# Abstract Nonsense

## Maps of Extensions and the Galois Group

Point of Post: In this post we discuss the morphisms in the “category of all extensions”.

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Motivation

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We have spent a fair amount of time talking about the structures that are field extensions, but we have yet to mention what the “structure preserving maps” between such extensions “should” be. Well, of course, it’s not exactly obvious how to define such maps, but if we stop thinking about an extension $k/F$ as being a field $k$ and a field $F$ with $k\supseteq F$ and instead think about it as an $F$-algebra $k$ which also happens to be a field then we know what to do. Namely, we already know what the morphisms in $F-\textbf{CAlg}$ (the category of commutative $F$-algebras) are. Namely, if $k,k'$ are two $F$-algebras then an $F$-algebra map $f:k\to k'$ is nothing more than a ring map $k\to k'$ which also respects the $F$-vector space structure (i.e. $f(\alpha k)=\alpha f(k)$ for all $\alpha\in F$).

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That said, despite the fact that this is really the reason that we define maps of extensions the way we do, this is rarely said in basic books on the subject. Instead such books like to define a map $f:k\to k'$ of extensions $k/F,k'/F$ to be a ring map such that $f_{\mid F}=\text{id}$.  Of course, it’s easy to see that such definitions are equivalent. If $f$ is a map of extensions as defined in the first paragraph then necessarily $f(\alpha)=\alpha f(1)=\alpha$ for all $\alpha\in F$. Conversely, if $f$ is a map of extensions as just defined then $f(\alpha x)=f(\alpha)f(x)=\alpha f(x)$ for all $\alpha\in F$ and $x\in k$ (since $f(\alpha)=\alpha$ for all $\alpha\in F$ by assumption).

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Being a map of extensions is a pretty strong condition. For example, we shall see that if $k/F$ is a finite extension then every map of extensions $k\to k$ is necessarily invertible. Moreover, in general mathematics one can construct a lot of information from an object by examining its automorphism group. This is (hyperbolically) nowhere more true than in the study of field extensions. We shall see that examining the automorphism group of an extension shall enable us to (more so in certain nice cases) read off certain extension-theoretic properties of our extension–this is the so-called field of Galois theory.

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Maps of Extensions

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Suppose that we have a given field $F$ and we have two extensions $k/F$ and $k'/F$. A map of extensions, or just map when we are being lazy, $k\to k'$ (or, probably, more correctly $k/F\to k'/F$) is an $F$-algebra map. Namely, it’s a ring map $k\to k'$ which is also $F$-linear. Thus, we can define the category of extensions of $F$ as the full subcategory of $F\text{-}\mathbf{CAlg}$ where the objects are themselves fields.  We shall denote the set of all maps of extensions $k/F\to k'/F$ as either $\text{Hom}(k/F,k'/F)$ or more commonly $\text{Hom}_F(k,k')$.

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The existence of a map $k\to k'$ is a very strong condition. Indeed, since such a map is a ring map from a field we know it’s an injection, and since it’s $F$-linear we may then conclude that $\dim_F k\leqslant\dim_F k'$.  For example, there does not exist a map $\mathbb{Q}(\sqrt[3]{2})\to\mathbb{Q}(\sqrt{2})$ as extensions of $\mathbb{Q}$ since $\dim_\mathbb{Q}\mathbb{Q}(\sqrt[3]{2})=3>\dim_\mathbb{Q}\mathbb{Q}(\sqrt{2})$ [in fact, there doesn’t exist a ring map such any such ring map will necessarily preserve $\mathbb{Q}$].

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Another piece of evidence that maps of extensions are well-behaved beasts comes from looking at endomorphisms (maps from an extension to itself) of finite extensions. Namely, suppose that $k/F$ is finite and that we have a map $f:k\to k$. From the above discussion we know that $f$ is injective and so $\dim_F f(k)=\dim_F k$ and since $\dim_F k$ is finite we know from basic vector space theory that $f(k)=k$. Thus, we see that every endomorphism of the extension $k/F$ is necessarily an automorphism (a bijective endomorphism).

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Although it makes total sense to denote the group of automorphisms of the extension $k/F$ as $\text{Aut}(k/F)$ (as is done in some books, such as Dummit and Foote) for historical reasons we call this group the Galois group of the extension $k/F$ and denote it $\text{Gal}(k/F)$. We, of course, say group since the composition and inverse of any invertible ring/$F$-linear map is a ring/$F$-linear map and the identity map is an automorphism. Thus, $\text{Gal}(k/F)$ is a group under composition.

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We begin our discussion of maps of extensions by noting two obvious properties:

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Theorem: Suppose that $k,k'$ are two extensions of $F$ with $k=F(X)$ for some set $X\subseteq k$. Then, any map $k\to k'$ is determined by its value on $X$ (i.e. if two maps agree on $X$ they are necessarily equal).

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and

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Theorem: Let $k,k'$ be two extensions of $F$ and $\sigma:k\to k'$ a map of extensions. Then, for any algebraic $\alpha\in k$ one has that $\sigma(\alpha)$ is a root of $m_{\alpha,F}$

Proof: The only obvious thing to note is that $m_{\alpha,F}(\alpha)=0$ and so $\sigma(m_{\alpha,F}(\alpha))=0$. But, distributing $\sigma$ (using the fact that it’s both a ring map and that it fixes elements of $F$, in particular the coefficients of $m_{\alpha,F}$) we find that $\sigma(m_{\alpha,F}(\alpha))=m_{\alpha,F}(\sigma(\alpha))$ from where the conclusion follows. $\blacksquare$

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With this we can compute a couple examples of Galois groups.

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For example, we claim that $\text{Gal}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})$ is trivial. Indeed, if $\sigma:\mathbb{Q}(\sqrt[3]{2})\to\mathbb{Q}(\sqrt[3]{2})$ is any map of extensions then $\sigma(\sqrt[3]{2})$ is necessarily sent to another root of $m_{\sqrt[3]{2},\mathbb{Q}}=x^3-2$ in $\mathbb{Q}(\sqrt[3]{2})$. That said, besides $\sqrt[3]{2}$ no other root of $x^3-2$ is real, and thus not in $\mathbb{Q}(\sqrt[3]{2})$. Thus, we must necessarily have that $\sigma(\sqrt[3]{2})=\sqrt[3]{2}=\text{id}(\sqrt[3]{2})$. Since $\sigma$ equals $\text{id}$ on a generating set of $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ we may conclude that they are equal, and thus the assertion of the triviality of the Galois group follows.

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This previous shows that the Galois group can pick up on the failure of an extension to be “whole” in the sense that it contains some but NOT ALL roots of some rational polynomials.

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We next show that $\text{Gal}(\mathbb{R}/\mathbb{Q})$ is likewise trivial. Of course, since any automorphism of $\mathbb{R}$ as a ring necessarily preserves $\mathbb{Q}$ we shall actually be proving that $\mathbb{R}$ has trivial automorphism group as a ring. Not surprisingly (since $\mathbb{R}$ is a fundamentally topological object) we need some topology to prove this. The basic idea is that since any automorphism $\sigma:\mathbb{R}\to\mathbb{R}$ necessarily maps squares to squares we have that it maps positive real numbers to positive real numbers–i.e. we have that if $x>0$ then $\sigma(x)>0$. Now, if $x>y$ then $x-y>0$ so $\sigma(x)-\sigma(y)=\sigma(x-y)>0$ and so $\sigma(x)>\sigma(y)$. Thus, we see that $\sigma$ is order preserving. We now claim that $\sigma$ is continuous. Indeed, if $q$ is any positive rational number we have that $-q implies that $\sigma(-q)<\sigma(x)-\sigma(y_0)<\sigma(q)$ but since $\sigma$ fixes $\mathbb{Q}$ this says that $-q<\sigma(x)-\sigma(y_0). Thus, for any $x\in\mathbb{R}$ we see that $\sigma(B_q(x))\subseteq B_q(\sigma(x))$ for any $q\in\mathbb{Q}^+$ which clearly proves continuity. We are basically done then. Since $latex\ mathbb{Q}$ is dense in $\mathbb{R}$ (and $\mathbb{R}$ is Hausdorff) and $\sigma$ fixes $\mathbb{Q}$ we must necessarily have that $\sigma=\text{id}$.

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As a last example (for now) we claim that $\text{Gal}(\mathbb{C}/\mathbb{R})=\{\text{id},\sigma\}$ where $\sigma$ is the complex conjugation map. Clearly both indicated maps are automorphisms, and so it suffices to prove the reverse inclusion. Let $\tau$ be any automorphism of $\mathbb{C}/\mathbb{R}$. We know that $\mathbb{C}=\mathbb{R}(i)$ and so $\tau$‘s action on $\mathbb{C}$ is determined by its action on $i$. But, since $\tau(i)$ is a root of $m_{i,\mathbb{R}}=x^2+1$ we have that $\tau(i)=\pm i$. If $\tau(i)=i$ then necessarily the identity map and if $\tau(i)=-i$ then $\tau$ is necessarily complex conjugation.

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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

April 24, 2012 -

## 3 Comments »

1. […] Maps of Extensions and the Galois Group (Pt. II) Point of Post: This is a continuation of this post. […]

Pingback by Maps of Extensions and the Galois Group (Pt. II) « Abstract Nonsense | April 24, 2012 | Reply

2. […] extensions are created equal. Obviously this could be interpreted as saying not all extensions are isomorphic, but this is too strong. Instead of going head-long at the isomorphism problem for extensions […]

Pingback by Normal Extensions « Abstract Nonsense | April 30, 2012 | Reply

3. […] one where there are multiple roots which become indistinguishable (inseparable, if you will) to the Galois group of the extension. For example, consider the finite field and the rational function field . Then, […]

Pingback by Separable Extensions (Pt. I) « Abstract Nonsense | May 4, 2012 | Reply