Thread: subfield

Let K be a field an let L be a finite field extension of K.
Let K be a subset of R and R a subset of L, so that for all a,b in R a+b, a*b in R.
Show: R is a subfield of L.

I still have to show the existence of inverse elements of the multiplication. The rest is all clear. But I have absolutely no real idea...

Maybe a proof by contradiction???
Can I somehow use that the field extension is algebraic?

Originally Posted by Icarus0

Let K be a field an let L be a finite field extension of K.
Let K be a subset of R and R a subset of L, so that for all a,b in R a+b, a*b in R.
Show: R is a subfield of L.

I still have to show the existence of inverse elements of the multiplication. The rest is all clear. But I have absolutely no real idea...

Maybe a proof by contradiction???
Can I somehow use that the field extension is algebraic?

every element of R is algebraic over K. now look at the minimal polynomial of a non-zero element of R over K.

Originally Posted by NonCommAlg

every element of R is algebraic over K. now look at the minimal polynomial of a non-zero element of R over K.

What has this to do with multiplicatice inverse elements? The minimal polynomial of a in R is just a irreducible polynomial f in K[X] with f(a)=0 ????

Originally Posted by Icarus0

What has this to do with multiplicatice inverse elements? The minimal polynomial of a in R is just a irreducible polynomial f in K[X] with f(a)=0 ????

let be the minimal polynomial of then and now let

it's clear that and