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Let $P(X,Y)\in \mathbf Z[X,Y]$ be an irreducible polynomial and let $A$ denote the quotient ring $\mathbf Z[X,Y]/(P)$.

What is known about the group of units of $A$? It's not even clear to me that why it is a finitely generated group. If so, can we say something about its rank?

This question is motivated by Dedekind's theorem, stating that with one variable less, the group of units of $\mathbf Z[X]/(P(X))$ is finitely generated and expressing its rank in terms of the number of real and complex embeddings.

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The group of units of any finitely generated $\mathbb{Z}$-algebra is finitely generated: this is a result of Samuel, À propos du théorème des unités. Bull. Sci. Math. (2) 90 (1966), 89-96. In such generality it is of course impossible to give a meaningful statement about its rank; I don't know if one can do better in your specific case.

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    $\begingroup$ One can give an upper bound for the rank in terms of the places where poles are allowed (poles in the sense of valuations) by adapting the classical proof for rings of dimension 1. To deal with poles at geometric places, one can use as inspiration the analysis in J. Denef, The Diophantine problem for polynomial rings of positive characteristic. Logic Colloquium '78 (Mons, 1978), pp. 131–145 $\endgroup$
    – Pasten
    Apr 23, 2014 at 20:29
  • $\begingroup$ Note that this is true only for reduced algebras. The "standard" counterexample is: $\mathbf Z[X,Y]/(X^2)$. $\endgroup$
    – Oblomov
    Jul 1, 2014 at 14:33

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