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So as I have it, a toric variety is a complex variety $X$ with an open embedding of a torus $T^n$ with dense image, and morphism: $$a:X\times_{\mathbb C}T^n\longrightarrow X$$ which extends the multiplication action: $$m:T^n\times_{\mathbb C} T^n\longrightarrow T^n$$ where $\times_{\mathbb C}$ denotes the fibre product over $\operatorname{Spec}\mathbb C$.

However, I take a complex variety to be a scheme $X\rightarrow \operatorname{Spec}\mathbb C$, which is integral (i.e reduced, and irreducible), separated, and of finite type. Since $X$ is an irreducible topological space, isn't every open subset of $X$ dense? So doesn't that make the condition that the open embedding of the torus has dense image automatic?

Disclaimer: I never took a course in classical algebraic geometry, so I really only know how to think about varieties as schemes, and only know how to do stuff with them using the language of schemes.

If it is redundant, why do we include in the definition then?

ETA: I saw this definition in my lectures and it matches the one found in Fulton's Introduction to Toric Varieties

Chris
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  • As shown in the linked duplicate, any nonempty open subset of an irreducible topological space is dense. So the answer to your question is yes. – KReiser Feb 11 '24 at 16:47
  • @KReiser I know any open subset of an irreducible topological space is dense, so since I’m not missing something, my actual question is why is there a redundancy in the definition? – Chris Feb 11 '24 at 16:56
  • It depends on where you get your definition! I've seen it defined often without the redundant condition. (Hint: adding the place where you saw your definition to your post in an [edit] would probably help...) – KReiser Feb 11 '24 at 17:06

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