Busy. Please wait.
or

Forgot Password?

Don't have an account?  Sign up
or

taken

why

Make sure to remember your password. If you forget it there is no way for StudyStack to send you a reset link. You would need to create a new account.

By signing up, I agree to StudyStack's Terms of Service and Privacy Policy.

Already a StudyStack user? Log In

Reset Password
Enter the associated with your account, and we'll email you a link to reset your password.

Remove ads
Don't know
Know
remaining cards
Save
0:01

 Flashcards Matching Hangman Crossword Type In Quiz Test StudyStack Study Table Bug Match Hungry Bug Unscramble Chopped Targets

Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.

Normal Size     Small Size show me how

# AA2 Chap 20

### Field Extensions

QuestionAnswer
Extension Field A field E is an EXTENSION FIELD of a field F, if F (subset of) E and the operations of F are the same as the operations of E.
Examples of Extension Fields R is an extension of Q C is an extension of R
Fundamental Theorem of Field Theory Let F be a field and f(x) is a non-constant polynomial in F[x], there there exists an extension field E in which f(x) has a root
Fundamental Theorem of Field Theory Example f(x)=x^2-2 in Q. f has no roots in Q but has a root in R
~field of quotients theorem If R is an integral domain, then it is contained in it's field of quotients. EX. {a/b|a,b in Z, b~=0}
Def: Split Let f(x) be in F[x]. We say f(x) splits in an extension field E of F if we can write f(x)=(x-a1)(x-a2)...(x-an) where ai is in E not necessarily distinct. i.e. f can be factored into a product of linear factors in E[x].
Def: Splitting field E is called a SPLITTING FIELD for f(x) over f, if f(x) splits in E[x] but has no proper subfield of E. The splitting field for f(x) is the smallest extension of F in which f(x) splits.
Theorem about splitting fields Splitting fields always exist and are unique up to isomorphism
Irreducible Theorem Let F be a field and p(x) in F[x] is irreducible over F. If a is a root of p(x) in some extension E of F, then F(a) is isomorphic to F[x]/<p(x)>.
Created by: csebald