A monomorphism from a ring to a direct sum

Question

Let $R$ be a ring with a family of ideals $A_i$'s ($i\in I)$. We could consider a well-defined $R$-monomorphism from $R/∩A_i$ to the direct product of $R/A_i$'s sending $r+∩A_i$ to the tuple $(r+A_i)_{i\in I}$. But, how could we define an $R$-monomorphism from $R/∩A_i$ to the direct sum of $R/A_i$'s? Of course, when $I$ is finite, this map is the same as that above. Thanks for any cooperation.

Answer 1

For a general collection of ideals $\{A_i\}$, no.

The problem is that the tuple $(r+A_i)_{i\in I}$ is only a member of $\oplus_{i\in I} R/A_i$ when all but finitely many of the elements are zero.

In other words, we can define such a map exactly when our collection of ideals has the following property: for all $r\in R$, $r\in A_i$ for all but finitely many $A_i$.

But this last property is always very boring when $R$ is a ring with $1$, because $1$ does not lie in any ideal except $R$ itself. So this map is defined when all but finitely many of the ideals are equal to $R$—if the ideals are distinct, this means that $I$ is already finite.

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Here is an example of a situation where we cannot possibly hope to get a monomorphism into the direct sum.

Let $R = \mathbb{C}[X]$, and let $I = \{(X-c) \mid c\in \mathbb{C}\}$. Importantly, we have $R/(X-c)\cong \mathbb{C}$ via the epimorphism $f(X)\mapsto f(c)$.

We have $\bigcap_c (X-c) = 0$, and a very nice monomorphism $R\to \prod_c R/(X-c)$, namely $f(X) \mapsto (f(c))_c$. In other words, polynomials are determined by their values.

But $S = \oplus_c R/(x-c)$ is a different story. It's a direct sum of copies of $\mathbb{C}$, and this means that any nonzero element of $S$ generates a $\mathbb{C}$-algebra that is finite-dimensional as a vector space. Since $X\in R$ clearly generates an algebra that is infinite-dimensional as a vector space, there is nowhere for it to be sent by a monomorphism $R\to S$.

(if you would rather work with rings instead of $\mathbb{C}$-algebras, the same example works with $R=\mathbb{Z}[X]$—just replace "dimension" with "rank")