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Excuse me, in a course of linear algebra, our assistant stated that, if $\mathbb{V}$ is a finite-dimensional vector space, and $\mathbb{W}$ its double dual, $\mathbb{V}$ and $\mathbb{W}$ are actually equal to each other; I am wondering if this has anything to do with the viewpoint in algebraic number theory that realizes elements, in algebraic number fields, as functions?
In any case, thank you very much.

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up vote 6 down vote accepted

Regarding your second question, it is true in some informal sense that when we view elements of a commutative ring $R$ as functions on $\text{Spec } R$, we are also viewing the points $\text{Spec } R$ as functions on $R$; in fact they are precisely the morphisms $R \to k$ where $k$ is a field, up to a certain equivalence relation. So I would say that this is not completely unrelated to double duals of vector spaces, although there isn't a direct formal connection since in this case the dual of an object is a different kind of object. This is sometimes summarized in the slogan "algebra is dual to geometry."

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This is a great answer, thanks very much. – awllower May 7 '11 at 3:26
Now I don't know which answer I am supposed to accept; they all are excellent. – awllower May 7 '11 at 8:46
Should anyone give me some advice in which one to accept, I will take it. I really do not know how to choose, but I want to do so, so as not to spare all your efforts. – awllower May 8 '11 at 2:12
It really doesn't matter that much. For the sake of not having this question get bumped, just pick one. – Qiaochu Yuan May 8 '11 at 2:35

Firstly, a vector space $V$ and its double dual are never equal. They may or may not be isomorphic, depending on whether $V$ is finite dimensional. Also, the answer to your original question is no; this is not related to viewing elements of a number field $K$ as rational functions on $\text{Spec}(\mathcal{O}_K)$.

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@Zev Chonoles: Yes, that is exactly my point, when I talked to the assistant, but he insisted that they are actually equal; in fact he announced that some books do not explicitly prove that they are equal, but they are actually so; moreover, he said that a subspace of $\mathbb{V}$ and its annihilator are equal! Besides, I am sorry about the dimension of the vector space; they all should be finite. – awllower May 7 '11 at 1:40
And thank you for such an elaborate explanation. – awllower May 7 '11 at 1:44
I think "equal" was a bit of an overstatement. But nonetheless, a vector space is always canonically embedded in its bidual by seeing a vector $v$ as the function $|v \rangle : l \in V^* \mapsto l(v) \in K$. In the finite dimensional case, the dimension being equal, you get a canoncial isomorphism (roughly no arbitrary choice involved in the definition). In a categorical point of view, that's as close as you can get to being equal. – Joel Cohen May 7 '11 at 1:49
@awllower: I'm not sure, but it could be that there was a basic misunderstanding: You can define the annihilator $U^{\perp} \subset V^{\ast}$ of a subspace $U \subset V$ (the elements in $V^{\ast}$ sending $U$ to zero) and you can define the annihilator $W_{\perp} \subset V$ of a subspace $W \subset V^{\ast}$ (the elements of $V$ that are sent to zero by all elements of $W$). Then in fact you always have equality $(V^{\perp})_{\perp} = V$, because $V^{\perp} = \{0\} \subset V^{\ast}$. – t.b. May 7 '11 at 6:58
@awllower: I see. You have canonical identifications $U \cong U^{\ast\ast} \cong (U^{\perp})^{\perp} \subset V^{\ast\ast}$, so Zev's answer and some of the comments explain why the first and the second isomorphisms aren't equalities. However, in the notation of my previous comment, if $U$ is a finite dimensional subspace of $V$ then $(U^{\perp})_{\perp} = U$ (really equality, this time). I fully second Zev's and Joel's comments. – t.b. May 7 '11 at 8:52

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