- #1
mathmari
Gold Member
MHB
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Hey! ![Eek! :eek: :eek:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
$K \leq E$ 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 subfield 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??
$K \leq E$ 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 subfield 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??