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Let $\mathcal{F}$ be the proper nonprincipal ideals. A chain in $\mathcal{F}$ has a nonprincipal union: if its union were $(a)$, then $a$ would belong to one member of the chain, forcing that member to equal the union and be principal. The union is also proper. Thus Zorn lemma gives a maximal member whenever $\mathcal{F}$ is nonempty.

The family satisfies the condition from part (a). Indeed, supposeWrite $a=bd$ and $b=i+ra$, where $i\in I$. Every $x=bs\in I$ has $as=dx\in I$, so $s\in (c)$ and $I\subseteq (bc)$. Conversely $ac\in I$ because $c\in (I:a)$, while $ic\in I$ because $i\in I$. Thus $bc=ic+rac\in I$, proving $(bc)\subseteq I$. Hence $I=(bc)$ is principal.

If a nonprincipal ideal existed, part (a) would therefore produce a nonprincipal prime ideal, contrary to the hypothesis. Every ideal is principal, so the integral domain is a principal ideal domain.

Let $A$ be a principal ideal domain that is not a field. A PID is Noetherian and is a unique factorization domain. Every nonzero prime ideal is generated by a prime element. Ifthen $a\mid p$, so primality of $p$ makes $a$ a unit or an associate of $p$. The proper alternative is $(a)=(p)$, and every nonzero prime is therefore maximal. Since $A$ has a nonzero prime ideal, its Krull dimension is exactly one.

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.

For every matrix $A$ over a principal ideal domain, there are invertible matrices $U,V$ such thatThe nonzero diagonal entries are unique up to multiplication by units. Over the integers, $U$ and $V$ are unimodular and the $d_i$ may be chosen positive; their products $d_1\cdots d_k$ are the greatest common divisors of the $k$-rowed minors.

Suppose the ring of integers of a number field is a principal ideal domain, $a,b$ are integral, and $b$ is a unit. On $y^2=x(x^2+ax+b)$ every nonzero $x$-coordinate has even valuation at every finite prime: positive valuation gives $2v(y)=v(x)$, and negative valuation gives $2v(y)=3v(x)$. Thus the two-torsion square-class homomorphism takes values among unit square classes. For the Gaussian integers, their unit group modulo squares has two elements, represented by $1,i$. This can reduce a two-isogeny descent to a very small calculation even though the field contains no ordering.