Flatness and Faithful Flatness — Examples

As an $R$-module, $R[x]={⨁}_{n\ge 0}R{x}^n,$ so it is free on the monomial basis. By [L1], it is flat over $R$.

If $I⊊R$, then every polynomial in $IR[x]$ has all coefficients in $I$, so $1\mathrm{\notin }IR[x]$. Thus $IR[x]$ is proper. By [L2], the map $R\to R[x]$ is faithfully flat.

Therefore polynomial algebras give basic faithfully flat examples.

By [L1], $\mathbb{Z}[1\mathrm{/}2]$ is flat over $\mathbb{Z}$.

The proper ideal $(2)\subset \mathbb{Z}$ becomes the unit ideal after localization, since $2$ is invertible in $\mathbb{Z}[1\mathrm{/}2]$. Thus $(2)\mathbb{Z}[1\mathrm{/}2]=\mathbb{Z}[1\mathrm{/}2]$. By [L2], the map is not faithfully flat.

So a proper localization can be flat without being faithfully flat.

Since $\mathbb{Q}$ is the localization of $\mathbb{Z}$ at the nonzero integers, [L1] gives that $\mathbb{Q}$ is flat over $\mathbb{Z}$.

Assume the Axiom of Choice. If $\mathbb{Q}$ were projective, [L3] would make it a direct summand of a free abelian group. Every direct summand of a free abelian group is reduced, while $\mathbb{Q}$ is nonzero and divisible. Hence $\mathbb{Q}$ is not projective.

Thus a fraction field can be flat without being projective.

The element $e=(1,0)\in A\times B$ satisfies $e^2=e$, and $I=Re$.

Therefore [L1] applies and shows that $R\mathrm{/}I$ is flat. Concretely, $R\mathrm{/}I\cong B$ as the second factor.

This is the standard idempotent-quotient example.

Here $I^2=({\epsilon }^2)=0$, while $I\mathrm{\ne }0$. So $I\mathrm{\ne }{I}^2$.

By [L1], the quotient $R\mathrm{/}I$ cannot be flat. Equivalently, the ideal criterion [L2] fails for the inclusion $I\hookrightarrow R$ after tensoring with $R\mathrm{/}I$.

Thus quotients by nonidempotent ideals need not be flat.

Each factor $R_{f_i}$ is flat by [L1], so the product ring ${\prod }_i{R}_{f_i}$ is flat over $R$ because finite direct products are finite direct sums as modules.

Let $I⊊R$ be proper. If $I{\prod }_i{R}_{f_i}={\prod }_i{R}_{f_i}$, then for every $i$ some power of $f_i$ lies in $I$, because $1\in I{R}_{f_i}$ implies $f_i^{m_i}\in I$ for some $m_i$. Since the $f_i$ generate the unit ideal, so do the powers $f_i^{m_i}$, forcing $1\in I$, contradiction. Thus the extended ideal is proper.

By [L2], the product map is faithfully flat.

By [L1], the module $M$ is free of rank $r$, because the residue vector-space dimension equals the rank of a free module.

The chosen lifts generate $M$ by Nakayama, and a generating set of size equal to the rank of a free module is automatically a basis. Therefore $x_1,\dots ,x_r$ is an $R$-basis of $M$.

So residue-field bases lift to actual bases in the finite flat local case.