Flatness of $R/(x)$ with $R$ being local

Question

Let $R$ be a commutative local ring and let $x \in R$ be a non-unit. Suppose that for all exact sequences $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ the following sequence is also exact $0 \rightarrow A/xA \rightarrow B/xB \rightarrow C/xC \rightarrow 0$. Prove or disprove $x=0$.

All I found out is that $0 \rightarrow A/xA \rightarrow B/xB \rightarrow C/xC \rightarrow 0$ is the same as $0 \rightarrow A \otimes R/(x) \rightarrow B \otimes R/(x) \rightarrow C \otimes R/(x) \rightarrow 0$. So if $R/(x)$ would be flat for all $x$ not a unit then the statement is false.

In general $R/(x)$ is not a flat $R$-module. For example choose $R=\mathbb{Z}$ then $ \mathbb{Z}/2\mathbb{Z}$ is not a flat $\mathbb{Z}$-module. I haven't found any example for this where $R$ is a local ring.

Can anyone help me out here how this is done?

Answer 1

Let $(R,m)$ be a local ring and $x\in m$ such that $R/(x)$ is $R$-flat. Then $x=0$.

From this answer we get $(x)=(x^2)$, so $x=ax^2$ for some $a\in R$. Then $x(1-ax)=0$ and since $1-ax\notin m$ we get $x=0$.

Answer 2

Take $R=\Bbb{Z}_{(3)}$, the localization of $\Bbb{Z}$ at the prime ideal $(3)$, and $x=3$. Let $$0 \to R \to R \to R/3R \to 0$$ be exact, where the second arrow is $x \mapsto 3x$, the third arrow is projection.

Tensoring with $R/3R$ we get $$0 \to R/3R \to R/3R \to R/3R \otimes_R R/3R \to 0$$ which is not exact since the second arrow is the zero map (not injective).

Answer 3

Flat implies torsion-free. In particular $R/xR$ is never flat when $x \neq 0$.

Yes, the problem is that $R/xR$ is indeed torsion-free if $x$ is a zero divisor. So the right solution should be the following:

Flat + finitely presented implies projective. In the local case projective and free are the same notions.

We deduce: If $R$ is local, then flatness of $R/xR$ implies that it is even free. Since it is generated by one element, we would have $R/xR \cong R$, which is wrong for $x \neq 0$, since $x$ annihilates a generator of $R/xR$ but does not annihilate any generator of $R$.