Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Deveno

Math Help - Subrings of Algebraic Extentions

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    29

    Subrings of Algebraic Extentions

    Let L:K be an algebraic extension. Suppose that R is a subring of Lthat contains K. Show that R is a field. Give an example to show that the algebraic restriction is necessary.

    Take an  r\in R, r \neq 0 and r \notin K. As R is a subring only need to show that r has an inverse also in R to show R is a field. As r is algebraic over K, there exists a minimal poly, p say which is irreducible over K. So p(r)= a_{n}r^n+...+ar_{1}+a_{0} = 0 with a_{i}\in K.

    As p is irreducible a_{0}\neq 0 otherwise p(r)= r(a_{n}r^{n-1}+...+a_{1})=0 but as r \neq 0 \Rightarrow a_{n}r^{n-1}+...+a_{1}=0 so p would not be the minimal polynomial.

    Now p(r) \Rightarrow 1= r( -a_{0}^{-1}) (a_{n}r^{n-1}+...+a_{1}) and so -a_{0}^{-1} (a_{n}r^{n-1}+...+a_{1}) \in R is the inverse of r thus R is a field.

    I think this is all OK and correct but an (annoying) friend of mine was saying oh but you haven't defined how many degrees p has. He's annoying as thats all he said and refused to tell me how to work it out, surely as nothing is known about degrees of K you can't really know?

    Also I'm struggling to think of an example...

    I'm assuming I'm missing something simple but if anyone could help with these two wee bits or show I've gone wrong somewhere else I'd appreciate it!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,397
    Thanks
    760

    Re: Subrings of Algebraic Extentions

    your proof seems fine to me. one caveat, there is nothing saying L might not EQUAL K, so you need a one-line dispatch for that. the degree of L over K isn't relevant, and without it given, you can't say much about the degree of the minimal polynomial of r.

    as for a counter-example, consider \mathbb{Q}[\pi], which is isomorphic to \mathbb{Q}[x]. this is certainly a subring of \mathbb{R}, which contains \mathbb{Q}. however, it is NOT a field, because (for one), it doesn't contain \frac{1}{\pi}
    Thanks from leshields
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    29

    Re: Subrings of Algebraic Extentions

    Thanks, that's great!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Subrings
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: February 17th 2011, 06:14 PM
  2. Rings and subrings
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 7th 2010, 12:30 PM
  3. Field extentions...
    Posted in the LaTeX Help Forum
    Replies: 1
    Last Post: May 13th 2009, 06:21 AM
  4. subrings
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: December 8th 2008, 04:12 PM
  5. Algebra....subrings
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 28th 2006, 05:23 PM

Search Tags


/mathhelpforum @mathhelpforum