# What are stalks of Spec $Z/(60)$?

What are stalks of (structure sheaf of) affine scheme Spec $\mathbb{Z}/(60)=\{(2),(3),(5)\}$? What are its global sections?

• Schemes don't have stalks; sheaves have. – Georges Elencwajg Nov 16 '12 at 8:53
• ?? I mean the structure sheaf of the affine scheme. – Tom Nov 16 '12 at 9:08
• and what have you tried? – user27126 Nov 16 '12 at 9:34

Hint: By the Chinese Remainder Theorem, we have an isomorphism $\mathbb{Z}/60 \cong \mathbb{Z}/4 \times \mathbb{Z}/3 \times \mathbb{Z}/5$. The prime ideals are generated by $(2,1,1),(1,0,1)$ and $(1,1,0)$, so localization is easy.
To find the stalk at $(2)$, that is, the local ring $(\mathbb{Z}/4 \times \mathbb{Z}/3 \times \mathbb{Z}/5)_{(2,1,1)}$, we are allowed to divide by anything of the form $(1,a,b)$ and $(3,a,b)$. This annihilates the second and third factor and leaves the first factor unchanged. So we are left with $\mathbb{Z}/4$. (the two last factors perish because we are allowed to divide by zero inside them)
• Thank you. In Vakil's FOAG notes (5.4.11,p131,line18), they are $Z/4,Z/3,Z/5$. It is wrong, isn't it? – Tom Nov 16 '12 at 13:30
• Second question(though it is minor): The prime ideals should be generated by $(2,1,1), (1,0,1),(1,1,0)$, shouldn't they? – Tom Nov 16 '12 at 14:01
• Thank you, I have understood! And the global sections are $\mathbb{Z}/(60)$, aren't they? – Tom Nov 17 '12 at 12:08
• @Tom: Yes, the global sections are $\mathbb{Z}/60$. Think of them as the product of the stalks. I've edited my answer again: it's clearer and more correct. – Fredrik Meyer Nov 17 '12 at 13:46