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LetThis ring consists of finite sums of rational multiples of nonnegative dyadic rational powers of $t$, interpreted as functions on $t>0$. Every pair of elements lies in one polynomial ring $\mathbb{Q}[u_n]$, so $R$ is an integral domain. The constant term defines a surjective ring homomorphism $R\to \mathbb{Q}$. Its kernel of a ring homomorphism $\mathfrak{m}$ is nonzero and is a maximal ideal, by the maximal ideal quotient criterion.

Every $f\in \mathfrak{m}$ has the form $u_{n}g(u_n)$ for some $n$ and some polynomial $g$. The identityexpresses it as a product of two elements of $\mathfrak{m}$. Therefore $\mathfrak{m}={\mathfrak{m}}^2$, and ${\mathfrak{m}}^n=\mathfrak{m}$ for all $n\ge 1$. This shows that strict descent of powers of an ideal can fail in an integral domain, even for a nonzero maximal ideal. By the determinant trick, such an ideal cannot be finitely generated, so this ring is not a Noetherian ring.

For a prime ideal $P$, suppose the kernel of a ring homomorphism $K=\ker (R\to R_P)$ is a finitely generated ideal, say $(a_1,\dots ,a_t)$. Choose $s_i\notin P$ killing $a_i$ and let $f={\prod }_i{s}_i$. Then $f\notin P$ and $fK=0$. If a fraction $a/f^n$ maps to zero in $R_P$, then $a\in K$, so $fa=0$ and the fraction was already zero in $R_f$. Thus the canonical ring homomorphism $R_f\to R_P$ is injective. The empty product $f=1$ handles $K=0$. Every Noetherian ring satisfies the finite generation hypothesis.

A kernel records the information lost by a homomorphism. For a group homomorphism it is the preimage of the identity element, giving the kernel of a group homomorphism; for a ring homomorphism or linear map it is the preimage of zero, giving the kernel of a ring homomorphism or kernel of a linear map. More generally, the relation $a\sim b⟺f(a)=f(b)$ records which inputs the homomorphism identifies. In the group case, its equivalence classes are cosets of the group kernel, as explained by fibres of a group homomorphism are cosets of its kernel.

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.

In this localization of a ring, a nonzero fraction $a/b$ with odd $b$ is a unit exactly when $a$ is odd: its reciprocal then has odd denominator; if $a$ is even, an inverse would equate an even integer with an odd one after clearing denominators. Consequently every nonzero element has a unique formHere $r$ is the exponent of $2$ in the numerator of the reduced fraction, its nonnegative 2-adic valuation.

Every nonzero ideal has the form $dR$ by part (c). Write the nonzero integer $d=2^{r}v$ with $v$ odd; since $v$ is a unit, $dR=2^{r}R$. Hence the nonzero proper ideals are exactlyThe zero ideal and $R=2^{0}R$ complete the full list. These ideals are distinct: $2^r$ does not belong to $2^{r+1}R$, since $1/2\notin R$.

The reduction map $a/b\mapsto a\mathrm{mod}2$ is a well-defined surjective ring homomorphism $R\to {\mathbb{F}}_2$ with kernel of a ring homomorphism $2R$; odd denominators reduce to $1$. Thus $R/2R\cong {\mathbb{F}}_2$ is a field, so $2R$ is a prime ideal and $2$ is a prime element. Every nonzero nonunit is associated to $2^r$ with $r\ge 1$. If $r\ge 2$, the factorization $2^r=2\cdot 2^{r-1}$ is into two nonunits, so it is not an irreducible element and therefore not a prime element. We concludeThe failed step in the argument of part (b) is the guarantee that the product plus one is a nonunit and hence has an irreducible factor. Here every prime element is an even fraction, so a product of prime elements plus $1$ is an odd fraction, therefore a unit. Already $2+1=3$ has no prime divisor in this ring. In $k[t]$, positive degree prevented that failure.

Take $I=(2,t)\subseteq \mathbb{Z}[t]$. The ring homomorphismis surjective. Its kernel of a ring homomorphism is exactly $I$: a polynomial maps to zero precisely when its constant coefficient is even, in which case it has the form $2a+th(t)$ with $a\in \mathbb{Z}$ and $h\in \mathbb{Z}[t]$. By the first isomorphism theorem for rings,This quotient ring is a field, so $I$ is a maximal ideal and in particular a prime ideal.

Suppose $I=(h)$ were a principal ideal. Because $2\in I$, the nonzero polynomial $h$ divides the constant $2$, forcing its polynomial degree to be zero: $h=d$ for a nonzero integer $d$. But $t\in I$ also makes $d$ divide the coefficient $1$ of $t$, so $d=\pm 1$. This would give $I=\mathbb{Z}[t]$, contradicting the properness of the kernel of a ring homomorphism $\ker \theta$. ThereforeThere is no conflict with part (c): $I\cap \mathbb{Z}=2\mathbb{Z}\ne \{0\}$. The zero-intersection hypothesis is exactly what excludes this example.

An ideal is an additive subgroup of a ring closed under multiplication by arbitrary ring elements; in a potentially noncommutative ring, a two-sided ideal must absorb multiplication on both sides. For a ring homomorphism $\varphi$, its kernel of a ring homomorphism $K=\{r:\varphi (r)=0\}$ contains zero, is closed under subtraction, and satisfies $\varphi (rk)=\varphi (r)\varphi (k)=0$ and $\varphi (kr)=0$. Thus $K$ is an ideal.

Define $\tilde{\varphi }:R/K\to S$ by $r+K\mapsto \varphi (r)$. It is well-defined because two representatives differ by an element of $K$, preserves addition and multiplication, is surjective by hypothesis, and is injective because its zero class is precisely $K$. Therefore $\boxed{R/\ker \varphi \cong S}$, the first isomorphism theorem for rings.

Apply this to the evaluation homomorphism $\mathbb{C}[z]\to \mathbb{C}$, $p\mapsto p(1)$. Constants prove surjectivity, and division by $z-1$ shows that its kernel of a ring homomorphism is $(z-1)$. Hence $\boxed{\mathbb{C}[z]/(z-1)\cong \mathbb{C}}$.

The evaluation homomorphism $\mathbb{R}[X]\to \mathbb{R}$, $p\mapsto p(0)$, is surjective and has kernel of a ring homomorphism $J_1=(X)$. Hence $J_1$ is an ideal, and the first isomorphism theorem for rings gives $\boxed{\mathbb{R}[X]/J_1\cong \mathbb{R}}$. Since this quotient is a field, $\boxed{J_1\text{ is maximal}}$.

Here $J_4=(X^2+1)$ is a principal ideal. The evaluation homomorphism $p\mapsto p(i)$ from $\mathbb{R}[X]$ onto $\mathbb{C}$ has this kernel of a ring homomorphism: division by $X^2+1$ leaves a remainder $a+bX$, and $a+bi=0$ with real $a,b$ forces both to vanish. Surjectivity follows because every complex number is $a+bi$. ConsequentlyEquivalently, $X^2+1$ is an irreducible polynomial over the real numbers.

The Chinese remainder theorem for two comaximal two-sided ideals $I,J$ of a ring states that the natural ring homomorphism inducesIndeed, the map $r\mapsto (r+I,r+J)$ has kernel of a ring homomorphism $I\cap J$. Choose $i\in I$, $j\in J$ with $i+j=1$. Given target representatives $a,b\in R$, the element $aj+bi$ is congruent to $a$ modulo $I$ and to $b$ modulo $J$, proving surjectivity. The first isomorphism theorem for rings gives the result. For a commutative ring, $I\cap J=IJ$: if $x\in I\cap J$, then $x=xi+xj\in IJ$, while the reverse inclusion follows from the ideal properties.

Set $X=0$ to define a surjective ring homomorphismIts kernel of a ring homomorphism is exactly $(X)$: the constant coefficient in $X$ must vanish, so every remaining monomial is divisible by $X$. The first isomorphism theorem for rings therefore givesThe polynomial ring $\mathbb{Q}[Y]$ is an integral domain, because the product of two nonzero polynomials has nonzero leading coefficient. It is not a field: $Y$ has no polynomial multiplicative inverse, since a nonzero product $YQ(Y)$ has positive degree of a polynomial. Consequently $(X)$ is prime and is not maximal. Concretely, $(X)⊊(X,Y)⊊\mathbb{Q}[X,Y]$ is a strict chain of ideals.

A maximal ideal is a proper ideal $I$ of $R$ such that an ideal $J$ with $I\subseteq J\subseteq R$ is either $I$ or $R$. Suppose first that $I$ is a maximal ideal. If $a\notin I$, then $I+(a)=R$, so $1=i+ra$ for some $i\in I$ and $r\in R$. In the quotient ring, $(r+I)(a+I)=1+I$. Every nonzero element of $R/I$ therefore has a multiplicative inverse, so $R/I$ is a field. Conversely, if $R/I$ is a field and $J$ properly contains $I$, choose $a\in J\setminus I$. The element $a+I$ has a multiplicative inverse $r+I$, whence $1-ra\in I\subset J$, giving $1\in J$ and $J=R$. ThusThe squares in ${\mathbb{F}}_7$ are $0,1,2,4$, which do not include $-1=6$. Hence $X^2+1$ has no root in ${\mathbb{F}}_7$. A polynomial of degree two or three over a field is an irreducible polynomial exactly when it has no root: any proper polynomial factorization of such a degree has a factor of degree one. The polynomial ring over a field is a Euclidean domain, so the ideal generated by an irreducible polynomial is a maximal ideal. It follows thatis a field. Every coset has a unique representative $a+bX$, by polynomial division, so $|F|=7^2=49$.

Let $j$ be the coset of $X$ in this field. Since $j^2=-1$, the mapis a ring homomorphism. It is surjective by the representative description. Since $1,j$ are linearly independent over ${\mathbb{F}}_7$, its kernel of a ring homomorphism isThere is also a field with $343$ elements. The cubes in ${\mathbb{F}}_7$ are $0,1,-1$, so $X^3-2$ has no root and is an irreducible polynomial. Thereforeis a field whose unique representatives have degree below three. There are $7^3=343$ such representatives.

Use ordinary matrix addition and multiplication on $R=K[A]$. Sums and products of polynomials in $A$ remain such polynomials, and they commute because powers of the same matrix commute. The identity is $A^0=I$. Evaluation $\varphi (\sum c_i{X}^i)=\sum c_i{A}^i$ is a surjective unital algebra homomorphism over $K$. It is unique among maps fixing the scalar matrices and sending $X$ to $A$.

The scalar condition is needed for the word “unique”: as an unrestricted ring-homomorphism assertion over an arbitrary field, prescribing only the image of $X$ does not suffice. For example, over $\mathbb{C}$ with a real scalar matrix $A$, conjugating every coefficient gives another unital ring homomorphism with the same image of $X$. The intended polynomial-evaluation map is the $K$-algebra map just defined.

The kernel of a ring homomorphism is an ideal of the principal ideal domain $K[X]$. It is nonzero because finitely many powers of $A$ must become linearly dependent in the finite-dimensional matrix space. Its unique monic generator is precisely the minimal polynomial $m$: division by $m$ shows that a polynomial killing $A$ has zero remainder, by minimality. Hence $\ker \varphi =(m)$ and $R\cong K[X]/(m)$.

If $\gcd (f,m)=1$, Bézout's identity gives $uf+vm=1$; evaluation gives $u(A)f(A)=I$. Conversely an inverse in $R$ has form $g(A)$, so $fg-1\in (m)$, which forces $\gcd (f,m)=1$. Thus invertibility in this ring is exactly coprimality with the minimal polynomial.

The zero ideal is a prime ideal because $R$ is an integral domain. Each of $(2),(3),(5)$ is a prime ideal: for $p$ among these three, reduction $a/b\mapsto \overline{a}{\overline{b}}^{-1}$ is a surjective ring homomorphism $R\to {\mathbb{F}}_p$ with kernel of a ring homomorphism $(p)$, so its quotient ring is a field.

Any other nonzero proper ideal in (c) has generator $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }$ with total exponent at least two. Choose one prime factor $p$ and put $e=d/p$. Then $pe=d\in (d)$ but neither $p$ nor $e$ belongs to $(d)$, as seen from the required prime exponents. Such an ideal is not prime. Thus with the standard proper-ideal convention the answer isThe definition printed in this paper omits properness. Literally under that definition $R$ itself also satisfies the implication, so its literal complete list is the preceding four ideals together with $\boxed{R}$. Both interpretations are thereby accounted for.

No. A domain with a nonzero idempotent maximal ideal is a counterexample. ConsiderThis can be realized concretely as a ring of functions on $t>0$: its elements are finite sums of rational multiples of nonnegative dyadic rational powers of $t$. Each stage $\mathbb{Q}[u_n]$ is a polynomial ring, because a nonzero polynomial cannot vanish for all $u_n>0$. Any two elements lie in one stage, which is an integral domain, so their product is nonzero if both are nonzero. Thus $R$ is an integral domain.

The constant term defines a surjective ring homomorphism $ϵ:R\to \mathbb{Q}$. It is well defined because replacing $u_n$ by $u_{n+1}^2$ preserves the constant term. Its kernel of a ring homomorphismis a maximal ideal, since $R/\mathfrak{m}\cong \mathbb{Q}$ is a field, by the maximal ideal quotient criterion. Also $u_0=t\in \mathfrak{m}$ is nonzero.

If $f\in \mathfrak{m}$, choose a stage in which $f=u_{n}g(u_n)$ for a polynomial $g\in \mathbb{Q}[z]$. ThenBoth factors belong to $\mathfrak{m}$, so $f\in {\mathfrak{m}}^2$. The reverse inclusion holds for every ideal, proving that $\mathfrak{m}$ is an idempotent ideal. HenceThe Noetherian ring hypothesis in c is therefore essential even for a maximal ideal.

Consider the kernel of a ring homomorphismSince $R$ is a Noetherian ring, this ideal has a finite generating set of an ideal, say $K=(a_1,\dots ,a_t)$. For each $i$, choose $s_i\notin \mathfrak{p}$ with $s_i{a}_i=0$, and putThen $f\notin \mathfrak{p}$ by the prime ideal condition, and $fK=0$. If $K=0$, take the empty product $f=1$.

If $a/f^n$ maps to zero under the ring homomorphism ${\theta }_f$ from a, then $a/1=0$ in $R_{\mathfrak{p}}$ because $f/1$ is a unit there. Hence $a\in K$, so $fa=0$. The vanishing criterion in a module localization now gives $a/f^n=0$ already in $R_f$. ThereforeThis is the finite generation clears localization kernels argument: the finite generating set of an ideal permits a single element outside $\mathfrak{p}$ to annihilate the entire kernel of a ring homomorphism.

For $n\ge 1$, let $K_n=\ker ({\varphi }^n)$, where the power denotes repeated composition of the ring endomorphism. These ideals form an ascending chain:The ascending chain condition for a Noetherian ring gives $K_N=K_{N+1}$ for some $N\ge 1$. Take $a\in \ker \varphi$. Since $\varphi$ is surjective, its iterate ${\varphi }^N$ is surjective, so choose $b\in R$ with ${\varphi }^N(b)=a$. Thenso $b\in K_{N+1}=K_N$. Therefore $a={\varphi }^N(b)=0$.

Thus the kernel of a ring homomorphism $\varphi$ is zero. This proves that a surjective endomorphism of a Noetherian ring is an automorphism, and in particular

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 $e_{n+1}\notin I_n$.

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 quotient. Evaluation $\mathbb{Z}[t]\to \mathbb{Z}[i]$, $t\mapsto i$, is a surjective ring homomorphism whose kernel of a ring homomorphism is $(t^2+1)$. Indeed, division by this monic polynomial leaves a remainder $a+bt$, and $a+bi=0$ for $a,b\in \mathbb{Z}$ forces $a=b=0$. Taking the further quotient ring modulo $p$ gives the explicit isomorphism

All fibers, including the generic fiber. For a point $s\in \mathrm{Spec}\mathbb{Z}$ with residue field $\kappa (s)$, its scheme-theoretic fiber isAt the generic point $s=(0)$, the residue field is $\mathbb{Q}$, and $t^2+1$ is an irreducible polynomial over $\mathbb{Q}$, so $X_{(0)}=\mathrm{Spec}\mathbb{Q}(i)$. For a closed point $(p)$ the residue field is ${\mathbb{F}}_p$. The permitted criterion for a square root of $-1$ gives three cases. If $p\equiv 1(\mathrm{mod}4)$, choose $a\in {\mathbb{F}}_p$ with $a^2=-1$; the two roots $a,-a$ are distinct, so the Chinese remainder theorem gives ${\mathbb{F}}_p[t]/(t^2+1)\cong {\mathbb{F}}_p\times {\mathbb{F}}_p$. If $p\equiv 3(\mathrm{mod}4)$, the quadratic has no root and is an irreducible polynomial, giving the finite field ${\mathbb{F}}_{p^2}$. Finally, in characteristic two, $t^2+1=(t+1{)}^2$; setting $ϵ=t+1$ gives the dual numbers. ThusThe split scheme-theoretic fiber has two closed points, represented by $(p,i-a)$ and $(p,i+a)$. The inert scheme-theoretic fiber has one closed point, represented by $(p)$. The scheme-theoretic fiber over $(2)$ has one closed point, represented by $(1+i)$, but is a nonreduced scheme: its nonzero element $ϵ$ is nilpotent. These descriptions retain the residue fields and the nilpotent structure, which a count of points alone would lose.

The topology. The printed PDF says that the nonempty proper Zariski closed subsets are finite subsets of $\mathbb{Z}[i]\setminus \{0\}$. Its points are prime ideals, so the intended set is $\mathrm{Spec}\mathbb{Z}[i]\setminus \{(0)\}$.

Put $R=\mathbb{Z}[i]$. Since $R$ is a principal ideal domain, every nonzero prime ideal is a maximal ideal. Each such prime ideal contains a nonzero Gaussian integer $z$, and therefore contains the positive integer $z\overline{z}$. Its contraction to $\mathbb{Z}$ is consequently a nonzero prime ideal $(p)$, so the scheme-theoretic fibers just computed list all the closed points of $X$. The sole remaining point is the generic point $(0)$.

Every proper Zariski closed subset is $V((a))$ with $a\ne 0$: if its defining ideal were zero, it would be all of $X$. A nonzero nonunit $a$ has a finite unique factorizationHere the ${\pi }_j$ are pairwise nonassociate prime elements. Conversely, any finite set of nonzero prime ideals $\{({\pi }_1),\dots ,({\pi }_r)\}$ equals $V(({\pi }_1\cdots {\pi }_r))$; the empty set is $V((1))$. ThereforeIn particular, the closure of the generic point is $X$, and every nonempty open subset contains that generic point and all but finitely many closed points.

A nonidentity automorphism. Complex conjugationis a ring automorphism fixing $\mathbb{Z}$. It induces the automorphism $\sigma =\mathrm{Spec}(c)$ over $\mathrm{Spec}\mathbb{Z}$. Since $c(i)=-i\ne i$, this is not the identity morphism of schemes. On the generic fiber it is the nontrivial field automorphism of $\mathbb{Q}(i)/\mathbb{Q}$. On a split scheme-theoretic fiber it swaps the two closed points, since $i-a$ maps to $-(i+a)$. On an inert scheme-theoretic fiber it fixes the only point but induces the nonidentity finite-field Frobenius automorphism $z\mapsto z^p$ on its residue field, because $i^p=-i$. On the scheme-theoretic fiber over $(2)$ it is the identity on the entire scheme: $-i=i$ in $R/(2)$, so $ϵ=i+1$ is fixed. Thus