# Abstract Nonsense

## 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] 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