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Thread: Application of "Tower Theorem"

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    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 $\displaystyle F \subset E \subset K$. 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 $\displaystyle Z_2[x]$"
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    Quote Originally Posted by Amanda1990 View Post
    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 $\displaystyle F \subset E \subset K$. I can't see how this relates to the last part of the question - how do we go about it?
    Say that $\displaystyle F\subseteq E\subseteq K$ then $\displaystyle [K:F] = [K:E][E:K]$. However, $\displaystyle [K:F]$ is prime which forces $\displaystyle [K:E]=1 \text{ or }[E:K]=1$ therefore $\displaystyle E=K\text{ or }E=F$. Hence, there is no $\displaystyle F\subset E\subset K$.

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