Commutative algebra 19 Affine schemes

This lecture is part of an online course on commutative algebra, following the book
"Commutative algebra with a view toward algebraic geometry" by David Eisenbud.

In this lecture we give an introduction to the affine scheme of a ring, that is heavily used in algebraic geometry. We check the sheaf property (at least for integral domains) and give a dictionary relating algebraic properties of the ring R to geometric properties of Spec R (with its sheaf of rings).

Reading: Lectures 18 and 19 do not correspond directly to a section of the textbook.
Exercise: 2.19, 2.26
Exercise for the ambitious: Check the sheaf property discussed in the lecture for rings with zero divisors. (This is tricky.)

[nourbielal]

this is the most beautiful lecture that I have ever seen in this serie. I study this serie lecture by lecture thank you very much for your efforts.

[jeffreyhowarth]

I'm glad some Europeans speak English and math very well. I've been watching your videos for a couple weeks in awe only to realize that you are a recipient of the fields metal. Thanks for "commutive algebra made easier".

[hausdorffm]

6:27 For any elements g, a, b of ring R satisfy a + b =1, ga^2=gb^2=0, then g =0.
In fact, 1=1^3= (a+b)^3 = a^3+3a^2b+3ab^2+b^3 and hence g =


11:19 Now, by step (2), R is replaced by R[1/f], so what we should find is element r of R, not R[1/f]

By step (1), R is integral domain, so the relation of locarization is rs’-r’s=0 instead of s”(rs’-r’s)=0.

18:40 The correspondence of ideals and closed sets is NOT 1:1 because radical of ideals gives an example.

18:09 Typo? For f of R, there is a hyper-surface {p|p dose NOT contain f} …. I guess hypersurface should be hypersurface {p|p DOSE contain f}…

20:35 If multiplicative set S is generated by two element f, g of R, then Spec R[1/S] = U(fg)= intersection of U(f) and U(g).

20:35 I do not understand why idempotent of R corresponds to clopen set.

23:03 In Spec C[x], for f = (x-1)(x-4), U(f) dose not contain the prime ideal (x-1) and (x-4) because f is in (x-1) and f is in (x-4)

25:01 Z[1/14] = Z[1/2, 1/7].

[robinbalean]

Thank you so much. This is a brilliant lecture.