Well, I agree, it’s a somewhat odd thing to write and I’m not quite sure what the point is supposed to be, unless it’s that associativity algebras are *a fortiori* possibly non-commutative rings and here “ring” always means commutative ring. The business boils down to “the category of commutative $k$-algebras and their homomorphisms is equivalent to the undercategory $k/CRing$”, and that’s what I’d write.

Edit: I took a quick look, and didn’t see anything wrong.

]]>Thank you, Todd. That’s what I thought. And in the preceding sentence: “category of associative $k$-algebras which are rings”. Is that redundant? Are there some associative algebras which are not rings? Couldn’t we just say “category of associative $k$-algebras and $k$-linear homomorphisms”, and then remark that this is the standard undercategory $k/Ring$? Maybe I’ll just make the edit, can you see if you like it?

]]>It should have said, “with $f$ a (commutative) ring homomorphism”.

With that in place, $R$ acquires a $k$-module structure by means of the formula $s \cdot x \coloneqq f(s)x$ where $s \in k$ and $x \in R$, and we are using the ring multiplication on $R$ to define the right side of the formula. Moreover, $R$ becomes a $k$-algebra assuming, as we are, that the rings in question are commutative. In other words, each element $r \in R$ induces a scalar multiplication $x \mapsto r x$ which preserves the $k$-linear structure on $R$, since $r (s \cdot x) = r f(s) x = f(s) (r x) = s \cdot (r x)$.

I’ll go in and correct, since I need to do something there anyway.

]]>At scheme, the definition of a $k$-ring notates the category as $k/Ring$ and says it’s a pair $(R,f\colon k\to R),$ with $f$ a $k$-algebra homomorphism. Is this correct? What does this mean? We can obviously view $R$ as a $k$-algebra *by means of the action by $f$*. How can we say that $f$ is itself a $k$-linear map? With respect to what $k$-action on $R$? It’s somehow circular.

Perhaps does it mean to say something more like “a $k$-ring is an object in the undercategory $k/Ring$, so objects are *all* pairs $(R,f\colon k\to R)$ (no restriction on $f$), and morphisms are $k$-algebra homomorphisms”?