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