Abstract Nonsense

Crushing one theorem at a time

Maps of Extensions and the Galois Group (Pt. II)


Point of Post: This is a continuation of this post.

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The above illustrates a general principle concerning simple extensions:

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Theorem: Let F be any field. Then, algebraic \alpha one has that \text{Gal}(F(\alpha)/F) is precisely the automorphisms \sigma:F(\alpha)\to F(\alpha) sending \alpha to \beta where \beta is another root of m_{\alpha,F} in F(\alpha). In particular, |\text{Gal}(F(\alpha)/F)| is the number of roots of m_{\alpha,F} in F(\alpha).

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The fact that all of the described mappings are the only possible mappings is clear, and the fact that there really do exist automorphisms with the desired \alpha as image we merely define \sigma_\beta:F(\alpha)\to F(\alpha) by mapping f(\alpha)\mapsto f(\beta) for any f\in F[x] (this is well-defined because m_{\alpha,F}(\beta)=0–then \sigma is an automorphism with the desired properties.

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This automatically could have told us why \text{Gal}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}) is trivial.

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An interesting corollary of the above is that for simple extensions one has that |\text{Gal}(F(\alpha)/F)|\leqslant \deg m_{\alpha,F}=[F(\alpha):F].  This actually turns out to be true in more generality, but we need to develop an important lemma first.

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Let M be a monoid and k a field. A character is a monoid homomorphism \chi:M\to k^\times. Of course we have a k-vector space k^M of all mappings M\to k. Then, we claim that if \{\chi_1,\cdots,\chi_n\} is any distinct set of characters, they are necessarily linearly independent in k^M. More precisely:

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Theorem(Dedekind’s Lemma): Let \chi_1,\cdots,\chi_n:M\to k^\times be characters. Then, if \alpha_1\chi_1+\cdots+\alpha_n\chi_n=0 (as functions) then \alpha_1=\cdots=\alpha_n=0

Proof: Suppose not, then we may assume without loss of generality that no smaller subset of \{\chi_1,\cdots,\chi_n\} is linearly dependent. Now, there necessarily exists \alpha_1,\cdots,\alpha_n\in k such that \alpha_1\chi_1+\cdots+\alpha_n\chi_n=0 as functions. But, since \chi_1\ne\chi_2 there exists m\in M such that \chi_1(m)\ne\chi_2(m). But, we clearly have that

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\chi_1(m)\alpha_1\chi_1(x)+\cdots+\chi_1(m)\alpha_n\chi_n(x)=0

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for all x\in M. Now, we must also have that

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\chi_1(m)\alpha_1\chi_1(x)+\cdots+\chi_n(m)\alpha_n\chi_n(x)=\alpha_1\chi_1(mx)+\cdots+\alpha_n\chi_n(mx)=0

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for all x\in M. Subtracting these two expressions gives

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\chi_2(m)\alpha_1\chi_2(x)+\cdots+\chi_n(m)\alpha_n\chi_n(x)=0

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for all x\in M which contradicts that \{\chi_2,\cdots,\chi_n\}\subsetneq\{\chi_1,\cdots,\chi_n\} is linearly independent. \blacksquare

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With this we can prove the fundamental result:

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Theorem: Let k/F be finite. Then, |\text{Gal}(k/F)|\leqslant[k:F].

Proof: The idea is simple. Suppose that \{\sigma_1,\cdots,\sigma_m\} are the elements of \text{Gal}(k/F) (we clearly know that \text{Gal}(k/F) is finite, since k/F is finite and so looks like F(\alpha_1,\cdots,\alpha_n) and \text{Gal}(k/F) acts faithfully on the union of the roots of m_{\alpha_1,F},\cdots,m_{\alpha_n,F}). Suppose then that [k:F]<m with basis x_1,\cdots,x_n, then necessarily there exists \alpha_1,\cdots,\alpha_m such that \displaystyle \sum \alpha_i\sigma_i(x_j) for all j. This clearly contradicts Dedekind’s lemma with M=k^\times. \blacksquare

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References:

[1] Morandi, Patrick. Field and Galois Theory. New York: Springer, 1996. Print.

[2] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[3] Lang, Serge. Algebra. New York: Springer, 2002. Print.

[4] Conrad, Keith. Collected Notes on Field and Galois Theory. Web. <http://www.math.uconn.edu/~kconrad/blurbs/&gt;.

[5] Clark, Pete. Field Theory. Web. <http://math.uga.edu/~pete/FieldTheory.print

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April 24, 2012 - Posted by | Algebra, Field Theory | , , , , , ,

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