suppose that $E \subset D \subset F$,Where $E,F$ are fiels,and the $D$ is a Domain,how to show $D$ is a field ? I am trying to show $\forall d \in D$,$d^{-1}\in D$,but I have no idea how to show it
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What is your definition of domain? – asdf Dec 19 '17 at 12:02
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integral domain – Bruce Dec 19 '17 at 12:03
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$F$ must be finite field extension of $E$. Check your question? – 1ENİGMA1 Dec 19 '17 at 12:04
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Yes ,sorry for my mistake, $F$ is algebraic extension of $E$ – Bruce Dec 19 '17 at 12:08
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1By the way, the request of $D$ being a domain is redundant, because it follows automatically from it being a subring of a field. – Dec 19 '17 at 12:17
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@Bruce, please fix your question. Also make it clear that $F/E$ can be an infinite extension. – lhf Dec 19 '17 at 12:25
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Possible duplicate of https://math.stackexchange.com/questions/2002147/show-that-if-f-subset-d-subset-e-then-d-is-a-field – lhf Dec 19 '17 at 12:37
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As stated, the claim is false: $ \mathbb Q \subset \mathbb Q[x] \subset \mathbb Q(x) $ is a counterexample.
The claim is true if $F/E$ is finite. In this case, $D$ is a finite dimensional vector space over $E$ and so the map $D \to D$ given by $x \mapsto ax$ for $a \in D$ is surjective because it is injective when $a\ne0$.
More generally, the claim is true if $F/E$ is algebraic but not necessarily finite. In this case, take $a \in D$. Then $E[a]$ is a finite dimensional vector space over $E$ and the argument above still works.
lhf
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What is the difference between $\Bbb{Q}[x]$ and $\Bbb{Q}(x)$? – gen-ℤ ready to perish Dec 19 '17 at 12:35
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I assumed that $F$ is a finite extension of $E$.
Because $F$ is a finite extension of $E$ we know that $d$ is algebraic over $E$.
Let $degf_{E}^d(x)=n$. Then we have
$$E(d)=\{a_0+a_1d+a_2d^2+\cdots+a_{n_{-1}} d^{n-1}\mid a_i\in E, i=0,1,...,n-1\}$$ Since $D$ is an integral domain, $a_0+a_1d+a_2d^2+\cdots+a_{n_{-1}} d^{n-1}\in D$ and so $E(d)\subset D$. In particular $d^{-1}\in E(d)$, so $d^{-1}\in D$
1ENİGMA1
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