Why there is no free-forgetful adjunction on radical ideals? Given a ring $R$ we can build two poset categories: the category of the ideals of $R$ ordered by inclusion $Ideals(R)$ and the category of radical ideals of $R$: $RadId(R)$.
It is obvious we have an inclusion functor $$For:RadId(R)\to Ideals(R),$$ making one a subcategory of the other. The problem is that we have another functor in the other direction: $$Rad:Ideals(R)\to RadId(R),$$ which sends each ideal to its radical.
By the characterization of the radical ideal as the intersection of all prime ideal containing it: $$Rad({\bf a})= \bigcap_{{\bf p}\in Spec(R),{\bf a \subseteq p}}{\bf p},$$ it becomes obvious that it is a functor, and that ${\bf a\subseteq} Rad({\bf a})$.
From this we get the Free-Forgetful adjunction:
$$Rad({\bf a})\subseteq {\bf b} \iff {\bf a} \subseteq For({\bf b})\hspace{3mm} \forall {\bf b}\in RadId(R),{\bf a}\in Ideals(R).$$
Knowing that the category $Ideals(R)$ is complete and cocomplete and knowing that $Rad$ is a left adjoint we should trivially get that $Rad$ commutes with colimits and in particular: $$Rad(\oplus_{i\in I}{\bf a}_i)=\oplus_{i\in I}Rad({\bf a}_i).$$ The problem is that we know that this is false even in the simplest case seeing that $(x^2-y)$ and $(y)$ are radical ideals in $R=K[x,y]$ but $(x^2-y)+(y)=(x^2,y)$ is not radical.
I should understand that probably $RadId(R)$ is not cocomplete, but where is the problem with this reasoning? We have an isomorphism of functors: $$\prod_{i\in I}Hom(Rad({\bf a}_i),-)\simeq Hom(Rad(\oplus_{i\in I}{\bf a}_i),-),$$ so the colimit should exist.
If someone could pinpoint the problem I would be very thankful.
 A: In a poset, the coproduct of two objects $X$ and $Y$ is the smallest object $Z$ which is bigger than both $X$ and $Y$.
You have your functor $\sqrt{-} : \mathsf{Id}(R) \to \mathsf{RadId}(R)$. You want to check that it commutes with colimits for $a_1 = (x^2-y)$ and $a_2 = (y)$ as object of $\mathsf{Id}(R)$. Their coproduct in $\mathsf{Id}(R)$ is indeed $a_1 + a_2$, the smallest ideal which contains both $a_1$ and $a_2$. Thus on the one hand you have $\sqrt{a_1 + a_2}$.
On the other hand you have the two ideals $\sqrt{a_1}$ and $\sqrt{a_2}$ (never mind that they are equal to $a_1$ and $a_2$ for a moment), and you want to compute their colimit in $\mathsf{RadId}(R)$. In other words, you're looking for the smallest radical ideal which contains $\sqrt{a_1}$ and $\sqrt{a_2}$. It's easy to check that this is $\sqrt{a_1 + a_2}$: it is a radical ideal which contains $\sqrt{a_1}$ and $\sqrt{a_2}$, and if $I$ is another such ideal, then it must contain $a_1 + a_2$ (because it contains $a_1$ and $a_2$) and since it's radical it must contain $\sqrt{a_1+a_2}$.
Thus you do get the equality:
$$\sqrt{a_1 \oplus a_2} = \sqrt{a_1} \oplus \sqrt{a_2}$$
where the first colimit is taken in $\mathsf{Id}(R)$, but the second colimit is taken in $\mathsf{RadId}(R)$.
