# Necessary & Sufficient condition for Algebraic Extension

The question is :

Let $L/K$ be a field extension. Then $L/K$ is algebraic extension IFF every sub-ring of $L$ , containing $K$, is a field.

My solution: Firstly, I must say I am unable to show both way implication. I have shown only ONE way & hope that it is correct.

CLaim:- Let $L/K$ be a field extension. Then if $L/K$ is algebraic extension & $\exists$ a sub-ring $S$ such that: $K \subseteq S \subseteq L$ ; THEN $S$ is a field .

Justification:- Let $s \neq 0 \in S$ . Since: $s \in L$ & $L/K$ is algebraic; so we have some minimal polynomial: $x^{n}+a_{n-1}x^{n-1}+....+a_{0}$ with coefficients in $K$ -which is satisfied by $s$ . By minimality, the coefficient $a_{0}$ must be non-zero; i.e. it has an inverse $a_{0}^{-1}$ in $K$. Then $s(-a_{0}^{-1})(s^{n-1}+a_{n-1}s^{n-2}+....+a_{1}) = 1$ .So, $s^{-1} = (-a_{0}^{-1})(s^{n-1}+a_{n-1}s^{n-2}+....+a_{1})$ .Now, since: each of $a_{i}$ & $s$ $\in S$, thus $s^{-1} \in S$. Consequently, $S$ is a field.

My query:

1) Please check the solution & rectify if necessary.I think it's okay!So, just have a look.

2) What about the converse part?? Please give a detailed solution for that part!

• The converse is the easy statement. What if the extension is not algebraic? Isn’t there then a nonalgebraic element? What sort of ring does it generate? – Lubin Apr 27 '14 at 19:45
• method of contradiction???...wait wait... let me think... !! Actually I was trying to prove it directly!! Is my part of the solution ok??..Btw, what do you mean by "what sort of ring" ?? @ Lubin – user86511 Apr 27 '14 at 19:46
• Your half looked fine to me on a quick scan. – Lubin Apr 27 '14 at 19:48
• okkay... so, if the extension is NOT algebraic, then $\exists$ an element, say $l \in L$ which is NOT algebraic over $K$ . Now, what "sort" of ring it generates??.. this I have to find..right?? – user86511 Apr 27 '14 at 19:59
• So if the element $\alpha$ is nonalgebraic, each polynomial expression $\sum_jc_j\alpha^j$ with coeffs in your smallest field $K$ is different! So you have an isomorphism between $K[X]$ and the smallest ring containing $K$ and $\alpha$. – Lubin Apr 27 '14 at 21:47

Assume every ring between $$K$$ and $$L$$ is a field. Then every element $$a\in L$$ is algebraic: Let $$a\in L$$. There exists a unique ring homomorphism $$\phi\colon K[X]\to L$$ that is the identity on $$K$$ and sends $$X\mapsto a$$. The image of $$\phi$$ is an intermediate ring, hence is a field. As $$K[X]$$ is not a field ($$X$$ is not invertible), $$\phi$$ has nontrivial kernel. If $$0\ne f\in\ker \phi$$ then $$f(a)=0$$.