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False: existence holds, but uniqueness fails. First, the Lasker–Noether theorem supplies existence; here is an argument from the preceding part. If some proper ideal in a Noetherian ring were not a finite intersection of irreducible ideals, the ascending chain condition would let us choose such an ideal $I$ maximal under inclusion. It is not irreducible, so $I=J\cap K$ with both $J$ and $K$ strictly larger than $I$. By maximality, both are finite intersections of irreducible ideals, and therefore so is $I$, a contradiction. Part (d) makes each factor a primary ideal. The whole ring is represented by the empty intersection if that convention is needed.

For a counterexample to uniqueness, work in $R=k[x,y]$ over a field $k$ and put $I=(x^2,xy)$. The Hilbert basis theorem makes $R$ a Noetherian ring. There are two distinct minimal primary decompositions:The ideal $(x)$ is a prime ideal, hence a primary ideal. Each of the other factors has quotient ring isomorphic to $k[x]/(x^2)$, by substituting respectively $y=0$ and $y=x$. As in part (b), each is $(x,y)$-primary.

To check both intersections, let $Q_c=(x^2,y-cx)$, for $c=0,1$. Substitution identifies $R/Q_c$ with $k[x]/(x^2)$. The class of $xg(x,y)$ is zero exactly when $g(0,0)=0$, equivalently when $g\in (x,y)$. Therefore $(Q_c:x)=(x,y)$ andThe two $Q_c$ are different, since $y$ belongs to $Q_0$ but has nonzero class $x$ modulo $Q_1$. Both decompositions are irredundant: $x\notin Q_c$, while $y-cx\notin (x)$. Applying the radical of an ideal operation to the two components in either decomposition gives $(x)$ and $(x,y)$, respectively.

The second uniqueness theorem for primary decomposition explains the distinction: the component at an isolated prime of a primary decomposition is unique, but embedded primary components need not be. The counterexample varies an embedded component.

A ring is a Noetherian ring if it satisfies the ascending chain condition on ideals: every chaineventually becomes constant. Equivalently, every ideal has a finite generating set of an ideal. To see this equivalence, if an ideal were not finitely generated, one could repeatedly choose an element outside the ideal generated by the preceding choices, producing a strictly ascending chain. Conversely, the union of an ascending chain is an ideal; if it is generated by finitely many elements, all of them lie in one term of the chain, and every later term equals that term.

A ring is an Artinian ring if it satisfies the descending chain condition on ideals: every chaineventually becomes constant. These are stabilization conditions on all ideals, not merely on principal ideals.

No. For example, take $R=k$, any field. It is an Artinian ring, since its only ideals are zero and itself. In the formal power series ring $k[[x]]$, however,is a strictly descending chain of ideals. The inclusion is strict because $x^n$ has a nonzero coefficient of $x^n$, whereas every element of $(x^{n+1})$ has that coefficient zero. Thus $k[[x]]$ fails the descending chain condition and is not Artinian.

The same argument works for any nonzero coefficient ring $R$: the coefficient $1$ in $x^n$ cannot be produced by a multiple of $x^{n+1}$. Therefore $R[[x]]$ is never Artinian when $R\ne 0$. The zero ring is the harmless exception.

No. Take $R=k[t]$ over a field $k$, and inside $R[x]=k[t,x]$ takeThis is a subring containing $R$: its elements have arbitrary constant coefficient in $k[t]$ as polynomials in $x$, and every positive-degree coefficient is divisible by $t$; that condition is preserved by sums and products. The ring $R$ is Noetherian. Indeed, a nonzero ideal of $k[t]$ has an element $g$ of least degree, and division by $g$ shows that every element of the ideal is a multiple of $g$. Thus $k[t]$ is a principal ideal domain, and its ideals are finitely generated.

In $S$, let $J_n=(tx,t{x}^2,\dots ,t{x}^n{)}_S$. These form an ascending chain. We show that $t{x}^{n+1}\notin J_n$. If it belonged, we could writeCancel $t$ in the integral domain $k[t,x]$, then apply the evaluation homomorphism $t\mapsto 0$. Every $s_j\in S$ becomes a scalar in $k$, so this would givewhich is impossible by comparison of degrees. Hence $J_n⊊J_{n+1}$ for every $n$, andThis is a constant-plus-ideal non-Noetherian subring construction: the restricted coefficients prevent a finite list of positive powers of $x$ from generating all the required powers.

TakeWe justify that this is a Dedekind domain. It is an integral domain inside the number field $K=\mathbb{Q}(\sqrt{-5})$. As an additive group it is free of rank two over $\mathbb{Z}$. The subgroup theorem for finitely generated abelian groups says that every subgroup of such a group is finitely generated. Thus every ideal has finitely many additive generators, and those same elements generate it as a $D$-ideal. This proves that $D$ is Noetherian.

We use the standard formula for the ring of integers of a quadratic field: for squarefree $d$, it is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Since $-5\equiv 3(\mathrm{mod}4)$, $D$ is the ring of integers of a number field $K$, meaning all elements of $K$ integral over $\mathbb{Z}$. We also use transitivity of integral extensions. If an element of $K$ is integral over $D$, it is integral over $\mathbb{Z}$ because $D$ is integral over $\mathbb{Z}$, and therefore belongs to $D$. Thus $D$ is an integrally closed domain.

Every nonzero prime ideal $P$ is maximal. Indeed, choose $0\ne \alpha =a+b\sqrt{-5}\in P$. Multiplying by its algebraic conjugate puts the nonzero integer $N(\alpha )=a^2+5b^2$ in $P$. Hence the integral domain $D/P$ is a quotient of the finite ring $D/N(\alpha )D$, and is finite. Every finite integral domain is a field, since multiplication by a nonzero element is injective and therefore surjective. The element $2$ has no inverse in $D$, since $1/2\notin \mathbb{Z}[\sqrt{-5}]$, so $D$ is not a field. Consequently $(0)$ is not maximal, and a maximal ideal above it is nonzero. Together with the preceding result, this gives Krull dimension exactly one.

Finally, consider the idealReduction modulo $2$ with $\sqrt{-5}\mapsto 1$ identifies $D/I$ with ${\mathbb{F}}_2$, so the additive index $[D:I]$ is $2$. If $I$ were the principal ideal $(a+b\sqrt{-5})$, multiplication by its nonzero generator would have integer matrixin the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, soThere are no integer solutions: $b\ne 0$ makes the right side at least $5$, while $b=0$ would require $a^2=2$. Thus $I$ is a nonprincipal ideal in the integers adjoined a square root of minus five, and $D$ is a Dedekind domain that is not a principal ideal domain.

The Krull dimension of a commutative ring $R$ isThe length counts strict inclusions, so a chain with $n+1$ prime ideals has length $n$. The supremum may be infinite. A field, for example, has just the prime ideal $(0)$ and therefore has Krull dimension zero. For the zero ring, the spectrum of a commutative ring is empty; under the convention $\sup \varnothing =-\infty$, its dimension is $-\infty$.

For a subring $S\subseteq R$ with the same identity, $R$ is an integral extension of $S$ if every $r\in R$ is an integral element over $S$: there exist $n\ge 1$ and $s_0,\dots ,s_{n-1}\in S$ such thatThe requirement that the polynomial be monic is essential. Merely satisfying some polynomial equation over $S$ is insufficient: for instance $1/2\in \mathbb{Q}$ satisfies $2T-1=0$ over $\mathbb{Z}$, but is not integral over $\mathbb{Z}$. To see this directly, a hypothetical monic equation for $1/2$, multiplied by $2^n$, would say that an odd integer $1$ plus a sum of even integers is zero.

The degree and coefficients may depend on $r$. Integrality does not by itself assert that the entire ring $R$ is a finite $S$-module; it does imply that the subalgebra $S[r]$ is generated as an $S$-module by $1,r,\dots ,r^{n-1}$.

No. In the Gaussian integers, chooseIndeed, substituting $i=-2$ gives $\mathbb{Z}[i]/(2+i)\cong \mathbb{Z}/5\mathbb{Z}$, so $P$ is a prime ideal and its contraction is $(5)$. Put $\beta =2-i$. Modulo $P$, $\beta$ becomes $4\ne 0$ in ${\mathbb{F}}_5$, so $\beta \notin P$. ThereforeWe show that $\alpha$ is not an integral element over ${\mathbb{Z}}_{(5)}$, using a direct reduction argument. Write $B={\mathbb{Z}}_{(5)}[i]$, viewed inside $\mathbb{Q}(i)$. There is a ring homomorphismwhich reduces rational coefficients modulo $5$. It is well defined because denominators in ${\mathbb{Z}}_{(5)}$ are not divisible by $5$, and $2^2+1=0$ in ${\mathbb{F}}_5$. In particular, $\rho (\beta )=0$ and $\rho (1)=1$.

If $\alpha$ satisfied a monic equation over ${\mathbb{Z}}_{(5)}$, saymultiplication by ${\beta }^n$ would give an equality in $B$:Applying $\rho$ gives $1=0$ in ${\mathbb{F}}_5$, a contradiction. ConsequentlyThis exhibits why localization of an integral extension at a prime need not be integral. Localizing both rings by the same multiplicative subset preserves an integral extension, but $\mathbb{Z}[i{]}_P$ inverts additional elements, including $2-i$, whose image at the other prime ideal $(2-i)$ is zero. The original extension is integral because $i$ satisfies the monic equation $T^2+1=0$: the subalgebra generated by finitely many integral elements property applies to $\mathbb{Z}[i]$.

All Topological K-theory groups below are complex and are indexed modulo two using Bott periodicity. For $n\ge 1$, Complex K-theory of a sphere givesFor an even-dimensional sphere, choose a generator $u$ of Reduced topological K-theory. Its square vanishes: the Chern character sends $u$ to a top-degree cohomology class, whose square is zero, and the Chern character is injective on this torsion-free abelian group. Thus the ring answers areThe exceptional zero-dimensional sphere consists of two points: $K^0(S^0)=\mathbb{Z}\times \mathbb{Z}$ with coordinatewise multiplication and $K^{-1}(S^0)=0$. In particular its reduced generator is idempotent rather than square-zero.

For the Euler characteristic identity, induct over the cells of a finite CW complex. The starting CW complex consisting of $v$ vertices has $K^0\cong {\mathbb{Z}}^v$, $K^{-1}=0$, and $\chi =v$. If $X$ is obtained from $A$ by attaching one $d$-dimensional cell, the quotient is $S^d$, and the K-theory six-term exact sequence isThese abelian groups are finitely generated by induction. Tensoring the exact sequence with $\mathbb{Q}$ preserves exactness. The alternating sum of dimensions in a cyclic six-term exact sequence is zero, soTherefore

There is a genuine convention issue in the nilpotence assertion. With the usual definition of Reduced topological K-theory as the kernel of restriction to one basepoint, the assertion needs $X$ to be connected. For $X=S^0$ based at its first point, $(0,1)$ lies in ${\tilde{K}}^0(X)$ and satisfies $(0,1{)}^m=(0,1)$ for every $m\ge 1$. Thus it is not a nilpotent element. For an arbitrary finite CW complex, the correct assertion uses the rank map in topological K-theory on every connected component:For a connected space, $I(X)={\tilde{K}}^0(X)$.

Here is an induction proving the corrected assertion. On the vertices, $I(X)=0$. For a positive-dimensional cell attachment $A\subset X$, the ideal $J=\ker (K^0(X)\to K^0(A))$ is square-zero. Indeed, lift two elements of $J$ to relative topological K-theory $K^0(X,A)$. Their product is induced by the reduced diagonalThis map is null-homotopic, since $d<2d$ and $S^{2d}$ is $(2d-1)$-connected. The relative product in topological K-theory is therefore zero, giving $J^2=0$. If $a\in I(X)$, its restriction belongs to $I(A)$, so induction gives $(a{|}_A{)}^m=0$ for some $m$. Then $a^m\in J$ and $a^{2m}=0$. We have proved every Topological K-theory class of rank zero on every connected component is a nilpotent element, and hence the requested result for connected $X$.

After the first of two equal-rate clocks rings, the residual lifetime of the other clock is, by memorylessness, a fresh exponential variable of rate $\lambda$, independent of the first ringing time and of which clock rang. Henceand these variables are independent.

Let $e^2=e$ with $e\ne 0,1$. Then $e(1-e)=0$, and the ideals $eR$ and $(1-e)R$ are nonzero rings with identities $e$ and $1-e$. DefineThis is a ring homomorphism. Its inverse is $(x,y)\mapsto x+y$, because $ex=x$, $(1-e)y=y$, and the cross terms vanish. Hence $R\cong eR\times (1-e)R$, proving (ii).

Forthe parity condition is preserved by componentwise addition, negation, and multiplication, and $(1,1)$ is the identity. Thus $R$ is a ring.

It is not an integral domain, since $(2,0)$ and $(0,2)$ are nonzero elements of $R$ whose product is zero. It is also not a product of two nontrivial rings. Indeed, an idempotent in ${\mathbb{Z}}^2$ has each coordinate in $\{0,1\}$, and the parity condition leaves only $(0,0)$ and $(1,1)$. The proved equivalence then excludes a nontrivial product decomposition.

In a Boolean ring, $(r+1{)}^2=r+1$. Using $r^2=r$ and $1^2=1$ givesso

If $R$ is also a nonzero integral domain, thenforces every $r$ to equal $0$ or $1$. Hence $R\cong {\mathbb{F}}_2$.

If $I$ is prime in a Boolean ring, then $R/I$ is a nonzero Boolean integral domain by part (i), and hence is ${\mathbb{F}}_2$. Since the quotient is a field,This is the prime ideals of a Boolean ring are maximal property.

An $R$-module is free if it has a basis: a subset $B$ such that every element has a unique expression as a finite $R$-linear combination of elements of $B$.

Choose a maximal ideal $\mathfrak{m}$ of the nonzero ring $R$. If $R^n\cong R^m$, quotienting by $\mathfrak{m}$ givesThese are vector spaces over the field $R/\mathfrak{m}$, so equality of dimension gives $n=m$. This proves invariant basis number for a commutative ring.

A direct summand of a free module need not be free. Take $R=\mathbb{Z}/6\mathbb{Z}$. Its ideals $P=(2)$ and $Q=(3)$ satisfyso $P\oplus Q\cong R$. But $P$ has three elements, whereas a finite-rank free $R$-module has $6^r$ elements; hence $P$ is not free.

An algebraic number is an algebraic integer when it is a root of a monic polynomial in $\mathbb{Z}[X]$.

If $\alpha$ and $\beta$ are algebraic integers, the ring $\mathbb{Z}[\alpha ,\beta ]$ is a finitely generated $\mathbb{Z}$-module. Multiplication by $\alpha \beta$ preserves this module, so the algebraic integer module criterion shows that $\alpha \beta$ is an algebraic integer.

The ring of integers of a quadratic field givessince $2\equiv 2(\mathrm{mod}4)$. For a direct verification, if $x=a+b\sqrt{2}$ with $a,b\in \mathbb{Q}$ is integral, then its trace $2a$ and norm $a^2-2b^2$ are integers. Writing $2a=m\in \mathbb{Z}$ and reducing the norm condition in lowest terms shows first that $b$ can have denominator at most $2$; a denominator $2$ with odd numerator would require $m^2\equiv 2(\mathrm{mod}4)$, which is impossible. Thus $b\in \mathbb{Z}$, and then the norm condition forces $m$ even, so $a\in \mathbb{Z}$. Conversely, every $a+b\sqrt{2}$ with $a,b\in \mathbb{Z}$ is integral because $\sqrt{2}$ is integral and algebraic integers form a ring.

Now let $f$ have degree $d$ and roots ${\alpha }_1,\dots ,{\alpha }_d$. Sinceand $|{\alpha }_1|>1$, the integer $|a_d|$ cannot be $2$; hence $|a_d|=1$. Thus $f$ is monic up to sign and ${\alpha }_1$ is an algebraic integer.

Let $B$ be the product of all roots with modulus greater than one, counting multiplicity. Complex roots occur in conjugate pairs, so $B$ is real up to the signs of the real roots andAs a product of algebraic integers, $B$ is an algebraic integer; the equality forces $B=\pm 2$. Thereforeis the absolute value of the product of the remaining conjugates, is rational, and is an algebraic integer. It is consequently a nonzero rational integer. Every remaining factor has modulus at most one, so this integer is at most one. It must equal one, and henceThis is the mahler-measure norm argument at measure two.