Let $E/K$ be a field extension. Show the following equivalence:
(i) E/K is algebraic
(ii) Every Intermediate Ring of $E/K$ is a Field.
Here is what I tried:
(i)$\Rightarrow$(ii) Let $E/K$ be algebraic. We need to show that for two Elements $a,b \in R$ the multiplicative Inverse also is in $R$. Let $R$ be a Intermediate Ring of $E/K$. Then there are is Set $A$ with $K(A)=R$. Let $a,b\in A$ with $a\neq0$. Let's consider $K(a,b)\subseteq R$. Since $a,b$ are algebraic elements, the extension $K(a,b)/K$ is finite $\Rightarrow$ all elements of $K(a,b)$ are algebaric over K, so $a+b, a-b, ab, 1/a \in K(a,b)\subseteq R$.
(ii)$\Rightarrow$(i)
Let $R$ be a Intermediate Ring of $E/K$ and $a\in E$. We know that $K[a]$ is a field, so $a^{-1}\in K[a]$. Now $K[a]:= \{\sum\limits_{i=1}^n x_ia^{i_1} : x_i\in K, i_j\in \mathbb{N}\}$, so there is a $\sum\limits_{i=1}^n y_ia^{i_1} = a^{-1} \Leftrightarrow \sum\limits_{i=1}^n y_ia^{i_1+1} = 1 \Leftrightarrow \sum\limits_{i=1}^n y_ia^{i_1+1} – 1 = 0 \Rightarrow$ a is algebraic.
Is this proof ok?
Best Answer
The second part of the proof is correct, though you don't need to consider an arbitrary intermediate ring $R$, only the specific $K[a]$ for a given nonzero element $a\in E$.
However, in the first part when you write $R=K(A)$, (at least the notation seems to suggest that) you implicitly assumed the conslusion, that $R$ is a field.
We only need that $R\subseteq E$ and so every nonzero element $a\in R$ is algebraic over $K$, implying $a^{-1}\in K[a]\subseteq R$.