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Given an algebraic set $Y=Z(T)$ and consider $\overline{f}\in A(Y)$. Can we safely conclude that $$Z(\overline{f})=\{P\in\mathbb{A}^n_k| f(P)=0,\forall f\in\overline{f}\}=Z(f)\cap Z(T)$$ I think we can because the set $\overline{f}$ contains all possibilies of $f+g$ where $g\in (T)$. Hence $Z(f+g)=Z(f)\cap Z(g)$. Is there any mistakes in the argument?
This question comes from my attempt in solving the exercise 1.1.8 from Hartshorne. Thank you!

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  • $\begingroup$ There's room for improvement here. Where does $Y$ live? Where do all of these intersections live? It's not really correct to say $Z(f+g)=Z(f)\cap Z(g)$, there's some missing explanation here. On the other hand, you're sort of on track to say something true - why don't you give this another pass while trying to be more precise? $\endgroup$
    – KReiser
    Aug 28, 2022 at 4:52

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