Lemma 110.36.1. There exists an epimorphism of local rings of dimension 0 which is not a surjection.
110.36 An epimorphism of zero-dimensional rings which is not surjective
In [Lazard-deux] 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.
Proof. See discussion above. \square
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