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Math Help - 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 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]"
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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 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.

    "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.
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