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Let $\mathsf{Artin}$ be the category of artinian rings, viewed as a full subcategory of the category $\mathsf{CRing}$ of rings. (Here "ring" means "commutative ring with one".)

Question 1. Does $\mathsf{Artin}$ admit finite limits?

As $\mathsf{Artin}$ has finite products, Question 1 is equivalent to

Question 2. Does $\mathsf{Artin}$ admit equalizers?

A closely related question is

Question 3. Let $A\to B$ be the equalizer in $\mathsf{CRing}$ of two morphisms from $B$ to $C$. Assume that $B$ and $C$ are artinian. Does this imply that $A$ is artinian?

Yes to Question 3 would imply yes to Questions 1 and 2. [Edit: I made a mistake when I claimed implicitly that yes to Question 3 would immediately imply yes to Questions 1 and 2. There was a subtlety I missed, but Jeremy's comment below his answer settles also Question 1 and 2. (I also missed that, sigh...)]

This answer of MooS implies that the category of noetherian rings does not admit finite limits.

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There's an example in

Gilmer, Robert; Heinzer, William, Artinian subrings of a commutative ring, Trans. Am. Math. Soc. 336, No. 1, 295-310 (1993). ZBL0778.13012.

of two artinian subrings of a commutative artinian ring whose intersection is not artinian. Since the intersection is the pullback of the inclusion maps, this answers the question.

Let $R=\mathbb{C}(x)[y]/(y^2)$. Then $R$ is a two-dimensional algebra over $\mathbb{C}(x)$, and is therefore artinian.

Let $R_1=\mathbb{C}(x^2)+y\mathbb{C}(x)\subset R$. Then $R_1$ is a three-dimensional algebra over $\mathbb{C}(x^2)$, and is therefore artinian.

Let $R_2=\mathbb{C}(x^2+x)+y\mathbb{C}(x)\subset R$. Then $R_2$ is a three-dimensional algebra over $\mathbb{C}(x^2+x)$, and is therefore artinian.

But $\mathbb{C}(x^2)\cap\mathbb{C}(x^2+x)=\mathbb{C}$, so $R_1\cap R_2=\mathbb{C}+y\mathbb{C}(x)$, which is not artinian, or even noetherian, since $y\mathbb{C}(x)$ is not finitely generated as an ideal.

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    $\begingroup$ Thanks! I realized that I made a mistake in the statement of the question when I wrote "Yes to Question 3 would imply yes to Questions 1 and 2", because we must exclude the possibility that the pullback does exist in Artin but does not coincide with the pullback in CRing. I'm planning to remove Questions 1 and 2 (and change the title), but before doing so I wanted to ask you if you see a way of showing that the pullback does not exist in Artin (or of showing that it exists but differs from the other). - I just asked a related question: math.stackexchange.com/q/3096154/660 $\endgroup$ – Pierre-Yves Gaillard Feb 1 '19 at 12:13
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    $\begingroup$ @Pierre-YvesGaillard Good point. No, in the example given, there is no pullback $A$ in Artin. If there were, there would be a map $\alpha:A\to R_1\cap R_2$ through which every map from an artinian ring factors. But every element of $R_1\cap R_2$ is in a fin.dim. and hence artinian $\mathbb{C}$-subalgebra, so $\alpha$ would have to be surjective. So $R_1\cap R_2$ would be a finitely generated $A$-module, and so would have to be artinian. $\endgroup$ – Jeremy Rickard Feb 1 '19 at 13:47
  • $\begingroup$ You're perfectly right! I should have thought of this... I edited the question. $\endgroup$ – Pierre-Yves Gaillard Feb 1 '19 at 14:23

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