# 解集函子可表示　`prop.solution-functor-representable`
命题：$A \mapsto \{a \in A^{n} : f_{i}(a) = 0\}$ 可表示
layer 13 · 命题 · 预层与米田 · 范畴论
[在星图里打开](../#n=prop.solution-functor-representable)　·　[节点索引](../index.md)

设 $f_{1}, \ldots, f_{r} \in k[t_{1}, \ldots, t_{n}]$。把交换 $k$-代数 $A$ 送到**解集** $S(A) = \{(a_{1}, \ldots, a_{n}) \in A^{n} : f_{1}(a) = \cdots = f_{r}(a) = 0\
> 陈述续见 [nodes/prop.solution-functor-representable.2.md](./prop.solution-functor-representable.2.md)


## 为什么成立（入边，证明在 proofs/）
- …另有入边，续页见 [nodes/prop.solution-functor-representable.3.md](./prop.solution-functor-representable.3.md)

refs: Le Stum, An introduction to condensed mathematics, Exercise 1.22

> 说明见 `notes/prop.solution-functor-representable.md`

引用：`prop.solution-functor-representable`（内容 v0.9.0）· 条款见 [LICENSE.txt](../LICENSE.txt)