# Doubt in Hartshorne's algebraic geometry book

I'm studying by myself Algebraic Geometry and I didn't understand this part in the Hartshorne's book:

I know that every polynomial $f$ in $\mathfrak a$ is written as $f=g_1f_1+\ldots + g_rf_r$, where $g_i\in A$, but I can't go further, I need help.

• If this is the first time you reach towards algebraic geometry and even more: if you're doing it by yourself, I'd say that reading from Hartshorne's book, which is a very, very hard book to read...*to say the least*, may not be the best of ideas . I wouldn't use this book even for a first course in the university, with lecturer, instructor, guided and stuff. As a reference it's great, as a text book there are, imo, much better and gentler books out there. – DonAntonio Aug 4 '13 at 4:11
• This is not even geometry really. This problem is really a commutative algebra problem that is covered in even the lightest of the well-known commutative algebra books (Atiyah and Macdonald), and to attempt Hartshorne without the commutative algebra necessary to do such a problem is close to suicidal. – PVAL-inactive Aug 4 '13 at 4:12
• What the above parraph is saying is twofold:(1) $\,Z(T)\,$ can be expressed as the set of zeros of any generating set of polynomials of an ideal in the polynomial ring $\,A[X_1,...,X_n]\;$ , and (2) Any ideal in this ring is finitely generated (and thus the generating set we were talking about in (1) can be taken to be finite). – DonAntonio Aug 4 '13 at 4:17
• You should not read Hartshorne, it is not a good idea. Even right now I am not using Hartshorne to learn similar material. – user38268 Aug 4 '13 at 4:27
• @BenjaLim unfortunately as I said below on ragibs answer I dont have choice I hope I can post my doubts here. Thank you all for trying to help me. – user42912 Aug 4 '13 at 4:32

The key is to prove $Z(f,g) = Z(f)\cap Z(g).$ The result then follows by applying this result inductively.

If $x\in Z(f,g)$ then $f(x)=0$ and $g(x)=0$ so $x\in Z(f)$ and $x\in Z(g)$ so $x\in Z(f)\cap Z(g).$ The reverse direction has similar ideas.

Just a comment: I would not recommend studying (especially self study) algebraic geometry for the first time from Hartshorne. Chapter 1 of Hartshorne is really a quick crash course that quickly goes through the material of an entire book. He mentions this in the preface and really intends the reader has seen this material before elsewhere, but he provides it for revision. I would recommend going through Fulton's "Algebraic Curves" (freely available online) or Reid's "Undergraduate Algebraic Geometry" before you read Hartshorne.

• Thank you for your advice, but my adviser said that I have to study the first chapter of Hartshorne before Fulton. I think because maybe I don't have so much time and I have to have as soon as possible an overview of algebraic geometry. – user42912 Aug 4 '13 at 4:03
• How does $Z(f,g)=Z(f)\cap Z(g)$ reasonably simplify a direct proof of $Z(S)=Z(<S>)$ which is the key equality the question needs. It seems since $S\ne<S>$ as sets you simply cant break these into their intersections and claim they are equal, and even if you do the extra work to show the intersections are equal it seems a direct proof is simpler. – PVAL-inactive Aug 4 '13 at 4:32
• @PVAL The OP underlined the last line of the paragraph and not the first line that states $Z(S) = Z(<S>),$ which makes it seem like they understand the first line but not how to conclude the last part. – Ragib Zaman Aug 4 '13 at 4:34
• @BenjaLim I need to understand only the first chapter – user42912 Aug 4 '13 at 4:41
• @RagibZaman If the OP understands this claim, then we simply have $(Z(S)=Z( <S>)=Z(<f_1,\dots,f_n>)=Z(f_1\dots,f_n))$ which has nothing to do with the decomposition into intersections. If the OP does not understand $Z( <S>)=Z(<f_1,\dots,f_n>)$, which is the only claim not directly from the first claim they can look at a proof of Hilbert's basis theorem. Yet none of this seems to have anything substantial to do with the fact that $Z(f,g) = Z(f)\cap Z(g)$. – PVAL-inactive Aug 4 '13 at 4:56

$Z(S)= Z( <S>)$, where $<S>$ is the ideal generated by $S$. This is easy to show by showing containment both ways (i.e. $x\in Z(S)$, then for $g \in <S>,g=f_1s_1+ \dots+ f_ns_n$ we have $g(x)= f_1(x)s_1(x)+\dots+f_n(x)s_n(x)=0$, since $x\in Z(S)$ etc). This is enough since any ideal is always equal to an ideal generated by finitely many polynomials (by Noetherian), which is enough for [H]'s claim $(Z(S)=Z( <S>)=Z(<f_1,\dots,f_n>)=Z(f_1\dots,f_n))$.

• These $<>$ are not brackets! Instead these $\langle\rangle$ are brackets! – user26857 Aug 4 '13 at 8:10