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Let $V$ be a finite-dimensional complex vector space. According to this Wikipedia page the ring of polynomial functions on $V$ is defined to be the subring of all functions $V\to\mathbb{C}$ generated by the dual space $V^*$.

I believe there is a problem with this definition: we only get polynomials whose constant coefficient (i.e. the degree $0$ coefficient) is in $\mathbb{Z}$. (Or even worse, polynomials of degree $\geq 1$, if rings are not considered unital.)

Indeed, how to you generate, for example, the constant function $V\to\mathbb{C},z\mapsto\sqrt{2}$ from $V^*$?

Is this observation right, and if yes, what is the correct definition?

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    You are right. You should replace ring by $\Bbb C$-algebra. – 2017-01-19

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Since you already have $\sqrt{2} \in \mathbb{C}$, $\sqrt{2}$ is present in any extension of $\mathbb{C}$ as well.

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    I don't understand what you mean. – 2017-01-19
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    I mean, I understand your sentence, but I don't get what it has to do with my question. – 2017-01-19
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    @Simon: $\mathbb{C}[V^*] \subseteq \mathbb{C}^V$ is an extension of $\mathbb{C}$. – 2017-01-19
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    Not according to the definition I gave. I didn't say the *algebra* generated by $V^*$, I said the *ring* generated by $V^*$. – 2017-01-19
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    @Simon: The $\mathbb{C}$-algebra generated by $V^*$ doesn't contain $\sqrt{2}$. The ring formed by adjoining the elements of $V^*$ to $\mathbb{C}$ does. – 2017-01-19
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    Why *adjoining to $\mathbb{C}$*? In ring theory, the subring generated by $V^*$ is the set of all finite sums of finite products of elements of $V^*\cup\{1\}$. This set doesn't contain $\sqrt{2}$. – 2017-01-19
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    @Simon: Because that's what it means when we ask for the ring of coordinate functions $\mathbb{C}[V]$. You could ask for $\mathbb{Q}[V]$, and that *wouldn't* have $\sqrt{2}$. $\mathbb{Z}[V]$ is weird, though; you only usually do this with fields. If you mean a non-field you probably have different things in mind. – 2017-01-19
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    It seems you didn't read the question carefully. I know what $\mathbb{C}[V]$ should be, but I think the ring theoretic definition in the Wikipedia page is bogus. You are answering with a different definition of $\mathbb{C}[V]$ than the Wikipedia page. – 2017-01-19
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    Please also read the sentence "In ring theory..." two comments above. – 2017-01-19
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    I didn't mean $\mathbb{Z}[V]$, but the ring of polynomials whose $0$th degree coefficient is in $\mathbb{Z}$ (and all others in $\mathbb{C}$). – 2017-01-19