$K \leq E$ an algebraic extension. I am asked to show that each subring of $E$ that contains $K$ is a field.
I have done the following:
$K \leq E$ algebraic $\Rightarrow \forall a \in E, \exists f(x)\in K[x]:f(a)=0$
A subring of $E$ that contains $K$ is $K[a], \forall a \in E$.
Since $a$ is algebraic over $K$,it stands that $K(a)=K[a]$.
Therefore, a subring of $E$ that contains $K$ is $K(a),\forall a\in E$, where $K(a)$ is a field since it is a ring and $1\in K(a)$ so $1=f(a)g^{-1}(a)\Rightarrow f^{-1}(a)=g^{-1}(a)\in K(a)$.
Is this correct??
Or is there something I could improve??
This is not totally correct. Actually, you're using a primitive element $a$ to generate the subring $L$, but this is only possible if the extension $K \hookrightarrow L$ is finite (which is stronger than just algebraic).
That doesn't affect the main idea of the proof much, as all you actually have to do is check that for any $a \in L$, $a$ has a multiplicative inverse $a^{-1}$ that also lies in the ring $L$. But this is an immediate consequence of $a$ being algebraic over $K$.
Indeed, if $P_a = \sum_{0\leq i\leq n} c_i X^i$ is the minimal polynomial of $a$ over $K$, then its constant coefficient $c_0$ is nonzero (by irreducibility) and hence $$1 = c_0^{-1}\left(\sum_{i=1}^n c_i a^i\right) = a \cdot \left(c_0^{-1}\sum_{i=1}^n c_i a^{i-1}\right).$$