The composite field of adjunctions is an adjunction

Let F be a field and let \alpha_1, \ldots, \alpha_n be algebraic over F. Prove that F(\alpha_1,\ldots,\alpha_n) is the composite of the fields F(\alpha_i).


Recall that the set K of all algebraic elements over F is a field, and we consider finite extensions to be subfields of K. Then the composite of the fields E_1,\ldots,E_n is the inclusion-smallest subfield of K containing all the E_i. (The intersection of subfields is a subfield, so this is just the intersection over the class of subfields containing the E_i.)

Certainly F(\alpha_1,\ldots,\alpha_n) is a subfield containing each of the F(\alpha_i). Now if E is a field containing the F(\alpha_i), then E contains F(\alpha_1,\ldots,\alpha_n) by definition. (F(A) is the smallest subfield containing F and A, for any set A.)

So F(\alpha_1,\ldots,\alpha_n) is indeed the composite of the fields F(\alpha_1),\ldots,F(\alpha_n).

Post a comment or leave a trackback: Trackback URL.

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: