A short question about the direct limits and direct sum of commutative rings.

Question

By viewing a ring as a $\mathbb{Z}$-module, it is possible to define the direct limit of rings following the same procedure for modules. Let's suppose to work with commutative rings with unity. It seems quite evident to me that the direct limit of rings is still a commutative ring, but could not have the neutral element respect to the multiplication. Am I missing something, or is it right? Furthermore, it is the same for an arbitrary direct sum of rings, isn't it?

Thanks in advance.

Cheers

Ps: roughly speaking, the reason that I see is that the multiplication in both cases is defined components by components, and because in a direct sum the number of non zero components is finite, "the element $(1,1,..1,...)$" cannot exists.

Answer 1

There is a unit element because all units in the different $A_i$ (identified with their image in the direct sum) satisfy the equivalence relation. In other words, $(\dots, 0,1\,(i\text{-th place}), 0, \dots)$ is equivalent to $(\dots, 0,1\, (j\text{-th place}), 0, \dots)$, so that, multiplying the class of an element $(x_i)_{i\in I}$ by $1$ amounts to multiply each $x_i$ (which are non-zero for a finite number of them) by the relevant unit.

Answer 2

You are correct that we can't form infinite direct sums in this way while remaining in the category of rings with unity.

However, the direct sum of rings is not supposed to be a direct limit in any case, because the (finite) coproduct is $\otimes$, not $\oplus$. The category of rings is very different from the category of $\mathbb{Z}$-modules, and we shouldn't use direct limits in one to compute direct limits in the other.