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The answer is no, consider $k[x,xy, xy^2, xy^3, \dots]$.

Some more details. This is basically a copy of A^2 where all the points of one axis (including the generic point of that axis) are all glued together (into the obvious maximal ideal of that ring).

EDIT: Your example may be right too, I'm not quite sure I see what's going on there.

EDIT2: (More information on the example) Call the ring $R$ and it obviously sits inside $k[x,y]$. The induced map $\mathbb{A}^2 \to Spec R$ contracts the axis $x = 0$ to a ($k$-valued) point, otherwise, it is an isomorphism (to check that, invert $x$). The ideal corresponding to the contracted axis is the maximal ideal $(x, xy, xy^2, xy^3, \dots)$.

Let me know if you have further questions.

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No

The answer is no, consider $k[x,xy, xy^2, xy^3, \dots]$.

I'll give some

Some more details. This is basically a copy of A^2 where all the points of one axis (including the generic point of that axis) are all glued together (into the obvious maximal ideal of that ring).

EDIT: Your example may be right too, I'm not quite sure I see what's going on there.

2 added 213 characters in body

No, consider $k[x,xy, xy^2, xy^3, \dots]$.

I'll give some more details. This is basically a copy of A^2 where all the points of one axis (including the generic point of that axis) are all glued together (into the obvious maximal ideal of that ring).

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