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The Chinese remainder theorem says that remainders modulo pairwise coprime positive integers can be specified independently. For $m_1,\dots ,m_r$ pairwise coprime and $M=m_1\cdots m_r$, every prescribed list $a_1,\dots ,a_r$ determines one congruence class modulo $M$ satisfying $x\equiv a_j(\mathrm{mod}{m}_j)$ for all $j$. It determines a class, rather than one unrestricted integer: adding a multiple of $M$ preserves every remainder.

For example, combine $x\equiv 2(\mathrm{mod}3)$ and $x\equiv 3(\mathrm{mod}5)$. Write $x=2+3k$ from the first condition. The second becomes $3k\equiv 1(\mathrm{mod}5)$. Since three has modular inverse two modulo five, $k\equiv 2(\mathrm{mod}5)$, giving $x\equiv 8(\mathrm{mod}15)$. The integers $8$, $23$ and $-7$ all work. The fifteen possibilities modulo fifteen correspond to the fifteen pairs in a three-by-five grid.

The theorem also preserves arithmetic: reducing an addition or a multiplication modulo each $m_j$ gives the same tuple as performing that operation on the corresponding congruence classes. In the language of quotient rings, reduction is a ring isomorphismOne large residue is thus an exact encoding of several smaller residues. This is useful for splitting arithmetic into smaller calculations and then reconstructing the result; MIT's modular arithmetic notes develop both the reconstruction and the product representation. ^[ref] For a concrete multiplication, $7$ corresponds to $(1,2)$ and $8$ to $(2,3)$ modulo $(3,5)$. Their componentwise product is $(2,1)$, which reconstructs to $11(\mathrm{mod}15)$, agreeing with $56\equiv 11(\mathrm{mod}15)$.

A constructive formula explains the independence. For $m_j>1$, put $M_j=M/m_j$ and choose $u_j$ with $M_j{u}_j\equiv 1(\mathrm{mod}{m}_j)$, using a modular multiplicative inverse. Then $e_j=M_j{u}_j$ is one modulo $m_j$ and zero modulo every other $m_i$. Thereforehas the desired remainders. A modulus equal to one imposes no restriction and can be omitted. Uniqueness follows because a difference divisible by every $m_j$ is divisible by their product when they are pairwise coprime. This short construction gives a reusable algorithm as well as an existence statement.

The coprime hypothesis explains why every cell in the left grid is attainable. With integer moduli six and eight, a pair must agree modulo their greatest common divisor, two. There are twenty-four compatible pairs, rather than forty-eight arbitrary pairs; each corresponds to a class modulo the least common multiple, twenty-four. The Chinese remainder theorem for noncoprime moduli states the resulting compatibility rule.

For pairwise comaximal ideals in a commutative ring, reduction has the displayed ring isomorphism. The kernel is their intersection, equal to their product. Choose $e_i\in {\prod }_{j\ne i}{I}_j$ with $e_i\equiv 1(\mathrm{mod}{I}_i)$; then ${\sum }_i{a}_i{e}_i$ lifts any prescribed residue tuple. The equality between intersection and product follows from $1=u+v$, $u\in I$, $v\in J$, which writes each $x\in I\cap J$ as $xu+xv\in IJ$.

An isomorphism is a morphism with a two-sided inverse that is also a morphism. Two objects are isomorphic when such a map exists between them. For groups, rings, and vector spaces, a bijective structure-preserving map has a structure-preserving inverse; in other settings the inverse condition needs separate attention, as for a homeomorphism between topological spaces.

For example, $f(k)=2k$ is a group isomorphism from the additive integers to the additive even integers. It is bijective onto that target, satisfies $f(a+b)=f(a)+f(b)$, and has inverse $2k\mapsto k$. The two groups have the same addition pattern even though they are different sets. It is not a ring isomorphism for their inherited multiplications, since $f(ab)=2ab$ while $f(a)f(b)=4ab$.

Name every integer and add a new constant asserted to be prime and unequal to every named integer. Every finite part is satisfied by an ordinary prime outside a finite list, using Euclid's theorem. Compactness theorem with the elementary diagram gives an elementary extension with the desired prime. A countable elementary substructure retaining that prime cannot be isomorphic to the integer ring, since any unital ring isomorphism fixes every standard integer.

By the first isomorphism theorem for rings, the given surjection identifies $S$ with $R/\ker \varphi$. Since $R$ is a principal ideal domain, write $\ker \varphi =(a)$.

If $a=0$, the quotient ring is $R$, an integral domain, so part (a)(ii) applies. If $a$ is a unit, the quotient ring is the zero ring; its unique element is associate to itself by multiplication by its identity, so it too is an associator ring.

In the remaining case, unique factorization giveswhere $\epsilon$ is a unit, the $p_i$ are pairwise nonassociate prime elements, and $k_i\ge 1$. The principal ideals $(p_i^{k_i})$ are pairwise comaximal: different prime factors have greatest common divisor $1$, and Bézout's identity gives a linear combination equal to $1$. The Chinese remainder theorem therefore gives a ring isomorphismEvery factor is an associator ring by part (c), and their finite direct product of rings is an associator ring by part (b). A ring isomorphism preserves units and principal ideals, so the property transfers to $S$. Hence

Take $R=\mathbb{Z}$, $I=a\mathbb{Z}$ and $J=b\mathbb{Z}$. Since $a,b$ are coprime integers, Bézout's identity gives $ra+sb=1$, so $I+J=\mathbb{Z}$. Their intersection is $ab\mathbb{Z}$: it consists of integers divisible by both $a$ and $b$, and coprimality makes this equivalent to divisibility by $ab$. Thus the Chinese remainder theorem givesas a ring isomorphism. This includes $a=1$ or $b=1$, where the corresponding factor is the zero ring.

The first isomorphism theorem for rings states that the mapis a ring isomorphism. It is well defined because two representatives differ by an element of the kernel exactly when their images agree. It preserves addition and multiplication, is onto its stated codomain, and has trivial kernel. In particular, if $f$ is surjective then $R/\ker f\cong S$.

Factorization over $\mathbb{Q}$ givesFor $R_1$, the factors $x-1$ and $x+1$ are coprime. The nonzero elements $(1,0)$ and $(0,1)$ multiply to zero, so the ring is neither an integral domain nor a field. Squaring coordinatewise shows it has no nonzero square-zero element.

For $R_2$, the polynomial $x^2+1$ has no rational root and is irreducible of degree two over $\mathbb{Q}$. The quotient is consequently a field, hence an integral domain, and has no nonzero square-zero element.

For $R_3$, the class $\epsilon$ is nonzero by its degree, but ${\epsilon }^2=0$. Thus this ring is neither an integral domain nor a field, and does have a nonzero square-zero element.No two are isomorphic. Being a field separates $R_2$ from the other two; existence of a nonzero square-zero element separates $R_1$ from $R_3$. Both properties are preserved by ring isomorphisms.

No. Let $k={\mathbb{F}}_2$ be the field with two elements, put $A=k[\epsilon ]/({\epsilon }^2)$, and take the closed immersioninduced by the ring homomorphism $q:A\twoheadrightarrow k$ with $q(\epsilon )=0$.

For every field $K$, a ring homomorphism $\alpha :A\to K$ must send $\epsilon$ to zero: $\alpha (\epsilon {)}^2=0$, and a field has no nonzero nilpotent elements. Consequently $\alpha$ factors uniquely through $q$. Passing from ring homomorphisms to morphisms of schemes between affine schemes gives a bijectionThis concerns all morphisms of schemes, without requiring them to be over $k$: if $K$ does not have characteristic two, both sets are empty; otherwise each contains exactly one element.

Nevertheless, $f$ is not an isomorphism of schemes. Its map on global sections is $q$, whose kernel of a ring homomorphism contains the nonzero element $\epsilon$, so $q$ is not a ring isomorphism. The target is the nonreduced scheme defined by the dual numbers, whereas the source is a reduced scheme. This illustrates why field-valued points do not detect nilpotents.

The immersions and gluing. For $p>0$, the prime ideals of the localization ${\mathbb{Z}}_{(p)}$ are $(0)$ and $(p){\mathbb{Z}}_{(p)}$. Inverting $p$ gives ${\mathbb{Z}}_{(p)}[1/p]=\mathbb{Q}$, sois an open immersion sending the generic point to the generic point. For $p=0$, ${\mathbb{Z}}_{(0)}=\mathbb{Q}$, and take ${\psi }_0$ to be the identity. Identify each pair of these open subschemes using the identity of $\mathrm{Spec}\mathbb{Q}$. The identities compose to the identity on every triple overlap, so gluing of schemes along open subschemes applies. No finiteness of the family is required.

The resulting scheme $Y$ has one common generic point $\eta$ and a distinct closed point $y_p$ for every positive prime number $p$. Its affine open subschemes $U_p=\{\eta ,y_p\}$, for $p>0$, intersect pairwise in $U_0=\{\eta \}$. A subset $W\subseteq Y$ is open precisely when its intersection with every $U_p$ is open. In the two-point topological space $U_p$, the open subsets are $\varnothing$, $\{\eta \}$ and $U_p$. Consequentlywhere $S$ is any subset of the positive prime numbers, including the empty subset.

The canonical morphism. The ring homomorphisms $\mathbb{Z}\to {\mathbb{Z}}_{(p)}$ give morphisms of schemes $U_p\to \mathrm{Spec}\mathbb{Z}$ agreeing on the overlaps, hence a morphism of schemes $f:Y\to \mathrm{Spec}\mathbb{Z}$. By uniqueness of the structure morphism, this is the required map. Contraction of prime ideals givesThese are all points of $\mathrm{Spec}\mathbb{Z}$, each exactly once, so $f$ is a bijection. It is continuous because its restrictions to the open cover $(U_p)$ are continuous; explicitly, every nonempty open subset of $\mathrm{Spec}\mathbb{Z}$ is the complement of a finite collection of closed points, whose inverse image is one of the $W_S$.

Sections on every open. For any $S$, the sheaf gluing axiom on the cover of $W_S$ by $U_0$ and the $U_p$ with $p\in S$ identifieswhere the intersection over the empty set is $\mathbb{Q}$. Indeed, a section on $U_p$ is an element of ${\mathbb{Z}}_{(p)}$; restrictions to every overlap are its inclusion into $\mathbb{Q}$. Compatibility means that all these elements are the same rational number. In lowest terms $a/b$, this rational number lies in the intersection exactly when no prime number in $S$ divides $b$. Equivalently, this is the integer localization at a set of primes obtained by inverting all positive prime numbers outside $S$.

Now let $V\subseteq \mathrm{Spec}\mathbb{Z}$ be nonempty and open. Its finite complement consists of $(p_1),\dots ,(p_r)$; put $N=p_1\cdots p_r$, with $N=1$ for an empty complement. Then $V=D(N)$ andThe equality follows by the denominator criterion above. The map $f^{\#}$ is this same equality of subrings of $\mathbb{Q}$, so it is a ring isomorphism. Over the empty open subset both rings of sections are the zero ring. These identifications respect every restriction map. Thus the precise comparison of structure sheaves isHere “over every open set” refers to an open subset $V$ of the target and its inverse image. The preceding intersection formula additionally describes the sections on every open subset of $Y$, including ones whose images are not open in the target.

Topology and finiteness. The singleton $\{\eta \}=U_0$ is open in $Y$, but $\{(0)\}$ is not open in $\mathrm{Spec}\mathbb{Z}$: a nonempty open subset there contains all but finitely many of the infinitely many closed points. Therefore $f$ is not a homeomorphism, despite being a continuous bijection and giving the indicated isomorphism of structure sheaves after direct image.

The open cover $(U_p{)}_{p>0}$ of $Y$ has no finite subcover, because $y_q$ belongs only to $U_q$. Thus $Y$ is not a quasi-compact scheme, and part (a) rules out an affine scheme. Each ${\mathbb{Z}}_{(p)}$ is a Noetherian ring, being a localization of the Noetherian ring $\mathbb{Z}$. Hence the $U_p$ form an affine open cover by spectra of Noetherian rings, proving that $Y$ is a locally Noetherian scheme. A Noetherian scheme is a quasi-compact scheme, so

Reducing coefficients modulo $p$ gives the ring isomorphismIf $d=\mathrm{deg}f>0$, its monic leading coefficient remains one modulo $p$, so $\mathrm{deg}\bar{f}=d$. Polynomial division gives a unique representative of degree less than $d$: existence follows by division, and the difference of two such representatives cannot be a nonzero multiple of a degree-$d$ monic polynomial. There are $p$ choices for each of the $d$ coefficients, provingIrreducibility of $\bar{f}$ is not needed for finiteness; without it the quotient ring need not be a field.

The projections $P_T$ in the Gelfand–Tsetlin algebra satisfyConsequently the map sending a diagonal element to its scalar on each Gelfand–Tsetlin basis vector is a ring isomorphismEach factor is a simple module over this algebra, and the algebra is their direct sum as a module over itself. This proves that $G{Z}_n$ is a semisimple algebra. The argument uses its explicit diagonal structure; being a subalgebra of a semisimple algebra by itself would not suffice.

The semicubical parabola has coordinate ringunder $x\mapsto t^3$, $y\mapsto t^2$. The isomorphism is valid in every characteristic: after division by the monic polynomial in $x$, a remainder $P(y)+xQ(y)$ maps to a sum of distinct even powers and odd powers at least three, and can vanish only when both polynomials are zero.

Its fraction field contains $t=t^3/t^2$, which is an integral element over $A$ because it satisfies $z^2-y=0$. However $t\notin A$, since a polynomial in $t^2,t^3$ has no term of degree one. Thus $A$ is not an integrally closed domain. The ring $k[t]$ of the affine line is a principal ideal domain and therefore an integrally closed domain. This property is preserved by ring isomorphism. By coordinate-ring functoriality,The parametrization is a bijection on points over an algebraically closed field, but that does not make its inverse a morphism at the cusp.

The Chinese remainder theorem says that if the positive integers $m_1,\dots ,m_r$ are pairwise coprime and $M={\prod }_i{m}_i$, then for arbitrary residue classes $a_i(\mathrm{mod}{m}_i)$ there is a unique residue class $x(\mathrm{mod}M)$ satisfyingEquivalently, reduction is a ring isomorphism

For existence, put $M_i=M/m_i$. Pairwise coprimality gives $\gcd (M_i,m_i)=1$, so Bézout's identity supplies an integer $u_i$ with $u_i{M}_i\equiv 1(\mathrm{mod}{m}_i)$. Thenhas the required residues, because every term except the $i$th is divisible by $m_i$. If $x$ and $y$ are two solutions, each $m_i$ divides $x-y$. Pairwise coprimality then implies that their product $M$ divides $x-y$, proving uniqueness modulo $M$.

Let $\mathfrak{p}\in \mathrm{Proj}T$. Some homogeneous $t\in T_{+}$ lies outside $\mathfrak{p}$. Choose $N$ with $N\mathrm{deg}t\ge d_0$. Since ${\phi }_{N\mathrm{deg}t}$ is surjective, $t^N=\phi (s)$ for some homogeneous $s\in S_{+}$; primality gives $t^N\notin \mathfrak{p}$. Thus $\mathfrak{p}\in U$, and $U=\mathrm{Proj}T$.

Fix a positive-degree homogeneous $s\in S$ and write $q=\phi (s)$. The mapis surjective: after multiplying the numerator and denominator of any degree-zero fraction by a sufficiently large power of $q$, its numerator has degree at least $d_0$ and therefore lifts through $\phi$. It is injective by the same device: if $a/s^n$ maps to zero, then $q^m\phi (a)=0$ for some $m$, and after increasing $m$ the injectivity of $\phi$ in degree $(m+n)\mathrm{deg}s$ gives $s^{m}a=0$. Hence the displayed map is a ring isomorphism.

The opens $D_{+}(q)$ obtained in this way cover $\mathrm{Proj}T$, and on every one of them $f$ is an isomorphism onto $D_{+}(s)$. The inverses agree on overlaps, so $f$ is an isomorphism of schemes. This proves the Invariance of Proj under an eventual graded isomorphism.

For $a,b\in I$, the square-zero ideal condition givesThus the square-zero unit subgroup $1+I$ is abelian, andis a group isomorphism from the additive group of $I$.

Use the specified ring isomorphism $R/I\cong \mathbb{Z}G$. For $x\in \mathbb{Z}G$, choose a lift $r\in R$ and define $x\cdot a=ra$ for $a\in I$. Two lifts differ by an element of $I$, whose product with $a$ vanishes, so this is well defined. Right multiplication is handled identically. The two actions commute by associativity, making $I$ a bimodule. If $u\in R$ lifts $g\in G$, then $u$ is a unit: a lift $v$ of $g^{-1}$ makes both $uv$ and $vu$ elements of $1+I$, hence units, and a ring element with both a left and a right inverse is invertible. Conjugation therefore definesChanging $u$ by an element of $I$ does not change this expression because $I^2=0$. Moreover,so $a\mapsto 1+a$ is an isomorphism of $\mathbb{Z}G$-modules for these conjugation actions.

Let $R^{\times }\to (R/I{)}^{\times }$ be reduction on unit groups, and define $U$ as the inverse image of the distinguished subgroup $G\subseteq (\mathbb{Z}G{)}^{\times }$. Every $g\in G$ has a unit lift by the preceding argument, and the kernel consists exactly of the units congruent to $1$, namely $1+I$. Multiplication in $R$ therefore gives the group extensionChoose a set-theoretic section $s:G\to U$ with $s(1)=1$. Its extension cocyclesatisfies the two-cocycle identity by associativity. A different section changes $c$ by a group coboundary, so second group cohomology classifies group extensions gives a well-defined classThe same construction for $(R_1,I_1)$ gives $U_1$, an extension cocycle $c_1$, and $x_1=[c_1]\in H^2(G,1+I_1)$.

The answer to the final question is no. An abstract ring isomorphism $R_1\cong R$ need not carry the distinguished ideal $I_1$ to $I$, need not induce the identity under the two chosen identifications of the quotient rings with $\mathbb{Z}G$, and need not induce the prescribed $\mathbb{Z}G$-module isomorphism $\theta$. Hence it need not give an isomorphism of the two displayed group extensions, so it imposes no equality $\varphi (x_1)=x$. That equality does hold if the ring isomorphism has all three compatibility properties, because it then carries one extension cocycle to the other up to a group coboundary.

Apply downward Löwenheim-Skolem to the extension from part (i), keeping all standard integers and the new prime $c$. The first-order language and this prescribed subset are countable, so there is a countable elementary substructure $N$ containing them. In particular $N\equiv \mathcal{Z}$ and contains an element outside the standard copy of $\mathbb{Z}$.

No ring isomorphism $N\to \mathcal{Z}$ can exist. Such an isomorphism fixes $0$ and $1$, hence every nonnegative integer numeral. It also fixes each negative integer, which is the unique solution of $x+\bar{n}=0$ for the corresponding positive $n$. Consequently it fixes the whole standard copy. The image of $c$ would be some standard integer $a$, already the image of that same named standard element, violating injectivity.

Thus $N$ and $\mathcal{Z}$ are two nonisomorphic countably infinite first-order models of $T$, so

Under the unital ring convention used elsewhere in this wiki, no such pair exists. In a two-element ring the elements are $0$ and $1$. Additive cancellation forces $1+1=0$, and the identity axiom forces $1\cdot 1=1$. The distributive property forces products involving zero to vanish. Every operation is therefore determined, giving the prime field ${\mathbb{F}}_2$ as the unique possibility up to ring isomorphism.

If the word ring allows a nonunital ring, there are instead two non-isomorphic examples: ${\mathbb{F}}_2$ and the zero-multiplication rng on the additive group $\mathbb{Z}/2\mathbb{Z}$. In the second example the unique nonzero element $u$ satisfies $u+u=0$ and $u^2=0$; every product is zero, so associativity and both distributive laws hold. An isomorphism would have to send $1$ to $u$ because it preserves zero and is bijective, but then it would send $1^2=1$ to both $u^2=0$ and $u$, which is impossible. These exhaust the rngs with two elements, since $u^2$ can only be $0$ or $u$. Specifying the identity convention is essential for this part.

The three answers, in the given order, are no, yes, yes.

For $\mathbb{Z}[x]$, consider the ideal $I=(2,x)$. Its members have even constant coefficient, so $1\notin I$ and the ideal is proper. If $I=(g)$, then $g$ divides the constant $2$, forcing $g$ to be constant by additivity of polynomial degree in this integral domain. Since $g$ also divides $x$, it must divide the coefficient $1$, giving $g=\pm 1$. That would make $I$ the whole ring, a contradiction. Thus $\mathbb{Z}[x]$ is not a principal ideal domain, and part (a) shows it cannot be a Euclidean domain. This is the nonprincipal ideal generated by a prime and an indeterminate obstruction.

Evaluation at $i$ gives the ring isomorphismIndeed, division by the monic polynomial $x^2+1$ leaves a unique remainder $a+bx$, and its value $a+bi$ is zero only when both integer coefficients vanish. Thus the quotient maps bijectively, with operations preserved, onto the Gaussian integers. These form a Euclidean domain for the Gaussian integer norm $N(a+bi)=a^2+b^2$, a standard Euclidean example allowed in this part. Geometrically, rounding the real and imaginary coordinates gives a quotient error with squared modulus at most $1/2<1$.

Finally,via $f(x,y)\mapsto f(0,y)\mathrm{mod}2$. Every polynomial over ${\mathbb{F}}_2$ occurs. The kernel consists exactly of polynomials whose $x$-independent part has even coefficients: write $f(x,y)=xg(x,y)+h(y)$ to see that this is precisely $(2,x)$. Hence the induced quotient map is a bijective ring homomorphism. Its target is a Euclidean domain with Euclidean function $\mathrm{deg}$: a polynomial ring over a field is Euclidean by the division algorithm in part (a).

Regard $\mathbb{R}[X,Y]$ as the polynomial ring $(\mathbb{R}[X])[Y]$. The coefficient ring $\mathbb{R}[X]$ is a Euclidean domain, hence a principal ideal domain. The element $X$ is a prime element: the quotient ring $\mathbb{R}[X]/(X)\cong \mathbb{R}$ is an integral domain.

The monic polynomial $Y^2+X$ has its other coefficients divisible by $X$, while $X^2\nmid X$. Part (a)(i), with $X$ playing the role of $p$ and $Y$ the indeterminate, proves that $Y^2+X$ is an irreducible polynomial.

Now use the automorphism of a polynomial ringIts inverse fixes $X$ and sends $Y$ to $Y-1$. A ring isomorphism preserves units and factorizations, hence preserves irreducible elements. Sincewe obtainApplying the Eisenstein criterion directly to the second expression with the prime element $X$ would fail, since $X\nmid 2$; the change of indeterminate supplies the required argument.