# What is mean by $R$-automorphism?

I'm confused about a definition and I've searched through books and the internet, but I couldn't find anything. I hope someone knowledgeable about the topic can explain it to me.

To begin, let's consider the following definition:

Let $$R$$ be a commutative ring. We say that $$f \in R[x_1, \cdots, x_n]$$ is a coordinate in $$R[x_1, \cdots, x_n]$$ if there exists an $$R$$-automorphism $$\phi$$ of $$R[x_1, \cdots, x_n]$$ such that $$\phi(f) = x_1$$.

I found this definition here.

Now, I understand that a morphism of $$R$$-modules is also called $$R$$-linear, and a morphism of rings is simply called a ring homomorphism. However, my question here is: what does the author mean by an $$R$$-automorphism of $$R[x_1, \cdots, x_n]$$? Is it a ring automorphism, a module automorphism, or something else?

Edit:

In addition to the aforementioned points, another definition is provided on page 2 of the same paper. It states:

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

Thus, one might wonder how we can demonstrate the equivalence of these two definitions.

• $R[x_1,\ldots,x_n]$ has both a structure of an $R$-module, and a ring structure. It is asking $f$ to be an $R$-module automorphism of $R[x_1,\ldots,x_n]$. May 8 at 19:37
• Not just that; $R$-linear means it is an $R$-module homomorphism. "Automorphism" means that it is an isomorphism from the object to itself. So $R$-automorphism means that it is an invertible $R$-module homomorphism (i.e., an $R$-module isomorphism) from the module to itself. May 8 at 19:43
• Okay great it's done. May 8 at 19:46
• To me the natural way to interpret this is that $\phi$ should be an $R$-algebra automorphism, in the same way that one talks about "$K$-automorphisms" in Galois theory for example. If it just meant $R$-module homomorphism, then it seems that every nonzero polynomial over a field would be a coordinate, which I don't think is true. But I don't really know much about coordinates. @ArturoMagidin, sorry to bother you - am I talking nonsense? May 8 at 20:55
• @IzaakvanDongen Actually you are probably right. Note that the "definition" is literally just a parenthetical comment in the paper linked. May 8 at 20:59

$$R$$-automorphism here means automorphism of $$R$$-algebras. So, it's a ring automorphism that preserves the $$R$$-algebra structure as well: $$\phi(r f) = r\phi(f)$$ for $$r \in R$$ and $$f \in R[x_1, \dots x_n]$$. Differently said, it's a ring automorphism that is also an $$R$$-module automorphism.