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In the quotient ring, the classes of $x-1$ and $x^2+x+1$ are nonzero: neither nonzero polynomial of degree less than three belongs to the ideal generated by $x^3-1$. But their product is the zero class, sinceThe quotient thus contains zero divisors, so it is not an integral domain.

The Hilbert basis theorem says that if $B$ is a commutative Noetherian ring, then $B[t]$ is Noetherian; iteration treats any finite number of variables. Here is a complete leading-coefficient proof. For an ideal $I\subseteq B[t]$, let $L_d$ be the set of degree-$d$ coefficients of its polynomials of degree at most $d$, including zero coefficients. These are ideals of $B$, and multiplication by $t$ gives $L_d\subseteq L_{d+1}$. The ascending chain condition makes the chain stationary from some $N$ onward. Choose finite generators for $L_0,\dots ,L_N$ and lift them to polynomials $f_{dj}\in I$ of degree at most $d$. If $f\in I$ has degree $m$, set $d=\min (m,N)$ and express its leading coefficient as ${\sum }_j{a}_{dj}{b}_j$, where $a_{dj}$ is the chosen degree-$d$ coefficient. Then $f-{\sum }_j{b}_j{t}^{m-d}{f}_{dj}$ has smaller degree. Induction proves that the finitely many lifts generate $I$. This proves the theorem, including coefficient rings with zero divisors.

Put $A=\mathbb{C}G$ and $L=\mathbb{C}[X,X^{-1}]$. The infinite dihedral group has normal forms $X^m$ and $X^{m}Y$. The relations move every $Y$ to the right and reduce its exponent modulo two; the action $X:n\mapsto n+1$, $Y:n\mapsto -n$ on $\mathbb{Z}$ shows that these forms are distinct. ThusThe Hilbert basis theorem makes $\mathbb{C}[X]$ Noetherian, and its localization $L$ is Noetherian: an ideal of a localization is generated by the images of the finite generators of its contraction. Since $A$ is free of rank two over $L$ on the left, it is a Noetherian module over $L$. Every left ideal of $A$ is an $L$-submodule, so all its ascending chains stabilize. Therefore $\mathbb{C}G$ is left Noetherian.

For the center of the infinite dihedral group algebra, write an element as $f(X)+g(X)Y$. Commuting it with $X$ gives $g(X)(X^{-1}-X)=0$ in the Laurent integral domain, forcing $g=0$. Commuting $f(X)$ with $Y$ then gives $f(X)=f(X^{-1})$. A symmetric Laurent polynomial is a linear combination of $1$ and $X^j+X^{-j}$, $j\ge 1$. With $z=X+X^{-1}$, the recursionexpresses $S_j=X^j+X^{-j}$ as a polynomial in $z$. Conversely every polynomial in $z$ is central. The highest Laurent exponent of a nonzero polynomial in $z$ shows that $z$ is transcendental. HenceThe identities $X^{-1}=z-X$ and $X^2-zX+1=0$ give $L=R+RX$. Independence follows by applying $X\mapsto X^{-1}$ to a relation $f(z)+Xg(z)=0$ and subtracting: $(X-X^{-1})g(z)=0$, so $g=f=0$. Consequentlyis a free $R$-module of rank four.

Let $M$ be a nonzero simple module on the left. For any $m\ne 0$, $M=Am$, so it is finite over $R$. Every central $r\notin P={\mathrm{Ann}}_R(M)$ acts bijectively: its kernel and image are $A$-submodules, and simplicity forces kernel zero and image $M$. To prove that $P$ is maximal, not merely prime, put $B=R/P$; then $M$ is a faithful finite $B$-module. For $0\ne b\in B$, $bM=M$. If $m_1,\dots ,m_d$ generate $M$, write $m_i=b{\sum }_j{c}_{ij}{m}_j$. The adjugate matrix identity for $I-bC$ shows that $\det (I-bC)$ annihilates every generator and hence $M$. Faithfulness makes this determinant zero in $B$. Its constant term is one, so it has form $1-bq$, proving $bq=1$. Thus $B$ is a field, and $P$ is a maximal ideal of $R$. This proves the general principle for simple modules over an algebra finite over its center in this instance.

Every maximal ideal of $\mathbb{C}[z]$ is $(z-\lambda )$ for some $\lambda \in \mathbb{C}$, by factorization of complex polynomials. The module $M=Am$ is therefore a quotient of the four-dimensional algebra $A/(z-\lambda )A$. In particular ${\dim }_{\mathbb{C}}M\le 4<\infty$, as required.

The actual simple dimensions are at most two. If $\lambda \ne \pm 2$, choose $u\ne \pm 1$ with $u+u^{-1}=\lambda$. The matrices $X=\mathrm{diag}(u,u^{-1})$ and $Y=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ give a representation of $A/(z-\lambda )A$ onto $M_2(\mathbb{C})$: polynomial functions of $X$ give the two diagonal matrix units, and multiplication by $Y$ gives the off-diagonal ones. Both algebras have dimension four, so this is an isomorphism. If $\lambda =\pm 2$, the element $X\mp 1$ has square zero and is conjugated to its negative by $Y$; its generated two-sided ideal is square zero. It annihilates every simple module, since its action on a simple module is either zero or the whole module, and the latter contradicts its square being zero. The remaining quotient is $\mathbb{C}[C_2]$, whose two simple modules have $Y=\pm 1$. Thus the generic simple modules have dimension two, and the four exceptional one-dimensional characters have $X=\pm 1$, $Y=\pm 1$.

Use generators $x,D$ with $Dx-xD=1$. The ordered monomials $x^i{D}^j$ form a basis: construct the Weyl algebra as the differential Ore extension of $\mathbb{C}[x]$ with $Df=fD+f^{\prime }$, which has unique ordered polynomial expressions. Giving both generators degree one produces the Bernstein filtration, whose associated graded ring is $\mathbb{C}[x,\xi ]$. If $a,b\ne 0$, their nonzero leading symbols have nonzero product in this polynomial integral domain. Thus $ab$ has nonzero leading symbol and cannot vanish. Hence $A_1(\mathbb{C})$ has no nontrivial zero divisors.

Here is the left-handed Ore theorem. Let $S$ be a multiplicative set of regular elements of a ring, containing $1$ and excluding $0$. There is an injective Ore localization whose elements are left fractions $s^{-1}a$ if and only ifThis is the left Ore condition. For a noncommutative domain, taking $S=A\setminus \{0\}$ gives a division ring exactly when this condition holds. With zero-divisor denominators a further left reversibility condition is needed; the regular-denominator statement is the one used here.

Necessity follows by writing $a{s}^{-1}=t^{-1}b$ in the quotient and multiplying by $t$ and $s$, giving $ta=bs$ in the embedded original ring. For a construction proving sufficiency, use the left Ore fraction construction: pairs $(s,a)$ represent $s^{-1}a$, and $(s,a)$ and $(t,b)$ are equivalent whenfor suitable $u,v\in A$. Reflexivity and symmetry are immediate. For transitivity, combine $us=vt=w$ and $u^{\prime }t=v^{\prime }q=w^{\prime }$ by a common left multiple $hw=k{w}^{\prime }\in S$. Then $(hv-k{u}^{\prime })t=0$, so regularity gives $hv=k{u}^{\prime }$, which combines the numerator equalities and proves transitivity.

Addition uses the common denominator $w$, with numerator $ua+vb$. To multiply $s^{-1}a$ by $t^{-1}b$, choose $v\in S,c\in A$ with $va=ct$; setThe Ore condition supplies a common refinement whenever a representative or an Ore relation is changed, and regularity cancels the shared denominator, proving independence of choices. The same refinements verify the ring axioms; the formulas are simply addition and multiplication after clearing left denominators. The map $a\mapsto (1,a)$ is injective, since equivalence to zero would give $ua=0$ with $u\in S$. Each $s\in S$ becomes invertible. A homomorphism inverting $S$ extends uniquely by $(s,a)\mapsto f(s{)}^{-1}f(a)$, giving the universal property. For all nonzero denominators in a noncommutative domain, a nonzero fraction has inverse $(s^{-1}a{)}^{-1}=a^{-1}s$, so the quotient is a division ring. This is the requested proof sketch, with the fraction equivalence and multiplication made explicit.

It remains to check the left Ore condition for $A_1(\mathbb{C})$. Its graded polynomial ring is Noetherian by the Hilbert basis theorem. Lifting homogeneous generators of a graded left ideal and successively cancelling leading symbols proves that the filtered algebra is a left Noetherian ring: each subtraction lowers the nonnegative degree, so it terminates.

In any left Noetherian noncommutative domain, nonzero principal left ideals intersect. Otherwise $Ab\cap Aa=0$ with $a,b\ne 0$ would makean infinite direct sum of nonzero left ideals. To verify directness, reduce a finite relation modulo $Aa$ to make its first summand zero, then cancel $a$ on the right and repeat. Its increasing finite partial sums contradict the ascending chain condition. A nonzero element of $Aa\cap As$ therefore has the form $ta=bs\ne 0$, with $t\ne 0$, proving the required condition. Consequently $A_1(\mathbb{C})$ has a classical ring of left quotients, and it is a division ring.

Let $X$ be an affine variety over an algebraically closed field $k$, with reduced coordinate ring $A=k[X]$. On an open subset $U$, define ${\mathcal{O}}_X(U)$ to be the functions $h:U\to k$ which, near every point, have an expression $a/b$ with $a,b\in A$ and $b$ nowhere zero on that neighbourhood. Restriction is ordinary restriction of functions. Compatible local functions glue uniquely and remain locally quotients, so this defines the structure sheaf. On the principal open subset $D(f)$ it gives $A_f$, and at $P$ it gives the local ring $A_{{\mathfrak{m}}_P}$. The principal-open identification can also be seen by identifying $D(f)$ with the affine algebraic set obtained by adjoining a coordinate $t$ and the equation $tf=1$; its coordinate ring is $A[t]/(tf-1)\cong A_f$. This construction permits several irreducible components; no division by a function vanishing identically on a component is assumed.

An abstract algebraic variety is a locally ringed space locally isomorphic to such affine models, with a finite affine cover; we use the usual separated convention. Reducible varieties are allowed here. The separation condition says that the diagonal morphism is a closed immersion. Two varieties are related by a birational map with an inverse when some dense open subset of one is isomorphic as a locally ringed space to a dense open subset of the other.

Define the ring of rational functions on a reduced variety by pairs $(U,h)$, where $U\subseteq X$ is dense open and $h\in {\mathcal{O}}_X(U)$, identifying two pairs when their restrictions agree on a common dense open subset. Finite intersections of dense open subsets are dense, so addition and multiplication on these intersections are well defined. For an irreducible variety this gives the usual function field; for a reducible variety it need not be a field. Restriction to any dense open subset $V$ induces an isomorphismIndeed, intersecting domains with $V$ defines the restriction, and a dense open subset of $V$ is also dense open in $X$, which defines its inverse. An isomorphism between dense opens therefore induces an isomorphism of their rational rings. Thus birational varieties have isomorphic rings of rational functions, also in the reducible case.

For the finite-cover assertion, write the closed complement of $U$ as $V(I)$, where $I=(f_1,\dots ,f_r)$ is an ideal of the Noetherian ring $A$. ThenLet ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_t$ be the minimal prime ideals of $A$, corresponding to its irreducible components. Since $U$ is dense, it meets every component, so $I$ is contained in none of these primes. Prime avoidance supplies $s\in I$ outside their union. Consequently $D(s)\subseteq U$ meets a dense open subset of every component and is dense in $X$. Add $D(s)$ to the finite cover if necessary. This proves that a finite principal-open cover can be chosen with a dense member. It does not say that every principal-open cover already has such a member: for $X=V(xy)$, the dense open $X\setminus \{(0,0)\}=D(x)\cup D(y)$ has neither displayed member dense, although adding $D(x+y)$ supplies a dense member.

For clarity, the zero divisors in this reduced Noetherian ring are exactly ${\bigcup }_i{\mathfrak{p}}_i$. If $ab=0$ and $a$ avoids every ${\mathfrak{p}}_i$, reduction in the domain $A/{\mathfrak{p}}_i$ forces $b\in {\mathfrak{p}}_i$ for all $i$, hence $b=0$. Conversely, if $a\in {\mathfrak{p}}_i$, choose for every $j\ne i$ an element $b_j\in {\mathfrak{p}}_j\setminus {\mathfrak{p}}_i$, possible because distinct minimal primes are incomparable. Their product $b$ is not in ${\mathfrak{p}}_i$ but belongs to all the other primes. Then $b\ne 0$ and $ab\in {\bigcap }_j{\mathfrak{p}}_j=0$. For a single minimal prime, reducedness makes that prime zero and this argument uses $b=1$. It follows that

The total quotient ring $S^{-1}A$, where $S$ consists of these non-zero-divisors, maps to $\mathrm{Rat}(X)$ byThe equivalence relation for localization makes this a well-defined $k$-algebra homomorphism. For surjectivity, a rational class represented on $U$ restricts to the dense principal open $D(s)\subseteq U$ constructed above. Its section there belongs to $A_s$, so it is $a/s^r$ and lies in the image. For injectivity, if such a fraction is zero on a dense open subset, its numerator vanishes on that dense subset. The numerator's zero set is closed, so it vanishes everywhere on $X$ and is zero in the reduced coordinate ring. ThusIn particular $A\to S^{-1}A$ is canonically injective: $a/1=0$ implies $sa=0$ for some non-zero-divisor $s$, hence $a=0$.

Finally let $Y$ be the reduced closed union of exactly those irreducible components of $X$ which contain $P$. Removing all the other components gives an open neighbourhood of $P$ on which $X$ and $Y$ coincide, including their structure sheaves. Hence ${\mathcal{O}}_{X,P}\cong {\mathcal{O}}_{Y,P}$. Write $B=k[Y]$ and let ${\mathfrak{n}}_P$ be its maximal ideal at $P$. Every minimal prime of $B$ is contained in ${\mathfrak{n}}_P$, because every component of $Y$ contains $P$. If $b\notin {\mathfrak{n}}_P$, it avoids each such prime and is a non-zero-divisor. Therefore the map of localizationsis defined and injective: if its value is zero, a non-zero-divisor annihilates $a$, so $a=0$. We obtain the desired local ring embeds in rational functions on incident components:At an intersection point one must retain all incident components. For example, the local ring of $V(xy)$ at the origin has two nonzero germs $x,y$ with $xy=0$; it cannot embed into the function field of just one of its axes.

Use the standard convention that an integral domain has $1\ne 0$. Since $(n)=(-n)$, the sign of a nonzero $n$ makes no difference. For $|n|=p$ prime, every nonzero residue has an inverse by Bézout's identity, so the quotient ring is a field and hence an integral domain. If $|n|=ab$ is composite with $1<a,b<|n|$, the nonzero classes of $a$ and $b$ multiply to zero, giving zero divisors. For $n=\pm 1$, the quotient is the zero ring, excluded by the convention above. Finally, $n=0$ gives $\mathbb{Z}/(0)\cong \mathbb{Z}$, an integral domain. ThusIf a convention restricts this notation to positive moduli, the answer is simply the positive primes.

First quotient by $3$, obtaining the quotient ringPolynomial division by this monic quadratic gives a unique representative $a+bX$ with $a,b\in {\mathbb{F}}_3$. Uniqueness holds because a nonzero multiple of a monic quadratic cannot have degree less than two. Therefore $|R|=3^2=9$.

In characteristic three, $X^2+X+1=(X-1{)}^2$. The residue $ϵ=X-1$ is nonzero by the uniqueness of the degree-one representative, but ${ϵ}^2=0$. Thus $ϵ$ is both a nilpotent element and a zero divisor. Consequently $R$ is not an integral domain. Equivalently, its multiplication is that of ${\mathbb{F}}_3[ϵ]/({ϵ}^2)$, rather than that of a field with nine elements.

Hilbert's basis theorem: if $R$ is a commutative Noetherian ring with identity, then the polynomial ring $R[x]$ is Noetherian. Iteration gives the same assertion for every finite number of variables.

Here is a bounded-degree coefficient proof of Hilbert basis theorem. Let $I$ be any ideal of $R[x]$. For $n\ge 0$, define $J_n\subseteq R$ to be the coefficients of $x^n$ in the polynomials of $I$ of degree at most $n$, including the zero coefficient. Closure under addition and scalar multiplication makes $J_n$ an ideal of $R$. Multiplication of a polynomial by $x$ shows $J_n\subseteq J_{n+1}$. By the ascending chain condition, there is $N$ such that $J_N=J_{N+1}=\cdots$.

Each $J_n$ for $0\le n\le N$ is finitely generated. Choose generators $a_{n,1},\dots ,a_{n,r_n}$ and corresponding polynomials $f_{n,j}\in I$ of degree at most $n$ whose coefficient of $x^n$ is $a_{n,j}$. There are only finitely many chosen polynomials. We prove they generate $I$ by induction on degree.

Take a nonzero $f\in I$ of degree $d$ and leading coefficient $a$. If $d\le N$, express $a={\sum }_j{c}_j{a}_{d,j}$ and subtract ${\sum }_j{c}_j{f}_{d,j}$. If $d>N$, use $a\in J_d=J_N$, express $a={\sum }_j{c}_j{a}_{N,j}$, and subtract ${\sum }_j{c}_j{x}^{d-N}{f}_{N,j}$. In both cases the leading coefficient cancels, leaving a polynomial in $I$ of smaller degree. For degree zero the remainder is zero, and the induction completes the claim.

Thus every ideal of $R[x]$ is finitely generated, proving the Hilbert basis theorem. The proof uses no cancellation of nonzero elements of $R$, so it remains valid when $R$ has zero divisors.

For a commutative ring $R$ and a multiplicative subset $S$ containing $1$, the universal property of localization specifies a pair $(R_S,\iota )$ with $\iota :R\to R_S$ a unital ring homomorphism, such that every $\iota (s)$, $s\in S$, is a unit, and any unital ring homomorphism $f:R\to T$ into a commutative ring that sends $S$ to units factors uniquely through $\iota$. Thus there is a unique $\tilde{f}:R_S\to T$ satisfying $\tilde{f}\iota =f$. The property determines the pair up to a unique compatible isomorphism.

For existence, take pairs $(r,s)\in R\times S$ and declareThe extra multiplier is essential when $R$ has zero divisors. This is an equivalence relation: for transitivity, multiply the two cross-multiplication equations by appropriate denominators and add them, using the product of their annihilating multipliers. Write the class as $r/s$ and defineMultiplying the defining equivalence equations verifies that these operations are well defined. They give a commutative ring, and $s/1$ has inverse $1/s$. The required map is necessarilyIt is well defined because applying $f$ to $u(s^{\prime }r-s{r}^{\prime })=0$ and cancelling the units $f(u),f(s),f(s^{\prime })$ equates the two images. Its addition and multiplication laws follow directly from the fraction formulas. This establishes existence and the universal property of localization. If $0\in S$, this construction gives the zero ring, which is allowed as a localization.

For a prime ideal $P$, its complement $S=R\setminus P$ is multiplicatively closed, and localization at a prime ideal means $R_P=S^{-1}R$. The mapis well defined and surjective, with kernel of a ring homomorphism $P{R}_P$: a numerator has zero image exactly when it belongs to $P$. Since $R/P$ is an integral domain, its field of fractions is a nonzero field, so $P{R}_P$ is a proper maximal ideal. Every fraction $r/s$ with $r\notin P$ is a unit, with inverse $s/r$. Every fraction with $r\in P$ belongs to this proper ideal and cannot be a unit. Thus the nonunits are exactly $P{R}_P$, and every proper ideal lies within it. ConsequentlyNow suppose each $R_P$ is reduced. If $x\in R$ is nilpotent, its image $x/1$ is nilpotent in every $R_P$, hence is zero. If $x\ne 0$, its annihilator ${\mathrm{Ann}}_R(x)$ is a proper ideal, so choose a maximal ideal $P$ containing it. Maximal ideals in a commutative ring are prime ideals. The vanishing criterion in a module localization would give $sx=0$ for some $s\notin P$. But then $s\in {\mathrm{Ann}}_R(x)\subseteq P$, a contradiction. Hence $R$ is reduced. This is reducedness is detected by prime localizations; no finiteness hypothesis is needed.

The stronger local-domain condition does not force $R$ to be an integral domain. Take $R=k\times k$ for a field $k$. Its only prime ideals are $P_1=0\times k$ and $P_2=k\times 0$: a prime ideal must contain one of the orthogonal idempotents $(1,0)$ and $(0,1)$, and the remaining quotient is the field $k$. At $P_1$, inverting $(1,0)$ kills the second component, so $R_{P_1}\cong k$; similarly $R_{P_2}\cong k$. Both localizations are integral domains, butis a product of nonzero elements in $R$. Therefore the answer is no, as illustrated by domain prime localizations do not imply a domain.

The ring form of the Artin–Wedderburn theorem states that the following conditions on a unital ring $R$ are equivalent: $R$ is a semisimple ring; $R$ satisfies the descending chain condition on left ideals and $J(R)=0$; andwhere $s$ is finite, each $n_i\ge 1$, and each $D_i$ is a division ring. The factors are unique up to permutation and isomorphism. The zero ring may be included as the empty product. This theorem also implies the corresponding right-handed statements.

First prove the implication from the chain condition and zero Jacobson radical. Among finite intersections of maximal left ideals, choose a minimal one, say $L=L_1\cap \cdots \cap L_t$. Such a minimal intersection exists by the descending chain condition. Intersecting with any other maximal left ideal cannot make $L$ smaller, so $L$ is contained in every maximal left ideal. Thus $L\subseteq J(R)=0$. It follows that the map of left modulesis injective. Each quotient is a simple module.

For completeness, a submodule $N$ of a finite direct sum $V={⨁}_j{T}_j$ of simple modules is itself a finite direct sum of simple modules and is a direct summand of $V$. Choose a sum $W$ of coordinate summands, maximal among those with $N\cap W=0$. For any omitted coordinate summand $T_j$, simplicity gives $T_j\cap (N+W)=0$ or $T_j\subseteq N+W$. In the first case $N\cap (W\oplus T_j)=0$, contradicting maximality; hence every coordinate summand lies in $N+W$. Therefore $V=N\oplus W$. Projection onto $N$ shows $N$ is a finite sum of images of the $T_j$, each image zero or simple. Add these images successively, discarding any image already in the sum: simplicity makes each new intersection zero. This yields a direct-sum decomposition. Applying this argument to the displayed injection proves that the left regular module is a finite sum of simple modules.

Conversely, if the regular module is a direct sum of simple modules, that sum is finite because $1$ lies in a finite subsum and generates the whole regular module. Every left ideal, being a submodule, has the preceding decomposition and splitting property. Its chain lengths are bounded by the number of simple summands, so the descending chain condition holds. The Jacobson radical annihilates every simple summand of the regular module, hence annihilates $1$ and is zero. This identifies the first two conditions.

Write the finite decomposition using pairwise nonisomorphic simple left modules:Every simple module is a quotient of $R$, by sending $1$ to any nonzero vector, so this list includes every simple isomorphism class. The Schur lemma gives ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$, and $E_i={\mathrm{End}}_R(S_i)$ is a division ring: any nonzero endomorphism has zero kernel and full image by simplicity. ConsequentlyEvery endomorphism of the left regular module is right multiplication by its value on $1$. Composition reverses multiplication of those values, so the left side is $R^{\mathrm{op}}$. Taking the opposite ring, and using transpose to identify $M_n(E{)}^{\mathrm{op}}$ with $M_n(E^{\mathrm{op}})$, gives the theorem's decomposition with $D_i=E_i^{\mathrm{op}}$. This keeps the left/right conventions explicit.

Conversely, the standard column module $D^n$ over $M_n(D)$ is simple: a nonzero coordinate of a vector can be inverted to construct a matrix sending it to any prescribed vector. The left regular module decomposes as the sum of its $n$ column ideals, each isomorphic to $D^n$. A finite product of these matrix rings is therefore a semisimple ring. This proves all implications. For uniqueness, the isomorphism classes of simple modules and their multiplicities in the regular module determine the numbers $n_i$ and the division rings $E_i^{\mathrm{op}}$. The multiplicity can also be recovered as the dimension of ${\mathrm{Hom}}_R(S_i,R)$ as a right $E_i$-module, so it does not depend on the chosen decomposition.

Now let $A=\mathbb{C}[G]$ for the finite group $G$. It has dimension $|G|$, with the elements of $G$ as a basis, so its left ideals satisfy the descending chain condition by their dimensions. The permitted assumption $J(A)=0$ makes the Artin–Wedderburn theorem applicable. Its decomposition is a complex-algebra decompositionwith each $D_i$ finite-dimensional over the central copy of $\mathbb{C}$. Each such division algebra is just $\mathbb{C}$. Indeed, for $d\in D_i$, finite dimensionality gives a nonzero polynomial relation $p(d)=0$ with complex coefficients. Factor $p$ into linear factors over $\mathbb{C}$. Their product at $d$ is zero, so the absence of zero divisors forces $d-\lambda =0$ for some scalar root $\lambda$. Thus every element is scalar.

The pairwise nonisomorphic simple modules of ${\prod }_i{M}_{n_i}(\mathbb{C})$ are the column modules ${\mathbb{C}}^{n_i}$, with all other factors acting as zero. Hence ${\dim }_{\mathbb{C}}{S}_i=n_i$. Taking complex dimensions gives

Suppose $x^2+y^2=0$ in the field $F$. If $y\ne 0$, division by $y^2$ gives $(x/y{)}^2=-1$, contrary to the hypothesis. Hence $y=0$ and then $x^2=0$; the absence of nonzero zero divisors in a field gives $x=0$. ThereforeFor the stated operations on $F^2$, addition is an abelian group operation: associativity and commutativity hold componentwise, its identity is $(0,0)$, and the negative of $(x,y)$ is $(-x,-y)$.

To check multiplication without omitting associativity, associate to each pair the matrixThis is an injection into the two-by-two matrices over $F$, and direct matrix multiplication givesIt also preserves addition. The associativity and distributivity of matrix multiplication therefore imply the same laws for the pair multiplication. Its formula is symmetric in the pairs, so it is commutative. Its identity is $(1,0)$, which is different from $(0,0)$.

For a nonzero pair, the first result shows that $x^2+y^2\ne 0$. Its multiplicative inverse issince multiplying $(x,y)$ by $(x,-y)$ gives $(x^2+y^2,0)$. This verifies every field axiom, so the given operations make $F^2$ a field. The embedded copy of $F$ is $\{(x,0):x\in F\}$; the element $(0,1)$ has square $(-1,0)$, describing a quadratic extension of $F$.

The multiplicative norm first rules out zero divisors: if $xy=0$ and $x\ne 0$, then $0=|x||y|$ forces $y=0$. If the dimension is zero the claimed bound already holds, so suppose it is positive.

It is worth checking that the printed description of a ring with an underlying real vector space really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed $x$, the map $L_x(y)=xy$ is additive by distributivity, andIt is therefore continuous. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves $L_x(ty)=t{L}_x(y)$ for every real $t$. The same argument holds in the other variable. This is automatic real bilinearity from a multiplicative norm, so multiplication is a bilinear map between finite-dimensional vector spaces, and hence is smooth.

Let $S=\{x\in A:|x|=1\}$ be the unit sphere of the given positive-definite bilinear form. It is a copy of $S^{n-1}$. The squaring mapis well defined because $|x^2|=|x{|}^2=1$, and is smooth. Commutativity gives its derivativeFor $x\in S$ and $v\ne 0$, the norm identity gives $|D{Q}_x(v)|=2|v|$, so the derivative is injective, including on the tangent space $T_{x}S=x^{\perp }$. It takes tangent vectors into $T_{x^2}S$ since $Q$ maps the sphere to itself. Alternatively, polarization of $|xy{|}^2=|x{|}^2|y{|}^2$ gives $⟨xy,xz⟩=|x{|}^2⟨y,z⟩$, so $⟨x^2,2xv⟩=2⟨x,v⟩=0$ directly.

The only identifications under squaring are antipodal:The factorization uses commutativity, and the last implication uses the absence of zero divisors. Conversely $Q(-x)=Q(x)$. It follows that $Q$ induces an injective smooth mapThe antipodal quotient map is a local diffeomorphism, so the injectivity of $DQ$ implies injectivity of $Dq$. Since Real projective space is compact, the stated compact injective immersion is an embedding theorem makes $q$ a closed embedding.

For $n\ge 2$ its source and target have the same dimension $n-1$. The inverse function theorem makes its image open as well as closed in the connected sphere $S^{n-1}$. It is nonempty, so the image is the whole sphere; consequently $q$ is a diffeomorphism. But for $n\ge 3$ this is impossible:Indeed, the antipodal two-sheeted covering map $S^{n-1}\to {\mathbb{RP}}^{n-1}$ has simply connected total space in this range, and is its universal cover, with deck group $\mathbb{Z}/2$. A sphere of dimension at least two is simply connected. These fundamental groups cannot be those of homeomorphic spaces.

We have obtained the dimension bound for a commutative Euclidean normed algebra:The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map ${\mathbb{RP}}^1\to S^1$ is entirely compatible with the argument: the circle squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.

All rings below are nonzero and commutative with identity; the zero ring is trivially Artinian. A primary ideal is proper. An ideal $Q$ is primary precisely when every zero divisor of $R/Q$ is a nilpotent element. Its radical of an ideal $\mathfrak{p}=\sqrt{Q}$ is prime: if $ab\in \sqrt{Q}$ and $a\notin \sqrt{Q}$, apply the primary property to $(ab{)}^r\in Q$ to obtain a power of $b$ in $Q$. In a Noetherian ring, finite generation of $\mathfrak{p}$ gives ${\mathfrak{p}}^s\subseteq Q$ for some $s$. For generators $u_i$ with $u_i^{s_i}\in Q$, one can take $s=1+{\sum }_i(s_i-1)$.

We first prove existence of primary decomposition. The ascending chain condition implies that every proper ideal is a finite intersection of irreducible ideals. Otherwise choose a maximal counterexample $I$ under inclusion. It is not irreducible, so $I=J\cap K$ with both $J,K$ strictly larger; their finite irreducible decompositions give one for $I$, a contradiction.

An irreducible ideal in a Noetherian ring is primary. Pass to the quotient and suppose zero is irreducible. If $xy=0$ with $x\ne 0$, the ascending chain of annihilators of $y^j$ stabilizes, say at $j=r$. Any element of $(x)\cap (y^r)$ can be written $z=ax=b{y}^r$. Then $yz=0$, so $b\in \mathrm{Ann}(y^{r+1})=\mathrm{Ann}(y^r)$ and $z=0$. Irreducibility of zero implies $(y^r)=0$, since $(x)\ne 0$. Thus $y$ is nilpotent. This proves irreducible ideals are primary in Noetherian rings, and hence the Lasker–Noether theorem.

An intersection of finitely many primary ideals with the same radical $\mathfrak{p}$ is again primary: if $ab$ belongs to all of them and $b\notin \mathfrak{p}$, the primary property forces $a$ into every component. Combining such components and removing redundant ones yields a minimal primary decompositionThe radicals are uniquely determined, but the components need not all be unique. Here is a proof identifying the invariant radicals as the associated primes of a module $R/I$.

Every nonzero module over a Noetherian ring has a nonzero element with a prime annihilator: maximize the annihilator of a nonzero element using the ascending chain condition. If $abm=0$ and $bm\ne 0$, maximality gives $\mathrm{Ann}(bm)=\mathrm{Ann}(m)$ and hence $am=0$. This is the maximal annihilator of a module element is prime argument. For a $\mathfrak{p}$-primary quotient $R/Q$, the radical of the annihilator of every nonzero element is $\mathfrak{p}$: it is contained in $\mathfrak{p}$ by the primary property and contains a power of $\mathfrak{p}$ because ${\mathfrak{p}}^s\subseteq Q$. Thus the only possible associated prime is $\mathfrak{p}$, and it does occur.

The diagonal injection $R/I\hookrightarrow {⨁}_{i}R/Q_i$ shows that every associated prime of $R/I$ is one of the ${\mathfrak{p}}_i$. Indeed, for an element whose annihilator $P$ is prime, that annihilator is the intersection of the finitely many component annihilators, all containing $P$. Their product is contained in $P$, so primality forces one of them to equal $P$; its component element is nonzero and has associated prime ${\mathfrak{p}}_i$. Conversely, irredundancy gives $b\in {\bigcap }_{j\ne i}{Q}_j\setminus Q_i$. The nonzero cyclic submodule generated by $b+I$ embeds in $R/Q_i$. It has an associated prime, necessarily ${\mathfrak{p}}_i$, which is then an associated prime of $R/I$. We have proved the first uniqueness theorem for primary decomposition:Also the zero divisors on $R/I$ are exactly ${\bigcup }_i{\mathfrak{p}}_i$. For a scalar killing a nonzero element, extend its element annihilator to a maximal element annihilator containing it; the preceding argument gives an associated prime containing that scalar. The converse is immediate from the definition of an associated prime.

The minimal members of the set $\{{\mathfrak{p}}_i\}$ are the isolated primes of a primary decomposition. They are exactly the primes minimal over $I$: if a prime contains ${\bigcap }_i{\mathfrak{p}}_i$, it contains one ${\mathfrak{p}}_i$ by the product argument. A component belonging to an isolated prime is unique. Localize at ${\mathfrak{p}}_i$. Every other component becomes the whole ring, since its radical contains an element outside ${\mathfrak{p}}_i$ whose power lies in that component. A ${\mathfrak{p}}_i$-primary ideal contracts unchanged from this localization, because $sa\in Q_i$ with $s\notin {\mathfrak{p}}_i$ implies $a\in Q_i$. Hence the second uniqueness theorem for primary decomposition givesAn embedded primary component can vary. For example, in $k[x,y]$,Both second components are $(x,y)$-primary, since their quotients are dual-number algebras. For the second equality, reducing modulo $(x^2,y+x)$ makes $y=-x$; an element $xg$ vanishes exactly when $g$ has zero constant term, giving $(x^2,xy)$. The isolated component is $(x)$ and the embedded associated prime is $(x,y)$.

The relevance to Artinian rings is particularly sharp in Krull dimension zero. Suppose $R$ is Noetherian and all its primes are maximal. Apply a minimal primary decomposition to zero. Its radicals are distinct maximal ideals, so the components are pairwise comaximal ideals: each contains a power of its radical, and expanding $1=(a+b{)}^N$ for $a+b=1$ shows that powers of comaximal ideals remain comaximal. The Chinese remainder theorem gives the Artinian decomposition into local factorsEach factor has one maximal ideal, and that ideal is nilpotent. Its finite radical filtration has successive layers finitely generated over its residue field, hence of finite vector-space dimension. It follows that the factor, and therefore $R$, has a finite composition series of a module, so $R$ is Artinian. This proves the Noetherian dimension-zero criterion for an Artinian ring in this direction. Applied to $R/I$, it says that the quotient is Artinian exactly when all primes over $I$ are maximal. In that case no primary component is embedded, so all components are unique.

For completeness, the converse does not require initially assuming Noetherianity. In an Artinian ring, an Artinian domain is a field: stabilization of $(a^j)$ and cancellation gives an inverse to every nonzero $a$. Thus every prime is maximal. There are finitely many maximal ideals, since infinitely many distinct ones would give strictly descending finite intersections; comaximality guarantees strictness by the Chinese remainder theorem. Its nilradical $N$ is nilpotent. Indeed, its powers stabilize at $J=N^r$, with $J^2=J$. If $J\ne 0$, choose an ideal $C$ minimal among those satisfying $JC\ne 0$. Then $JC=C$, since $J(JC)=JC\ne 0$. Choose $x\in C$ with $Jx\ne 0$; minimality also gives $Rx=C$. Therefore $x=ax$ for some $a\in J\subseteq N$. But $1-a$ is a unit, contradicting $x\ne 0$. Hence $J=0$.

Now $R/N$ is a finite product of residue fields. Each $N^j/N^{j+1}$ is an Artinian module over that product, and therefore a finite direct sum of finite-dimensional vector spaces: an infinite-dimensional vector space would have a strictly descending chain of subspaces. The finite filtration by powers of $N$ makes $R$ a module of finite composition length, in particular Noetherian. Combining both directions yieldsThus primary decomposition separates the local pieces of a zero-dimensional ring, while their nilpotent maximal ideals record the multiplicities that the reduced set of primes alone does not detect.

The Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.