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The pullback is a subset of the cartesian product in the category of commutative rings with unit.

What is the pullback in the category of commutative $k$-algebras? Is it the same set as in rings?

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The forgetful functor to Set is representable, so it preserves limits (including pullback). Hence the pullback should be the set-theoretic pullback with the natural commutative k-algebra structure. –  Qiaochu Yuan Dec 9 '10 at 21:12
    
@Qiaochu I think this is most natural way to see this. –  BBischof Dec 9 '10 at 22:26
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1 Answer 1

Yes.

More generally, fix a category $\mathcal{C}$ that contains finite limits. Given an object $A \in \mathcal{C}$, we can construct the category $\mathcal{C}_A$ of objects "under $A$." That is, an object of this category $\mathcal{C}_A$ is a morphism $A \to B$ and a morphism is a commutative triangle.

The claim is that $\mathcal{C}_A$ admits finite limits, and the limits are the same as in $\mathcal{C}$. This is basically formal. Given a functor $F:I \to \mathcal{C}_A$ from a finite category $I$, we get a functor $G: I \to \mathcal{C}$ which has a limit by assumption. Moreover, for each $i \in I$, we have a morphism $A \to Gi$. By the universal property, this becomes a morphism $A \to \lim G$.

I claim that the object $A \to \lim G$ in $\mathcal{C}_A$ is a limit of $F$. This can be checked directly from the definitions.

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