TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31

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Holomorphic germs at a point form a local ring

Statement

Fix $a \in C$ , and let $O_{a}$ be the set of holomorphic germs at $a$ . Define

$[f]_{a} + [g]_{a} : = [f + g]_{a}, [f]_{a} [g]_{a} : = [f g]_{a} .$

Then $O_{a}$ is a commutative ring with identity $[1]_{a}$ . A germ $[f]_{a} \in O_{a}$ is a unit if and only if $f (a) \neq 0$ . Consequently

$m_{a} : = {[f]_{a} \in O_{a} : f (a) = 0}$

is the unique maximal ideal, so $O_{a}$ is a local ring.

Facts & Assumptions

Given: A point $a \in C$ and holomorphic germs $[f]_{a}, [g]_{a}, [h]_{a}$ .

[L1]

Two holomorphic functions define the same germ at $a$ exactly when they agree on some neighbourhood of $a$ , and then they have the same value at $a$ (Holomorphic germs at a point).

[L2]

Sums and products of holomorphic functions are holomorphic; if a holomorphic function is nonzero at a point, then it is nonzero on some neighbourhood of that point and its reciprocal is holomorphic there (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L3]

A local ring is a nonzero commutative ring with a unique maximal ideal (A local ring is a nonzero commutative ring with a unique maximal ideal).

Proof

technique · direct

1.1L1algebra

If $[f]_{a} = [f_{1}]_{a}$ and $[g]_{a} = [g_{1}]_{a}$ , then [L1] gives a neighbourhood of $a$ on which $f = f_{1}$ and a neighbourhood on which $g = g_{1}$ ; on their intersection one has $f + g = f_{1} + g_{1}$ and $f g = f_{1} g_{1}$ . Thus the displayed sum and product are well defined on germs.

1.2givenL2

If $f (a) \neq 0$ , [L2] gives a neighbourhood $U$ of $a$ on which $f$ never vanishes. For each $z \in U$ , the reciprocal rule in [L2] applies to $f$ at $z$ , so $1 / f$ is holomorphic on $U$ and $[f]_{a} [1 / f]_{a} = [1]_{a}$ . Hence $[f]_{a}$ is a unit.

1.3L1algebra

If $f (a) = 0$ and $[f]_{a} [g]_{a} = [1]_{a}$ , then [L1] lets us evaluate at $a$ and obtain $0 = f (a) g (a) = 1$ , impossible. So a germ vanishing at $a$ is not a unit.

2.1step 1.1L2algebra

Pointwise addition and multiplication on representatives give associative and commutative operations on $O_{a}$ , the constant germs $[0]_{a}$ and $[1]_{a}$ are additive and multiplicative identities, and $[- f]_{a}$ is an additive inverse of $[f]_{a}$ . So step 1.1 makes $O_{a}$ a commutative ring with identity $[1]_{a}$ .

2.2step 1.2step 1.3L1algebra

Steps 1.2 and 1.3 show that the nonunits are exactly the germs in $m_{a}$ . This set is an ideal because sums of germs vanishing at $a$ still vanish at $a$ , additive inverses still vanish at $a$ , and $h (a) f (a) = 0$ for every $[h]_{a} \in O_{a}$ and $[f]_{a} \in m_{a}$ . Also $[1]_{a} \notin m_{a}$ , so $m_{a}$ is proper.

3.1step 2.2L3∎

Every proper ideal consists entirely of nonunits, hence step 2.2 places it inside $m_{a}$ . Therefore $m_{a}$ is the unique maximal ideal, and [L3] makes $O_{a}$ a local ring.

Holomorphic germs at a point form a local ring

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources