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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.

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.

The printed domain assertion is false. For example, take $K=\mathbb{Q}$ and $I=(4)$. Then ${\mathcal{O}}_K/I=\mathbb{Z}/4\mathbb{Z}$ has the nonzero zero divisor $2+(4)$, so it is not an integral domain and cannot be a principal ideal domain. The valid conclusion is that ${\mathcal{O}}_K/I$ is a principal ideal ring. This error is present in the original PDF, not just its transcription.

To prove the corrected conclusion, factorinto powers of distinct nonzero prime ideals. These powers are pairwise comaximal ideals. Indeed, a maximal ideal containing two such powers would contain both underlying prime ideals, forcing them to be equal. The Chinese remainder theorem givesEvery factor is a principal ideal ring by the preceding part. Every ideal $H$ in a finite direct product of rings is a product $H_1\times \cdots \times H_s$ of ideals in the factors: multiplication by the coordinate idempotent elements separates the coordinates of every element of $H$. Write $H_j=(h_j)$. Thenso $R/I$ is a principal ideal ring. This proves the principal ideal property of quotients of a Dedekind domain in the present setting. In fact, this finite quotient ring is an integral domain precisely when $I$ is a prime ideal, in which case it is a field.

For the final deduction, choose any $0\ne a\in I$ and apply the result to the nonzero proper ideal $(a)$. The ideal $I/(a)$ of $R/(a)$ is generated by one residue class $b+(a)$, which can be represented by $b\in I$. For each $x\in I$, some $r\in R$ satisfies $x-rb\in (a)$. HenceThis proves two generators for an ideal in a Dedekind domain; moreover, its first generator can be any prescribed nonzero element of $I$.

Use the direct product of ringswith coordinatewise addition and multiplication and unit $(1,1,\dots )$. Define the shiftCoordinatewise operations show that $g$ preserves addition, multiplication and the unit, so it is a ring endomorphism. It is surjective because $(0,b_0,b_1,\dots )$ maps to any prescribed $(b_0,b_1,\dots )$. Its kernel of a ring homomorphism isso it is not injective. This is the noninjective surjective shift on a product of fields.

This ring fails the Noetherian hypothesis in c. Indeed, let $e_n$ have a single $1$ in coordinate $n$. The ideals $I_n=(e_0,\dots ,e_n)$ form the strictly ascending chainsince $e_{n+1}\notin I_n$.

A ring is reduced if its only nilpotent element is zero. Equivalently,The second expression says that its nilradical vanishes. A reduced ring may have zero divisors; for example, the direct product of rings $\mathbb{R}\times \mathbb{R}$ is reduced but $(1,0)(0,1)=0$.

Yes: $R$ can have nonzero zero divisors. Let $R=k\times k$, a direct product of rings, where $k$ is any field. The nonzero idempotents $e_1=(1,0)$ and $e_2=(0,1)$ satisfy $e_1{e}_2=0$, so both are zero divisors.

There are exactly two prime ideals:Indeed, any prime ideal must contain $e_1$ or $e_2$, since their product is zero. If it contains $e_1$, it contains $k\times 0$ and corresponds to a prime ideal of the quotient ring $k$, whose only prime ideal is zero; hence it equals $P_2$. The other case gives $P_1$.

For the localization at a prime ideal $P_1$, every denominator $(u,v)\notin P_1$ has $u\ne 0$, andis an isomorphism. It is surjective using constant first-coordinate fractions. It is injective because a numerator with first coordinate zero is annihilated by $e_1\notin P_1$, so its fraction is zero. Interchanging the coordinates proves $R_{P_2}\cong k$.

Thus the localization at a prime ideal is a field in every case, although $R$ has zero divisors. The obstruction to using the previous part's argument is that the two nonzero factors of a zero product can survive at different prime ideals.

Let $e_j$ be the sequence with $1$ in coordinate $j$ and zero elsewhere. In the direct product of rings $R=k^{\mathbb{N}}$, the idealsform a strict ascending chain: every element of $(e_1,\dots ,e_n)$ vanishes after coordinate $n$, whereas $e_{n+1}$ does not. Hence $R$ is not a Noetherian ring.

Nevertheless, each $a=(a_j)\in R$ has a coordinatewise generalized inverse $b=(b_j)$, defined by $b_j=a_j^{-1}$ when $a_j\ne 0$ and $b_j=0$ otherwise. It satisfiesThis is the defining property of a Von Neumann regular ring. Fix any prime ideal $P$, and take $a\in P$. Since $ab\in P$, $1-ab\notin P$; butConsequently $a/1=0$ in the localization at a prime ideal $R_P$. All elements of $P{R}_P$ therefore vanish. By the local ring description in part (a), $P{R}_P$ is the unique maximal ideal, so $R_P$ is a nonzero field. A field has only the ideals zero and itself, and thus is Noetherian.

We have proved the stronger conclusionThe argument applies to every prime ideal, without assuming that it comes from a coordinate projection.

Multiplication in a direct product of rings is coordinatewise. Thus $(r_1,\dots ,r_n)$ has a two-sided inverse exactly when each $r_j$ has a two-sided inverse in $R_j$, and thenConsequently its unit group is the direct product of groupsThe identification preserves multiplication because both sides use coordinatewise multiplication. This is the unit group of a direct product of rings formula.

For a Dedekind domain $R$ and a nonzero proper ideal $I$, the quotient ring $R/I$ is a principal ideal ring. First, the correspondence theorem for ideals and unique factorization of ideals in a Dedekind domain show that the ideals of $R/P^n$ are exactly $P^i/P^n$, $0\le i\le n$. Consecutive powers are distinct by cancellation of invertible fractional ideals. Choose $\pi \in P^i\setminus P^{i+1}$. There is no intermediate ideal between these consecutive powers, so $P^i=(\pi )+P^{i+1}$. Multiplication by $P^r$ and substitution show inductively that $P^i=(\pi )+P^{i+r}$ for every $r\ge 1$. Taking $r=n-i$ gives a single generator modulo $P^n$. Now factor $I$ into pairwise comaximal ideals $P_j^{n_j}$. The Chinese remainder theorem writes $R/I$ as a finite direct product of rings of the preceding type, which is a principal ideal ring. It need not be an integral domain, as $\mathbb{Z}/4\mathbb{Z}$ demonstrates. Applying the result to $(a)$ for any $0\ne a\in I$ makes $I/(a)$ cyclic, yielding two generators for an ideal in a Dedekind domain.

A principal ideal ring is a ring whose left ideals and right ideals each have one generator. For a commutative ring, this says that every ideal is a principal ideal. The examples here use commutative rings. Unlike a principal ideal domain, it may have zero divisors: $\mathbb{Z}/4\mathbb{Z}$ has this property, with ideals $(0)$, $(2)$, and $(1)$. A finite direct product of rings that are principal ideal rings is again a principal ideal ring. Indeed, the coordinate idempotent elements show that every ideal is a product of coordinate ideals; a tuple of their generators generates the product.

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.

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.

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.

A ring is von Neumann regular if every element $a$ has an element $b$ with $a=aba$. In a commutative ring, this is $a=a^{2}b$. An arbitrary direct product of rings whose factors are fields has this property by taking inverses in each nonzero coordinate. Every localization at a prime ideal of a commutative von Neumann regular ring is a field.