I have a question regarding the Abhyankar and Sathaye conjecture. Before stating the conjecture, let's begin with this definition (about which I have a question here). The definition states that:

$f\in R[x_1,\cdots, x_n] $ is a coordinate in $R[x_1,\cdots, x_n] $ if there exist $f_2,\cdots, f_n$ such that $R[f, f_2,\cdots, f_n] =R[x_1,\cdots, x_n]$.

Now, the conjecture states:

$f$ is a coordinate in $R[x_1,\cdots, x_n] $ if and only if $R[x_1,\cdots, x_n]/\langle f\rangle\cong R[x_1,\cdots, x_{n-1}] $.

The paper here asserts that the first direction of the equivalence is evidently true, and the other direction is the conjecture. So, I'm attempting to prove the first direction of the equivalence.

If we take $\phi: R[x_1,\cdots, x_n] \longrightarrow R[x_1,\cdots, x_{n-1}]$.

I'm contemplating constructing $\phi$ such that $\operatorname{ker}(\phi)=\langle f\rangle$ and $\phi$ is surjective. Then, we can utilize the first isomorphism theorem, implying that $R[x_1,\cdots, x_n]/\operatorname{ker}(f)\cong R[x_1,\cdots, x_{n-1}] $.

Does anyone have an idea on how to construct $\phi$, or anything else? Thanks in advance.


1 Answer 1


If $R[f_1,f_2,\dots,f_n] = R[x_1,\dots,x_n]$ then $f_1, f_2, \dots, f_n$ behave just like variables: every element of $R[f_1,f_2,\dots,f_n]$ can be written in a unique way as a polynomial in $f_1, f_2, \dots, f_n$. Note that this is evidently true, since $x_i \mapsto f_i$ is an isomorphism of $R$-algebras.

So the isomorphism from the conjecture is just $R[x_1,\dots,x_n]/(f) = R[f,f_2,\dots,f_n]/(f) \cong R[f_2,\dots,f_n] \cong R[x_2,\dots,x_n]$.

  • $\begingroup$ @ Magdiragdag Could you please explain the concept of isomorphism and its origin in the context of the expression $R[f,f_2,\dots,f_n]/(f) \cong R[f_2,\dots,f_n]$ $\endgroup$ Commented May 16 at 11:48
  • $\begingroup$ @Santa-claus In this context, isomorphism means isomorphism of $R$-algebras, i.e., an isomorphism of rings that preserves the $R$-algebra structure. $\endgroup$ Commented May 16 at 14:42
  • $\begingroup$ @ Magdiragdag My apologies for any confusion. While I acknowledge that it's a morphism of $R$-algebras, could you guide me on demonstrating their isomorphism? In other words, how can we construct such an isomorphism? $\endgroup$ Commented May 17 at 13:07
  • $\begingroup$ Because the $f_j$ are 'like variables', (the residue class of) $f_i$ just maps to $f_i$ in that isomorphism. It's really no different from $R[F,F_2,\dots,F_n]/(F) \cong R[F_2,\dots,F_n]$, the capital letters here meaning that it's really a polynomial ring in the indeterminates $F, F_2, \dots, F_n$. (Of course, you do need the assumption that $R[f,f_2,\dots,f_n] = R[x_1,\dots,x_n]$.) $\endgroup$ Commented May 17 at 14:15

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