1. ## 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...

Can I somehow use that the field extension is algebraic?

2. 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...

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.

3. 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 ????

4. 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 $f(x)=x^n + c_1x^{n-1} + \cdots + c_{n-1}x + c_n \in K[x]$ be the minimal polynomial of $0 \neq a \in R.$ then $c_n \neq 0$ and $a^n + c_1a^{n-1} + \cdots + c_{n-1}a + c_n=0.$ now let $b=-c_n^{-1}(a^{n-1} + c_1a^{n-2} + \cdots + c_{n-1}).$

it's clear that $b \in R$ and $ab=1.$