The Stacks project

Example 15.15.5. If the equivalent conditions of Lemma 15.15.4 hold, then it is not always the case that every injective map of free $R$-modules is a split injection. For example suppose that $R = k[x_1, x_2, x_3, \ldots ]/(x_ i^2)$. This is an auto-associated ring. Consider the map of free $R$-modules

\[ u : \bigoplus \nolimits _{i \geq 1} Re_ i \longrightarrow \bigoplus \nolimits _{i \geq 1} Rf_ i, \quad e_ i \longmapsto f_ i - x_ if_{i + 1}. \]

For any integer $n$ the restriction of $u$ to $\bigoplus _{i = 1, \ldots , n} Re_ i$ is injective as the images $u(e_1), \ldots , u(e_ n)$ are $R$-linearly independent. Hence $u$ is injective and hence universally injective by the lemma. Since $u \otimes \text{id}_ k$ is bijective we see that if $u$ were a split injection then $u$ would be surjective. But $u$ is not surjective because the inverse image of $f_1$ would be the element

\[ \sum \nolimits _{i \geq 0} x_1 \ldots x_ ie_{i + 1} = e_1 + x_1e_2 + x_1x_2e_3 + \ldots \]

which is not an element of the direct sum. A side remark is that $\mathop{\mathrm{Coker}}(u)$ is a flat (because $u$ is universally injective), countably generated $R$-module which is not projective (as $u$ is not split), hence not Mittag-Leffler (see Algebra, Lemma 10.93.1).

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