Absolutely Flatness (not finished)

~~啥都能水系列~~

Absolutely flatness 是使得任意module flat 的环, 并没有找到很多的absolutely flat 环, 常见的例子还是有field 和Boolean ring. Absolutely flatness 最大的特征是对任意极大理想局部化后是一个field. 同时absolutely flatness 和Zariski topology 的分离性有很强的关系.

Absolutely Flatness

Definition 1. A ring $R$ is absolutely flat if every $R$ -module is flat.

Now we give some property and criterion of absolutely flatness.

Theorem 1. For a ring $R$ , the followings are equivalent:

1. $R$ is absolutely flat.

2. Every finite generated ideal is direct summand of $R$ .

3. Every finite generated ideal is idempotent.

4. Every principal ideal is idempotent.

Proof

Proof. 2 $\Rightarrow$ 3: We have $R\cong \mathfrak{a}\oplus R /\mathfrak{a}$ , so $R /\mathfrak{a}$ is projective thus flat. Tensor $0\to \mathfrak{a}\to R\to R /\mathfrak{a}\to 0$ with $R/\mathfrak{a}$ , we will have $0\to R /\mathfrak{a}\otimes \mathfrak{a}\to R /\mathfrak{a}\otimes R\to R /\mathfrak{a}\otimes R /\mathfrak{a}\to 0$ . exact. Since $R /\mathfrak{a}\otimes \mathfrak{a}\to R /\mathfrak{a}\otimes R$ is obviously zero morphism, it’s also injective so $R /\mathfrak{a}\otimes\mathfrak{a}\cong \mathfrak{a} /\mathfrak{a}^{2}=0$ . Thus $\mathfrak{a}$ is idempotent.

3 $\Rightarrow$ 2: Let $\mathfrak{a}$ be a finite generated ideal. By Nakayama lemma, there’s $x\in \mathfrak{a}$ s.t. $(1-x)\mathfrak{a}=0$ . Thus $xa=a$ for all $a\in \mathfrak{a}$ . So $x$ generates $\mathfrak{a}$ and $x^{2}=x$ . So $(x)\cap (1-x)=0\Rightarrow R\cong (x)\oplus (1-x)$ .

1 $\Rightarrow$ 3: Notice that $R /\mathfrak{a}$ is flat and apply similar argument as 2 $\Rightarrow$ 3.

3 $\Rightarrow$ 4 is clear.

4 $\Rightarrow$ 3: Let $\mathfrak{a}=(a_{1},\dots, a_{n})$ . Then each $(a_{i})$ are idempotent. So $a_{i}=x_{i}a_{i}^{2}$ for each $i$ . Thus $\mathfrak{a}=\mathfrak{a}^{2}$ .

2 $\Rightarrow$ 1: For any finite generated ideal $\mathfrak{a}$ and an $R$ -module $M$ . $R=\mathfrak{a}\oplus R/\mathfrak{a}$ so $R /\mathfrak{a}$ is projective and thus flat. Consider the exact sequence $0\to \mathfrak{a}\to R\to R /\mathfrak{a}\to 0$ , the sequence $0\to \mathfrak{a}\otimes M\to R\otimes M\to R /\mathfrak{a}\otimes M\to 0$ is exact. So $0\to \mathfrak{a}\otimes M\to R\otimes M\to M /\mathfrak{a}M \to 0$ is exact. Thus $\mathfrak{a}M\cong \mathfrak{a}\otimes M$ , $M$ is flat. ◻

Theorem 2. $R$ is absolutely flat and $\mathfrak{a}$ is any ideal of $R$ , then $R /\mathfrak{a}$ is absolutely flat.

Proof

Proof. Take any principal ideal $(\bar{x})$ of $R /\mathfrak{a}$ . Then $\pi^{-1}((\bar{x}))=(x)$ is also a principal ideal. So $(x)$ is idempotent. Thus $(\bar{x})$ is idempotent. ◻

Theorem 3. For $R$ an absolutely flat ring, $x\in R$ is not a unit, then $x$ is a zero divisor.

Proof

Proof. $(x)$ is idempotent so $ax^{2}=x$ for some $a\in R$ . So $x(1-ax)=0$ , $x$ is not a unit, $1-ax\neq 0$ . Thus $x$ is a zero divisor. ◻

Theorem 4. For $R$ local absolutely flat ring. Then $R$ is a field.

Proof

Proof. Take $x\in R$ be a nonunit, $x\in \mathfrak{m}$ the maximal ideal of $R$ . $(x)$ is idempotent so $ax^{2}=x$ for some $a\in R$ . $x$ is in the Jacobson radical, so $1-ax$ is a unit, thus $x=0$ . ◻

Theorem 5. Let $R$ be a ring and $S$ is a multiplicative subset, $R$ is absolutely flat, then $S^{-1}R$ is absolutely flat.

Theorem 6. $R$ is absolutely flat if and only if $R_{\mathfrak{m}}$ is a field for any maximal ideal $\mathfrak{m}$ of $R$ .

Absolutely flatness has close relationship with the contructible topology, the Zariski topology and the constructible topology on the $Spec(R)$ are the same if and only if $R/\mathfrak{N}$ is absolutely flat.

Theorem 7. Let $R$ be a ring, the followings are equivalent:

1. Let $\mathfrak{N}$ be the nilradical of $R$ , then $R /\mathfrak{N}$ is absolutely flat.

2. Every prime ideal of $R$ is a maximal ideal.

3. $Spec(R)$ is $T_{1}$ space.

4. $Spec(R)$ is Hausdorff.