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I am having trouble solving the following problem.

Let $\mathcal B$ be a small filtered subcategory of the category of commutative $k$-algebras, and suppose that all objects in $\mathcal B$ are integral domains. Show that the colimit of this inclusion diagram is an integral domain.

I think I can show that, if $L$ is the colimit of this diagram with maps $\psi_B:B\to L$ for all $B\in\operatorname{ob}(\mathcal B)$, then $L\cong\cup_{B\in\operatorname{ob}(\mathcal B)} \operatorname{Im}(\psi_B)$. Now I suppose that $ab=0$ in $L$, then there exists $x\in B_1$ and $y\in B_2$ in $\mathcal B$ such that $\psi_{B_1}(x)=a$ and $\psi_{B_2}(y)=b$. Then, using the filtered condition, we get maps to a common object $B$ in $\mathcal B$ so $x\mapsto x'\in B$ and $y\mapsto y'\in B$. Now we have $\psi_B(x')=\psi_{B_1}(x)$ and $\psi_B(y')=\psi_{B_2}(y)$, so $0=ab=\psi_B(x')\psi_B(y')=\psi_B(x'y')$.

Now I am stuck, because I don't know that $x'y'=0$ in $B$. If I did, then I could use that $B$ is an integral domain to get the $x'=0$ or $y'=0$, and if $x'=0$ then $a=0$ and if $y'=0$ then $b=0$.

Any help on where I am stuck would be appreciated, or if anyone has an alternative solution then I would also like to see that.

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First, it will be useful to describe the colimit a little more precisely. We have $$ L = \coprod_{B \in \operatorname{Ob}(\mathcal{B})} B / \sim, $$ where the equivalence relation $\sim$ is defined as follows. For $a \in B_1$, $b \in B_2$ we have $a \sim b$ when there are maps $B_1 \xrightarrow{f} B \xleftarrow{g} B_2$ in $\mathcal{B}$ such that $f(a) = g(b)$. The coprojections $\psi_B: B \to L$ are then given by taking the equivalence class of an element.

This has the nice property that whenever we have finitely many elements in $a_1, \ldots, a_n \in L$, there are $a_1', \ldots, a_n' \in B$ for some $B$ in $\mathcal{B}$, such that $a_i = [a_i']$ for all $1 \leq i \leq n$ (this is because $\mathcal{B}$ is filtered). Here $[a_i']$ denotes the equivalence class of $a_i'$.

Then operations on $L$ are defined as follows. For $a, b \in L$, let $a', b'$ be representatives in some $B$. Then we define $ab = [a'b']$, and similar for other operations. Because the maps in the diagram are homomorphisms and the diagram is filtered, this is well-defined.

Now to answer your question: let $a,b \in L$ such that $ab = 0$. Then there are $a', b', z$ in some $B$ such that $a = [a'], b = [b'], 0 = [z]$ and $[a'b'] = [z]$, because of how operations are defined on $L$. Denote by $0_B \in B$ the zero in $B$, then $[0_B] = 0 = [z] = [a'b']$. So using the definition of $\sim$ and filteredness of $\mathcal{B}$ we find $f: B \to B'$ in $\mathcal{B}$ such that $$ f(a')f(b') = f(a'b') = f(z) = f(0_B) = 0_{B'}. $$ So because $B'$ is an integral domain, we see that either $f(a') = 0_{B'}$ or $f(b') = 0_{B'}$. That means that either $[f(a')] = [0_{B'}] = 0$ or $[f(b')] = [0_{B'}] = 0$, and so we conclude that $L$ is an integral domain.

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  • $\begingroup$ Thanks for the answer. It definitely helps when you see the colimit in this form. $\endgroup$ – Dave Oct 24 '19 at 7:19

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