E\subseteq K$ then $[K:F] = [K:E][E:K]$. However, $[K:F]$ is prime which forces $[K:E]=1 \text{ or }[E:K]=1$ therefore $E=K\text{ or }E=F$. Hence, there is no $F\subset E\subset K$.

Quote:

"Assuming that fields of order 32 exist, show that there are exactly 6 irreducible polynomials of degree 5 in the ring $Z_2[x]$"
Let $K$ be a field of order $32=2^5$, then $K$ must contain a subfield $F$ which is isomorphic to $\mathbb{Z}_2$. Notice that $[K:F]=5$. If $\alpha \in K - F$ then $F \subset F(\alpha) \subseteq K$, but since $5$ is prime it follows by above that $K = F(\alpha)$. Therefore, if $f(x)$ is the mimimal polynomial for $\alpha$ then $f(x)$ must be an irreducible monic polynomial of degree $5$ in $F[x]$. There are, $32 - 2 = 30$ elements in $K-F$ and for each one the mimimal polynomial has $5$ roots in $K$, therefore it means there has to be $\tfrac{30}{5} = 6$ elements in $K-F$ that give rise to each different polynomial of degree $5$.