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$K \leq E$ 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??

user26857
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Mary Star
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    You should say though that the subring contains $K(a)$. You need to start with an arbitrary subring and then show that it contains some finite extension of $K$. So far you have only shown that the subrings which are simple extensions of $K$ are fields, but this does not include every subring of $E$. – Bruce Zheng Oct 08 '14 at 14:08
  • @BruceZheng Ok!!! Thanks a lot!!! :-) – Mary Star Oct 25 '14 at 11:17

1 Answers1

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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 $$1 = c_0^{-1}\left(\sum_{i=1}^n c_i a^i\right) = a \cdot \left(c_0^{-1}\sum_{i=1}^n c_i a^{i-1}\right).$$

Chocosup
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  • I understand!!! Thank you very much!!! So, a necessary condition for this is that a is algebraic, right? That means that if the extension were not algebraic this wouldn`t stand right?? – Mary Star Oct 08 '14 at 14:51
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    Well, $a$ being algebraic is a necessary and sufficient condition for its inverse to lie in $K[a]$. If $a$ is transcendent over $K$ then $K(a) \neq K[a]$ and the latter is not a field.

    In the case of the statement of your problem, if the extension $K \hookrightarrow E$ is not algebraic then there is an element $a \in E$ transcendent over $K$, so $K[a]$ is a subring of $E$ containing $K$ that is not a field.

     – Chocosup Oct 08 '14 at 14:59
  • I understand!! Thanks a lot!!! :-) – Mary Star Oct 25 '14 at 11:16