## 109.35 An epimorphism of zero-dimensional rings which is not surjective

In and [Autour] one can find the following example. Let $k$ be a field. Consider the ring homomorphism

$k[x_1, x_2, \ldots , z_1, z_2, \ldots ]/ (x_ i^{4^ i}, z_ i^{4^ i}) \longrightarrow k[x_1, x_2, \ldots , y_1, y_2, \ldots ]/ (x_ i^{4^ i}, y_ i - x_{i + 1}y_{i + 1}^2)$

which maps $x_ i$ to $x_ i$ and $z_ i$ to $x_ iy_ i$. Note that $y_ i^{4^{i + 1}}$ is zero in the right hand side but that $y_1$ is not zero (details omitted). This map is not surjective: we can think of the above as a map of $\mathbf{Z}$-graded algebras by setting $\deg (x_ i) = -1$, $\deg (z_ i) = 0$, and $\deg (y_ i) = 1$ and then it is clear that $y_1$ is not in the image. Finally, the map is an epimorphism because

$y_{i - 1} \otimes 1 = x_ i y_ i^2 \otimes 1 = y_ i \otimes x_ i y_ i = x_ i y_ i \otimes y_ i = 1 \otimes x_ i y_ i^2.$

hence the tensor product of the target over the source is isomorphic to the target.

Lemma 109.35.1. There exists an epimorphism of local rings of dimension $0$ which is not a surjection.

Proof. See discussion above. $\square$

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