# Abstract Nonsense

## Algebraic Extensions (Pt. III)

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

The above illustrates an important fact:

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Theorem: Let $k/F$ be a finite extension. Then, every element of $k$ is algebraic over $F$.

Proof: By the same argument as above, if $[k:F]=n$ then $1,\cdots,\alpha^n$ are $n+1$ distinct elements of an $n$-dimensional space and so must be $F$-independence, and so there exists $a_0,\cdots,a_n\in F$ such that $a_0+\cdots+a_n\alpha^n=0$ or $\alpha$ satisfies $f(t)=a_0+\cdots+a_nt^n\in F[t]$. $\blacksquare$

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What we’d like to prove is that there is a partial converse to the above, which will roughly say that a finite extension of a given field is precisely a ‘small’ extension by algebraic elements, with whatever ‘small’ will mean. Why partial? Clearly not all algebraic extensions are finite since, for example, $\mathbb{C}/\mathbb{Q}$ is algebraic and certainly $[\mathbb{C}:\mathbb{Q}]$ is not finite (in fact, it’s $\aleph_0$)–but of course the difficulty here is that whatever ‘small’ will mean, we must have that $\mathbb{C}/\mathbb{Q}$ is ‘large’.

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To properly phrase this converse we need to discuss the notion of finitely generated algebras. Let $R$ be a commutative ring and $A$ an $R$-algebra. We say that $A$ is finitely generated if there exists $\alpha_1,\cdots,\alpha_n\in A$ such that $A=R(\alpha_1,\cdots,\alpha_n)$. If $A=R(\alpha)$ for some $\alpha\in A$ we call $A$ simple and call $\alpha$ a primitive element or just primitive.

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Of course, if $k/F$ is an extension we call it a finitely generated extension if $k$ is a finitely generated $F$-algebra and we similarly define a simple extension.

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This notion of ‘finitely generated’ is precisely the notion of “small” that we are looking for. To begin to show why then (substituting ‘finitely generated’ in for ‘small’ in our preceding paragraphs) finite extensions are precisely finitely generated algebraic extensions let’s show that the one way is true–let’s show that all finite extensions are algebraic and finitely generated. The first of these has already been done and since evidently $\text{span}_F(\alpha_1,\cdots,\alpha_n)\subseteq F[\alpha_1,\cdots,\alpha_n]\subseteq F(\alpha_1,\cdots,\alpha_n)$ for any $\alpha_1,\cdots,\alpha_n$ in some extension $k$ of $F$ we can clearly conclude that $[k:F]<\infty$ implies that $k$ is finitely generated by taking the generating set to be the basis for $k$ as an $F$-vector space. Thus, we may definitively conclude that

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Theorem: Let $k/F$ be a finite extension then $k/F$ is algebraic and finitely generated.

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So, how exactly should we go about concluding the converse? The idea is simple enough. Basically we are starting with a field $F$ sitting inside some extension $k$, and we know that if we take some algebraic $\alpha_1\in K$ then $F(\alpha_1)$ has finite dimension over $F$. Ok, so what happens when we add in another $\alpha_2$? Well, there is nothing a priori that we can say about the dimension of $F(\alpha_1,\alpha_2)$ since we only know things about appending roots to $F$ itself. The key in thinking (and this is an important idea, so keep it in your back pocket) is that instead of thinking about $F(\alpha_1,\alpha_2)$ as appending both $\alpha_1$ and $\alpha_2$ to $F$ let’s think about appending $\alpha_2$ to $F(\alpha_1)$ and now think about everything in sight as an extension of $F(\alpha_1)$. Why is this the right way to go? There are two reasons for why not only is this a good idea, but a natural one. First, all of our theorems up until this point have concerned us appending a single element to a given field. Second, thanks to our product formula we know that if we can cut the extension $F(\alpha_1,\cdots,\alpha_n)$ into a tower where each piece is finite then the top field is actually finite over the bottom one, and since analyzing a tower piece-by-piece should be easier, it should always be our first instinct.

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Ok, so because of the above we have decided that the first natural thing to do to decide whether or not $F(\alpha_1,\cdots,\alpha_n)/F$ is finite is to examine the tower

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$\begin{array}{c}F(\alpha_1,\cdots,\alpha_n)\\ \vert\\ F(\alpha_1,\cdots,\alpha_{n-1})\\ \vert\\ \vdots\\ \vert\\ F(\alpha_1,\alpha_1)\\ \vert\\ F(\alpha_1)\\ \vert\\ F\end{array}$

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and decide whether field is finite as an extension over the field immediately below it.  But this is easy. Why? Well, we know that $F(\alpha_1,\cdots,\alpha_m)=F(\alpha_1,\cdots,\alpha_{m-1})(\alpha_m)$ and so if we know that $\alpha_m$ is algebraic in $k/F(\alpha_1,\cdots,\alpha_{m-1})$ we’re golden. That said, it’s trivially algebraic since we know that $m_{\alpha_m,F}(t)\in F[t]\subseteq F(\alpha_1,\cdots,\alpha_{m-1})[t]$ and $m_{\alpha_m,F}(\alpha_m)=0$. In fact, we know that $[F(\alpha_1,\cdots,\alpha_m):F(\alpha_1,\cdots,\alpha_{m-1})]$ is $\deg m_{\alpha_m,F(\alpha_1,\cdots,\alpha_{m-1})}$ but since $m_{\alpha,F}$ annihilates $\alpha_m$ we have that $m_{\alpha_m,F(\alpha_1,\cdots,\alpha_{m-1})}\mid m_{\alpha,F}$ so that, in fact, $[F(\alpha_1,\cdots,\alpha_m):F(\alpha_1,\cdots,\alpha_{m-1})]\leqslant \deg m_{\alpha_m,F}=[F(\alpha_m):F]$. Thus we have proven the following:

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Theorem: Let $k/F$ be an extension and $\alpha_1,\cdots,\alpha_n$ algebraic in $k/F$. Then,

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$\displaystyle \left[F(\alpha_1,\cdots,\alpha_n):F\right]\leqslant\prod_{j=1}^{n}[F(\alpha_j):F]$

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Thus, we may definitively conclude the following:

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Theorem: Let $k/F$ be an extension. Then, $k/F$ is finite if and only if it is algebraic and finitely generated.

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While this theorem and the theorems leading up to it may have seemed to be mostly a digression from algebraic extensions/elements to a tangent about finite extensions it actually has a very important use. Namely, suppose that we have some extension $k/F$. Up until this point we have been looking at individual members of the set of all algebraic elements in $k/F$ instead of the conglomerate of all such elements. In other words, if we let  $L$ be $\left\{\alpha\in k:\alpha\text{ is algebraic over}\;F\right\}$, called the algebraic closure of $F$ in $k$, then we have been looking at individual elements of $L$ instead of $L$ itself. That said, it seems fairly natural to want to know things about $L$ itself. Since $L\supseteq F$ (obviously) probably the most natural of these natural questions is to whether $L$ is actually a subextension of $k/F$, or said differently, is $L$ a field?

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As is (unfortunately) often the case, this most natural of natural questions is not very obvious. Think about what might be the obvious route to proving this. We’d take some $\alpha\in L$ and some $\beta\in L$–good start–then we’d say that there exists $p,q\in F[t]$ such that $p(\alpha)=q(\beta)=0$–right, good, good– and then we just…. Just what? There is no obvious way to take the polynomials $p$ and $q$ and use them to make some new polynomial $P$ such that $P(\alpha+\beta)=0$. For example, the minimal polynomials of $\sqrt{2},\sqrt[5]{5}$, and $\sqrt[6]{2}$ over $\mathbb{Q}$ are $t^2-2$, $t^5-5$, and $t^6-2$ respectively. That said, one can verify (via, say, wolframalpha) that the minimal polynomial of $\sqrt{2}+\sqrt[6]{2}$ is $x^6-6x^2+18$ and the minimal polynomial of $\sqrt{2}+\sqrt[5]{5}$ is

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$x^{12}-12x^{10}+60x^8-170x^6-60x^4-792x^2+9$

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So, in general, it seems pretty hopeless to easily be able to prove that the sum of two algebraic elements is algebraic via the direct method of finding a polynomial which annihilates.  But, using our above machinery we can indeed prove that $L$ is a field very easily. Indeed, we want to prove that $\alpha+\beta\in L$, right? Well, we know from our previous theorem that $F(\alpha,\beta)$ is algebraic and so every element is algebraic including $\alpha+\beta$, $\alpha\beta$, and $\alpha^{-1}$. Thus, putting it all together we have the following:

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Theorem: Let $k/F$ be an extension. Then, the algebraic closure of $F$ in $k$ is a subextension.

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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.p

March 25, 2012 -