Abstract Nonsense

Crushing one theorem at a time

Field Extensions (Pt. II)


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

\text{ }

Let’s now discuss a way of generating (pun intended, as we shall see) new extensions of a given field if we are given an extension. In particular, let k/F be some extension and X\subseteq k. Then, by F(X) we mean the intersection of all subfields of k containing X\cup F. We can say this differently by defining a subextension of k/F to be a field k\supseteq L\supseteq F. We see then F(X) is the intersection of all subextensions of k/F containing X. When X=\{s_1,\cdots,s_n\} we will denote F(X) by F(s_1,\cdots,s_n). For reasons that shall soon become clear it will also behoove us to consider the ring F[X] defined to be the intersection of all subrings of k containing both F and X.

\text{ }

It’s not hard to prove that the notation for \mathbb{Q}(\sqrt{2}) was not misleading as \mathbb{Q}(\sqrt{2}) really is the intersection of all subextensions of \mathbb{C}/\mathbb{Q} (or \mathbb{R}/\mathbb{Q}) containing \sqrt{2}. This presentation for \mathbb{Q}(\sqrt{2}) (as linear \mathbb{Q}-polynomials in \sqrt{2}) is not an accident. We have the following theorem:

\text{ }

Theorem: Let k/F be an extension and X\subseteq F. Then, if T=\{t_x:x\in X\} denotes a set of indeterminates indexed by X then the image of the natural map F[T]\to k (where F[T] denotes the polynomial ring) is F[X]. Moreover, F(X) is the field of fractions of F[X] in k.

Proof: Since polynomial rings are the free objects in F\text{-}\mathbf{Alg} the set map f:T\to X\subseteq k extends to a F-algebra map \varphi:F[T]\to k such that \varphi_{\mid T}=f and which is universal with respect to this property. Now,  let R be any subring of k that contains X and let S=\{t_r:r\in R\} be a formal set of variables. We see then that the set map g:S\to R extends to an F-algebra map \psi:F[S]\to R such that \psi_{\mid S}=g. Note then that (\psi_{\mid F[T]})_{\mid X}=g_{\mid X}=f so that by the universal property of \varphi we have that  \psi_{\mid F[T]}=\varphi and thus we see that R contains \text{im }\varphi. Thus, \text{im }\varphi is a subring of k containing X and minimal with respect to this property, and thus \text{im }\varphi=F[X].

\text{ }

Now, clearly the field of fractions \text{Frac}(F[X]) of F[X] is a subextension of k/F containing X. Moreover, suppose that L is another subextension containing X. Since L is a subring of k containing X\cup F we know that L\supseteq F[X] and thus L=\text{Frac}(L)\supseteq\text{Frac}(F[x]). Thus, \text{Frac}(F[X]) is a subextension of k/F containing X and is minimal with respect to this property, and so \text{Frac}(F[X])=F(X). \blacksquare

\text{ }

Ok, so after writing the above proof I realize it was mostly an exercise in me trying to sound smart. But, that’s how things are proven–right? Regardless, what did it really say? First off, what is the image of the “natural map” F[T]\to k? It’s just F-polynomials in the variables X. In other words, to get F[X] we take every single possible polynomial we can find in F[T]=\{f(t_{x_1},\cdots,t_{x_n}):x_1,\cdots,x_n\in X\} and evaluate them at the appropriate values of X (e.g. f(x_1,\cdots,x_n)\in k). So, if X=\{x_1,\cdots,x_n\} this looks like \{f(x_1,\cdots,x_n):f\in F[t_1,\cdots,t_n]\}. That’s not so bad, right? What then does \text{Frac}(F[X])=F(X) look like? Well, since F[X] is already sitting inside a field we just have to literally take products of elements of F[X] and their inverses, which, in the finite case X=\{x_1,\cdots,x_n\}  looks something like

\text{ }

\displaystyle F(x_1,\cdots,x_n)=\left\{\frac{f(x_1,\cdots,x_n)}{g(x_1,\cdots,x_n)}:f,g\in F[t_1,\cdots,t_n]\text{ and }g(x_1,\cdots,x_n)\ne 0\right\}

\text{ }

\text{ }

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.pdf&gt;.

Advertisements

March 7, 2012 - Posted by | Algebra, Field Theory | ,

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: