Structure sheaf Affine variety

proof: inclusion ⊃ clear. opposite, let g in left-hand side ,



j
=
{
h
∈
a

|

h
g
∈
a
}


{\displaystyle j=\{h\in a|hg\in a\}}

, ideal. if x in d(f), then, since g regular near x, there open affine neighborhood d(h) of x such



g
∈
k
[
d
(
h
)
]
=
a
[

h

−
1


]


{\displaystyle g\in k[d(h)]=a[h^{-1}]}

; is, h g in , x not in v(j). in other words,



v
(
j
)
⊂
{
x

|

f
(
x
)
=
0
}


{\displaystyle v(j)\subset \{x|f(x)=0\}}

, hilbert nullstellensatz implies f in radical of j; i.e.,




f

n


g
∈
a


{\displaystyle f^{n}g\in a}

.



◻


{\displaystyle \square }

the claim, first of all, implies x locally ringed space since

o



x
,
x


=




lim




→






f
(
x
)
≠
0


⁡
a
[

f

−
1


]
=

a




m



x






{\displaystyle {\mathcal {o}}_{x,x}=\varinjlim _{f(x)\neq 0}a[f^{-1}]=a_{{\mathfrak {m}}_{x}}}

where






m



x


=
{
f
∈
a

|

f
(
x
)
=
0
}


{\displaystyle {\mathfrak {m}}_{x}=\{f\in a|f(x)=0\}}

. secondly, claim implies






o



x




{\displaystyle {\mathcal {o}}_{x}}

sheaf; indeed, says if function regular (pointwise) on d(f), must in coordinate ring of d(f); is, regular-ness can patched together.

hence,



(
x
,



o



x


)


{\displaystyle (x,{\mathcal {o}}_{x})}

locally ringed space.