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For the discrete valuation $v$ corresponding to $|\cdot |$, the valuation ring and its maximal ideal areThe first is a subring of the field $K$, hence an integral domain. An element of ${\mathcal{O}}_K$ is a unit exactly when its valuation is zero, so every nonunit lies in $\mathfrak{m}$ and $\mathfrak{m}$ is the unique maximal ideal.

Choose a uniformizer $\pi$ with $v(\pi )=1$. If $I\ne 0$ is an ideal, the set of valuations of its nonzero elements has a least member $n$. Choose $x\in I$ with $v(x)=n$. Then $x=u{\pi }^n$ for a unit $u$, so $(x)=({\pi }^n)$. Every $y\in I$ has $v(y)\ge n$, hence $y/x\in {\mathcal{O}}_K$, and therefore $I=(x)$. Thus ${\mathcal{O}}_K$ is a discrete valuation ring, in particular a principal ideal domain.

Two nonisomorphic punctual schemes areTheir underlying spaces each have one point, but the second has a nonzero nilpotent element and the first is reduced.

After translating their common support to the origin, punctual closed subschemes of ${\mathbb{A}}_k^1$ correspond to ideals of $k[t]$ whose radical is $(t)$. Since $k[t]$ is a principal ideal domain, each such ideal is $(t^r)$ for a unique $r\ge 1$. Its coordinate ring has basis $1,t,\dots ,t^{r-1}$ and hence dimension $r$. Equal dimensions force equal exponents, so $Z$ and $Z^{\prime }$ are in fact the same closed subscheme after the common coordinate choice, and in particular are isomorphic.

In ${\mathbb{A}}_k^2$, the idealsdefine distinct punctual closed subschemes supported at the origin. Both quotient rings have dimension two over $k$, with bases $1,y$ and $1,x$ respectively. Thus they have equal-dimensional global-section spaces despite being distinct embedded closed subschemes.

A morphism of schemes $f:X\to Y$ is a flat morphism when every local-ring map ${\mathcal{O}}_{Y,f(x)}\to {\mathcal{O}}_{X,x}$ makes ${\mathcal{O}}_{X,x}$ a flat module.

In (i), the coordinate map is $k[t]\to k[y]$, $t\mapsto y^2$, andIt is therefore a free module of rank two and the morphism is flat, including in characteristic two.

In (ii), $k[x]$ is finite over the cusp ring $R=k[x^2,x^3]$ and has generic rank one. Were it flat, finite flatness over the local ring at the cusp would make it free of rank one. Its fiber there is insteadwhich has dimension two, so this morphism is not flat.

In (iii), the base coordinate $t$ acts as $x$, and the nonzero element $y$ satisfies $ty=xy=0$. Thus the coordinate ring has torsion as a $k[t]$-module. Since $k[t]$ is a principal ideal domain and a module over it is flat exactly when it is torsion-free, this morphism is not flat. Consequently

An integral domain is a nonzero commutative ring with identity and no zero divisors. A ring is Noetherian when every ideal is finitely generated, equivalently every ascending ideal chain stabilizes. A principal ideal domain is an integral domain in which every ideal is generated by one element.

Yes. The integers form a principal ideal domain, and over a principal ideal domain a module is flat exactly when it is torsion-free. Thus the torsion-free modules $A$ and $B$ are flat modules. The functoris a composite of two exact tensor functors, so $A{\otimes }_{\mathbb{Z}}B$ is flat. Applying the converse direction of the same characterization shows that it is torsion-free. This is the torsion-free module over a principal ideal domain is flat criterion.

This can be false. In a valued field with value group $\mathbb{Q}$, such as the Hahn series field $k((t^{\mathbb{Q}}))$, the maximal idealof the valuation ring is not principal: if $v(a)=q>0$, an element of valuation $q/2$ belongs to $\mathfrak{m}$ but not to $(a)$. Thus the valuation ring need not be a principal ideal domain.

The Kummer-Dedekind theorem states the following. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral and $p\nmid [{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor the minimal polynomial modulo $p$ as $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ into distinct monic irreducible factors. ThenThe index restriction is essential.

For the proposed ideal relation, reflexivity uses $\alpha =\beta =1$ and symmetry interchanges $\alpha ,\beta$. If $(\alpha )I_1=(\beta )I_2$ and $(\gamma )I_2=(\delta )I_3$, multiplication gives $(\alpha \gamma )I_1=(\beta \delta )I_3$. All multipliers remain nonzero. Thus it is an equivalence relation, namely equality of ideal classes.

Put $s=\sqrt{-2}$. The minimal polynomial is $x^2+2$ and the given full power basis makes every prime eligible for the theorem. Modulo $3$ its roots are $\pm 1$; modulo $5$ and $7$ it is irreducible; modulo $17$ its roots are $\pm 7$. ConsequentlyThe two inert ideals have norms $25$ and $49$; each split factor has norm equal to its rational prime.

For $p_1\equiv 1(\mathrm{mod}8)$, quadratic reciprocity and its supplementary laws give $(-2/p_1)=1$. Thus $x^2+2$ has two distinct roots modulo $p_1$, giving distinct prime ideals $P,P^{\prime }$ withThe given principal ideal domain property makes $P=(x+ys)$ for some integers $x,y$. Taking norms, which are positive here, gives $p_1=N(P)=x^2+2y^2$.

Let $\omega =(1+\sqrt{-3})/2$, whose minimal polynomial is $x^2-x+1$. For $p_2\equiv 1(\mathrm{mod}3)$, the cyclic group ${\mathbb{F}}_{p_2}^{\times }$ has a nontrivial cube root of unity $t$. Then $-t$ and $-t^2$ are distinct roots of $x^2-x+1$. By the Kummer-Dedekind theorem,Since the given ring of integers is a principal ideal domain, write $Q=(a+b\omega )$. Its norm isMultiplication by $4$ completes the square:Thus $x=2a+b$ and $y=b$ supply the required integers.

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).

Since ${\omega }^2=-1-\omega$, products of integer linear combinations of $1,\omega$ remain of the same form. Thus the Eisenstein integers $E=\mathbb{Z}[\omega ]$ form a subring of $\mathbb{C}$, and hence an integral domain. Their Eisenstein norm isIt is a positive integer on nonzero elements and is multiplicative, because the complex modulus is multiplicative.

For $\alpha ,\beta \in E$ with $\beta \ne 0$, write $\alpha /\beta =u+v\omega$ with real $u,v$. Choose integers $m,n$ with $|u-m|,|v-n|\le 1/2$, and let $q=m+n\omega$. With $s=u-m$ and $t=v-n$,Consequently $r=\alpha -q\beta$ belongs to $E$ and satisfies $N(r)\le \frac{3}{4}N(\beta )<N(\beta )$ when $r\ne 0$. This proves $E$ is a Euclidean domain with Euclidean function $N$, the Euclidean norm on the Eisenstein integers.

For the representation of the prime $p$, first note that $3=N(1-\omega )$. Now suppose $p>3$ and $p\mid n^2+3$. The factorizationshows that the Eisenstein integer $p$ divides a product. It divides neither factor: an element $a+b\omega$ lies in $pE$ exactly when both integer coefficients are divisible by $p$, and the second coefficients here are $2$ and $-2$. Hence $p$ is not a prime element of $E$.

To justify the necessary factorization step, in any principal ideal domain an irreducible element $\pi$ is prime. If $\pi \nmid a$, write $(\pi ,a)=(d)$. Since $d\mid \pi$, irreducibility makes $d$ a unit or an associate of $\pi$. The latter would give $\pi \mid a$, so $d$ is a unit and $u\pi +va=1$ for some $u,v$. If $\pi \mid ab$, multiplication of this identity by $b$ gives $\pi \mid b$. This proves irreducible elements of a principal ideal domain are prime without assuming unique factorization.

By part (a), $E$ is a principal ideal domain. The integer $p$ has norm $p^2>1$, so it is a nonzero nonunit. Since it is not prime, the result just proved implies that it is reducible: $p=uv$ with $u,v$ both nonunits. A nonzero element of $E$ is a unit exactly when its norm is one, since $N(z)=1$ gives inverse $\overline{z}$ and multiplicativity proves the converse. ThusThe only possible positive integer factors exceeding one are $N(u)=N(v)=p$. Writing $u=a+b\omega$ givesThis is the prime representation by the Eisenstein norm argument.

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.

We prove the stated Eisenstein criterion directly. Assume $n\ge 1$. Suppose $f=gh$ in the polynomial ring $R[X]$. Since the principal ideal domain $R$ is an integral domain, polynomial degree is additive under polynomial multiplication, and the product of the leading coefficients of $g,h$ is the leading coefficient $1$ of $f$. These two coefficients are therefore units. Multiplying $g$ and $h$ by inverse units, without changing their product, makes both factors monic polynomials.

If either factor is a constant polynomial, that factor is now $1$, so the original factor was a unit. It remains to rule out $r=\mathrm{deg}g>0$ and $s=\mathrm{deg}h>0$.

The prime element $p$ generates a maximal ideal in the principal ideal domain $R$. Indeed, if $(p)\subseteq (d)$, then $p=de$, and the irreducible element property of $p$ forces either $(d)=R$ or $(d)=(p)$. Consequently $k=R/(p)$ is a field. Apply reduction modulo an ideal to every coefficient. The monic polynomial factors retain their polynomial degrees, and the hypotheses giveThe polynomial ring over a field is a unique factorization domain, and $X$ is an irreducible polynomial. Thus its only monic polynomial divisors in $X^n$ are powers of $X$. Hence $\overline{g}=X^r$ and $\overline{h}=X^s$. Both have zero constant term, so $p\mid g(0)$ and $p\mid h(0)$. This gives $p^2\mid g(0)h(0)=a_0$, a contradiction.

The monic polynomial $f$ has positive polynomial degree, so it is a nonzero nonunit. Every factorization therefore has a unit factor, proving

Every multiple of $dh$ is a multiple of both $d$ and $h$, so $dh\mathbb{Z}\subseteq d\mathbb{Z}\cap h\mathbb{Z}$.

For the reverse inclusion, let $z=du=hv$ lie in this intersection of ideals. Since $d,h$ are coprime integers, Bézout's identity supplies $r,s\in \mathbb{Z}$ with $rd+sh=1$. Multiplying by $u$ givesThus $h\mid u$, and $dh\mid du=z$. ConsequentlyHere Bézout's identity follows from the allowed principal ideal domain property of $\mathbb{Z}$: the ideal $d\mathbb{Z}+h\mathbb{Z}$ is generated by their greatest common divisor, which is $1$.

The polynomial ring ${\mathbb{F}}_2[X]$ is a euclidean domain under polynomial degree, hence a principal ideal domain. By part (b)(i), eachis a field. By polynomial division, every residue class has exactly one representative of polynomial degree less than $n_j$. Thus $1,\overline{X},\dots ,{\overline{X}}^{n_j-1}$ form a basis over ${\mathbb{F}}_2$, and $|K_j|=2^{n_j}$.

For $j\ne k$, the ideals $(f_j)$ and $(f_k)$ are distinct maximal ideals, so they are comaximal ideals: their sum is an ideal properly containing $(f_j)$ and must be the whole ring. The Chinese remainder theorem for ideals and part (b)(ii) therefore give the ring isomorphismIn particular, this is a product of finite fields of the required sizes. The map preserves multiplication as well as addition and is surjective by the Chinese remainder theorem for ideals.

Every multiple of $x_1\cdots x_r$ is a multiple of each $x_j$, giving one inclusion. For the reverse inclusion, a principal ideal domain is a unique factorization domain, so each irreducible element $x_j$ is a prime element.

Let $a\in {\bigcap }_{j=1}^r(x_j)$. The case $a=0$ is immediate. Otherwise, write $a=x_{1}b$. Since $x_2\mid a$ and the prime element $x_2$ cannot divide the non-associate irreducible element $x_1$, it divides $b$. Write $a=x_1{x}_2{b}_2$, and repeat. At stage $j$, the prime element $x_j$ divides none of $x_1,\dots ,x_{j-1}$, so it divides the remaining factor. Hence $x_1\cdots x_r\mid a$, proving

Since the irreducible element $x$ is not a unit, the principal ideal $(x)$ is proper. Suppose $(x)\subseteq J\subseteq R$. As $R$ is a principal ideal domain, write $J=(d)$. Then $x=dr$ for some $r\in R$. The definition of an irreducible element says that $d$ or $r$ is a unit.

If $d$ is a unit, then $J=R$. If $r$ is a unit, then $d$ and $x$ are associate elements, so $J=(x)$. Thus $(x)$ is a maximal ideal. By the permitted criterion that a quotient ring by a maximal ideal is a field,

Regard $A$ as a matrix over the rational numbers. Part (a)(ii), with $n=2$, gives its characteristic polynomial $X^2+1$. Therefore, writing $i$ for the imaginary unit,The matrix $M=iI-A$ is not zero: otherwise $A=iI$ would have noninteger diagonal entries. It consequently has rank of a matrix one over the fraction field $\mathbb{Q}(i)$ of the Gaussian integers.

By part (b)(i), $\mathbb{Z}[i]$ is a euclidean domain, hence a principal ideal domain, so apply Smith normal form. There exist $P,Q\in {\mathrm{GL}}_2(\mathbb{Z}[i])$ such thatIndeed, invertible multiplication preserves rank of a matrix, so exactly one diagonal entry is nonzero. For $e_2=(0,1{)}^{\mathsf{T}}$, multiplication on the right gives $P(iI-A)Q{e}_2=0$. Since $P$ is an invertible matrix,The Smith normal form supplies a kernel basis lemma ensures that the null vector $Q{e}_2$ is a column of an invertible matrix over $\mathbb{Z}[i]$; an arbitrary nonzero solution over $\mathbb{Q}(i)$ would not itself establish that requirement.

Over a principal ideal domain $D$, an $m$-by-$n$ matrix is in Smith normal form if all entries off the main diagonal vanish and its nonzero diagonal entries $d_1,\dots ,d_r$ occur first and satisfyAll remaining diagonal entries are zero. The zero matrix is allowed, with $r=0$. For an arbitrary matrix $M$, the Smith normal form theorem gives invertible matrices $P\in {\mathrm{GL}}_m(D)$ and $Q\in {\mathrm{GL}}_n(D)$ withwhere the right side is interpreted as a rectangular diagonal matrix when necessary. The $d_j$ are determined up to multiplication by units. Positivity can be required over $\mathbb{Z}$, but is not a condition over the Gaussian integers.

The prime ideals $P$ of $\mathbb{Z}[t]$ with $P\cap \mathbb{Z}=\{0\}$ are the zero ideal and the principal ideals $(f)$ generated by nonconstant primitive polynomials that are irreducible in $\mathbb{Q}[t]$.

To prove the description, invert the nonzero integers to obtain the localization $\mathbb{Q}[t]$. The extension of an ideal $P$ consists of the polynomials $f/n$ with $f\in P$ and $n>0$. Its contraction is $P$: if $nh\in P$ and $n\ne 0$, the defining condition puts $n\notin P$, so primeness gives $h\in P$. The prime ideal correspondence for localization makes this extension a proper prime ideal of $\mathbb{Q}[t]$.

If $P\ne 0$, its extension has a nonzero generator because $\mathbb{Q}[t]$ is a principal ideal domain. Dividing this generator by its rational polynomial content gives an integer-coefficient primitive polynomial $f$ generating the same extended ideal. Divisibility descent by Gauss lemma for polynomials then gives $P=(f)$ in $\mathbb{Z}[t]$. The proper prime ideal $(f)$ in $\mathbb{Q}[t]$ makes $f$ irreducible there. Conversely, any such polynomial generates a prime ideal in $\mathbb{Q}[t]$ whose contraction is $(f)\subset \mathbb{Z}[t]$ and whose intersection with $\mathbb{Z}$ is zero.

The intersection condition is essential: $(2,t)$ is a maximal ideal of $\mathbb{Z}[t]$, since its quotient ring is ${\mathbb{F}}_2$, but it is not principal. A common divisor of $2$ and $t$ must be a unit, which would generate the whole ring.

Let $p\ge 3$ be a prime number dividing $n^2+3$, and set $\omega =(-1+i\sqrt{3})/2$. The case $p=3$ is $N(1-\omega )=3$. For $p>3$, useThe Eisenstein integer $p$ divides this product, but divides neither factor: divisibility by the rational integer $p$ in the basis $1,\omega$ requires both integer coefficients to be divisible by $p$, whereas the second coefficients are $2$ and $-2$. Thus $p$ is not a prime element of $\mathbb{Z}[\omega ]$.

The Euclidean norm on the Eisenstein integers makes this ring a principal ideal domain. Since irreducible elements of a principal ideal domain are prime, $p$ admits a factorization $p=uv$ into two nonunits. The positive integral Eisenstein norm is multiplicative, and its value is one exactly on units: when $N(u)=1$, its complex conjugate is an inverse. Thus $N(u),N(v)>1$ and $N(u)N(v)=p^2$, forcing both norms to be $p$. Writing $u=a+b\omega$ proves the displayed representation.