Direct product of rings (${\prod }_{\lambda }{R}_{\lambda }$)

The direct product of a family of rings consists of all tuples of their elements, with coordinatewise addition and multiplication. Unlike a direct sum, tuples may have infinitely many nonzero coordinates.

For a commutative ring $k$ and a set $X$, the set $k^X$ of all functions $X\to k$ is a commutative ring under pointwise addition and multiplication. It is the direct product of rings with one copy of $k$ for each $x\in X$.

If $k$ is a field and $X$ is finite, every ideal is generated by some of the singleton indicator functions $e_x$. Indeed, whenever $f$ in the ideal has $f(x)\ne 0$, multiplication by $e_x/f(x)$ puts $e_x$ in the ideal. Every function in that ideal is a finite sum of its coordinate multiples of these $e_x$.

If $X$ is infinite and $k$ is a nonzero commutative ring, the functions with finite support form an ideal without a finite generating set of an ideal. A finite collection of such functions vanishes outside one finite subset, as does the ideal it generates, whereas a singleton indicator function outside that subset belongs to the ideal of all finitely supported functions. Thus, for a field $k$, $k^X$ is a Noetherian ring exactly when $X$ is finite.

For a field $k$, the shift on the direct product of rings ${\prod }_{n\ge 0}k$ preserves coordinatewise addition, multiplication and the unit. It is a surjective ring endomorphism because an arbitrary first coordinate can be prepended to a desired output. Its kernel of a ring homomorphism consists of the sequences supported in coordinate zero, so it is not injective. The ring is not Noetherian: the ideals generated by the first $n+1$ coordinate idempotents form a strictly ascending chain.

A quotient ring of a finite direct product of rings whose factors are fields is a product of a subset of those factors. By ideals of a finite product of rings, write $J={\prod }_j{J}_j$. Since a field has only the zero ideal and the whole field as ideals, every $J_j$ is one of these. The ring homomorphism $(x_j)\mapsto (x_j+J_j)$ is surjective with kernel of a ring homomorphism $J$, so the first isomorphism theorem for rings gives $({\prod }_j{K}_j)/J\cong {\prod }_j(K_j/J_j)$. Discard the zero ring factors. If no factors survive, the result is the zero ring.

Every two-sided ideal $I$ of a finite direct product of rings $T={\prod }_{j=1}^n{R}_j$ is ${\prod }_j{I}_j$, where $I_j$ is its image under the $j$th coordinate ring homomorphism. These images are ideals because the coordinate maps are surjective. To prove the reverse inclusion, lift $a_j\in I_j$ to $b_j\in I$ and multiply by the coordinate idempotent $e_j=(0,\dots ,1,\dots ,0)$. Then ${\sum }_j{e}_j{b}_j=(a_1,\dots ,a_n)\in I$. Finiteness matters: in ${\prod }_{j\ge 0}\mathbb{Z}$, the finitely supported sequences form an ideal whose every coordinate image is $\mathbb{Z}$, but which is not the whole product.

A tuple in a direct product of rings is a unit exactly when each coordinate is a unit. Indeed, multiplying by an inverse tuple gives an inverse in each coordinate; conversely, the coordinate inverses form an inverse tuple. This identifies the unit group with the direct product of groups of the individual unit groups, for finite or infinite families of unital rings.