# Need help in this proposition

I didn't understand this part in this proof in Hartshorne's algebraic geometry book:

I didn't understand why $W=Z(\mathfrak a)$ for some ideal $\mathfrak a$.

I'm starting to study algebraic geometry with this book which is very hard, I really need help in this proposition.

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The closed sets are defined as $Z(T)$ for some subset $T \subseteq A$. If $\mathfrak a$ is the ideal generated by $T$ in $A$, it is $Z(\mathfrak a) = Z(T)$: The inclusion $Z(\mathfrak a) \subseteq Z(T)$ is given by a), since $T \subseteq \mathfrak a$. On the other hand let $P \in Z(T)$ and $f \in \mathfrak a$. Then $f = \sum a_i \cdot t_i$ with $a_i \in A$ and $t_i \in T$. Then $f(P) = \sum a_i(P) \cdot t_i (P) = 0$ since $t_i(P) = 0$ because $P \in Z(T)$ and $t_i \in T$. Thus $P \in Z(\mathfrak a)$ and every closed set can be viewed as the common zeros of an ideal in $A$.

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At this point, closed sets are defined by the zero loci of a set of polynomials. On a set where these polynomials vanish, we can multiply them by any polynomial and still get a polynomial that vanishes there, since $0 \cdot a = 0$. Any linear combination of these also vanishes at our set. Thus we can just think of the set of polynomials that defines $W$ as an ideal.

Note, over a Noetherian ring, by Hilbert's basis theorem, this set can thus be taken to be finite.

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Closed sets in the Zariski topology are precisely those sets of the form $Z(\mathfrak{a})$.

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but why not $Z(T)$, where $T$ is not necessarily an ideal? –  user42912 Aug 11 '13 at 21:53
@user42912 Because $Z(T)=Z(\langle T \rangle )$, where $\langle T \rangle$ is the ideal generated by T. –  Fredrik Meyer Aug 14 '13 at 9:55