Prime Spectra and Radicals Examples

A proper ideal of a field cannot contain a nonzero element, because any such element is a unit and would force $1$ into the ideal. Hence the only proper ideal of $F$ is $(0)$, and $(0)$ is prime because $F$ has no zero divisors.

Let $I⊴\mathbb{Z}$. If $I\mathrm{\ne }(0)$, choose the least positive integer $n\in I$ using [L1]. For any $m\in I$, [L4] gives $m=qn+r$ with $0\le r<n$; since $r=m-qn\in I$ and $n$ was the least positive element of $I$, one must have $r=0$, so every element of $I$ is a multiple of $n$ and therefore $I=(n)$.

If $n=ab$ with $1<a,b<n$, then $(a+(n))(b+(n))=0+(n)$ in $\mathbb{Z}\mathrm{/}(n)$ while neither factor is zero, so [L2] shows $(n)$ is not prime. If $n=p$ is prime, then [L3] makes $\mathbb{Z}\mathrm{/}(p)$ a field and hence a domain, so [L2] shows $(p)$ is prime. Finally $(0)$ is prime because $\mathbb{Z}$ is an integral domain.

Steps 1.1, 1.2, and 1.3 prove the listed prime ideals of $F$ and $\mathbb{Z}$, and the inclusion order is immediate because every nonzero prime ideal of $\mathbb{Z}$ is maximal.

Let $P\in \mathrm{Spec}(R\times S)$. The idempotents $e_1=(1,0)$ and $e_2=(0,1)$ satisfy $e_1{e}_2=0\in P$, so [L1] gives $e_1\in P$ or $e_2\in P$. They cannot both lie in $P$ because then $1=e_1+e_2\in P$. If $e_1\in P$, then every $(r,0)=e_1(r,s)$ lies in $P$, so $P=R\times \mathfrak{q}$ where $\mathfrak{q}=\{s\in S:(0,s)\in P\}$. The same argument with $e_2$ gives the alternative form $\mathfrak{p}\times S$. In either case the factor ideal is prime because products in the factor ring are products in $R\times S$.

Conversely, if $\mathfrak{p}\in \mathrm{Spec}(R)$, then $\mathfrak{p}\times S$ is a prime ideal of $R\times S$: if $(r_1,s_1)(r_2,s_2)\in \mathfrak{p}\times S$, then $r_1{r}_2\in \mathfrak{p}$, so [L1] gives $r_1\in \mathfrak{p}$ or $r_2\in \mathfrak{p}$, which means $(r_1,s_1)$ or $(r_2,s_2)$ lies in $\mathfrak{p}\times S$. The same proof works for $R\times \mathfrak{q}$.

Therefore every prime ideal of the product ring comes from exactly one factor.

By [L1], prime ideals of $A$ correspond to prime ideals of $k[\epsilon ]$ containing $({\epsilon }^2)$. Any such prime contains $\epsilon$ because ${\epsilon }^2$ lies in it. The ideal $(\epsilon )$ itself is prime since $k[\epsilon ]\mathrm{/}(\epsilon )\cong k$ is a field. Therefore $\mathrm{Spec}(A)=\{(\bar{\epsilon })\}$.

The element $\bar{\epsilon }\in A$ is nilpotent, so the reduced quotient of $A$ is exactly $A\mathrm{/}(\bar{\epsilon })\cong k$. By [L2], this quotient has the same prime spectrum as $A$, which is the singleton from step 1.1.

Hence the dual numbers and their reduced quotient have the same prime set.

By [L1], prime ideals of $A$ correspond to prime ideals of $k[x,y]$ containing $(xy)$. If $\mathfrak{p}$ contains $(xy)$, then $xy\in \mathfrak{p}$, so [L2] gives $x\in \mathfrak{p}$ or $y\in \mathfrak{p}$. Therefore every prime of $A$ contains $(\bar{x})$ or $(\bar{y})$. Since $A\mathrm{/}(\bar{x})\cong k[y]$ and $A\mathrm{/}(\bar{y})\cong k[x]$ are integral domains, both $(\bar{x})$ and $(\bar{y})$ are prime, and no smaller prime can contain them. Thus they are the minimal primes of $A$.

If $\mathfrak{m}$ is maximal in $A$, then it is prime, so step 1.1 shows that it contains $(\bar{x})$ or $(\bar{y})$. The maximal ideal $(\bar{x},\bar{y})$ indeed contains both, so "at least one" is the correct boundary statement here.

This gives the minimal-prime picture of the node ring and the maximal-ideal boundary at the singular point.

By [L1], a prime ideal of $k[x,y]$ lies in $D(x)$ exactly when it avoids $x$.

By [L2], the same condition characterizes the prime ideals that survive in the localization $k[x,y{]}_x$. Therefore the contraction map from $\mathrm{Spec}(k[x,y{]}_x)$ identifies its image with $D(x)$.

So $D(x)$ is the prime-set avatar of localizing at $x$.

If $\mathfrak{p}$ is a prime ideal containing $I$, then $x^2\in \mathfrak{p}$, so [L2] gives $x\in \mathfrak{p}$. Conversely, any prime ideal containing $x$ also contains $x^2$ and $xy$, hence contains $I$. Thus the primes over $I$ are exactly the primes containing $x$.

The ideal $(x)$ is itself prime because $k[x,y]\mathrm{/}(x)\cong k[y]$ is an integral domain. Therefore [L1] and step 1.1 give $\sqrt{I}={\bigcap }_{\begin{array}{c}\mathfrak{p}\in \mathrm{Spec}(k[x,y])\\ I\subseteq \mathfrak{p}\end{array}}\mathfrak{p}={\bigcap }_{\begin{array}{c}\mathfrak{p}\in \mathrm{Spec}(k[x,y])\\ x\in \mathfrak{p}\end{array}}\mathfrak{p}=(x)\mathrm{.}$

Hence $\sqrt{(x^2,xy)}=(x)$, as claimed.

By [L1] and [L2], $\sqrt{(x^2,xy)\cap (y^3)}=\sqrt{(x^2,xy)}\cap \sqrt{(y^3)}=(x)\cap \sqrt{(y^3)}$. Since $y^3\in (y^3)$, [L3] gives $y\in \sqrt{(y^3)}$, so $(y)\subseteq \sqrt{(y^3)}$. Conversely, if $g\in \sqrt{(y^3)}$, choose $m\ge 1$ with $g^m\in (y^3)$ and write $g=a(x)+yh(x,y)$ with $a(x)\in k[x]$; if $a(x)\mathrm{\ne }0$, then $g^m=a(x{)}^m+yb(x,y)$ for some $b(x,y)\in k[x,y]$, so $g^m$ has a nonzero $y^0$ term and cannot lie in $(y^3)$, a contradiction. Hence $a(x)=0$, so $g\in (y)$. Thus $\sqrt{(y^3)}=(y)$.

Step 1.1 reduces the problem to $(x)\cap (y)$. A polynomial lies in both ideals exactly when it is divisible by both $x$ and $y$, hence by $xy$. Therefore $(x)\cap (y)=(xy)$.

Therefore $\sqrt{(x^2,xy)\cap (y^3)}=(xy)$.

Every element of $I=(x^2)$ is divisible by $x^2$, so no power $y^n$ lies in $I$.

Applying [L1] to step 1.1 yields a prime ideal containing $I$ but avoiding $y$. Concretely, $(x)$ works: it contains $x^2$ and does not contain $y$.

Thus this example exhibits the separating-prime conclusion directly.

In the zero ring one has $0=1$, so the only ideal is the whole ring. A prime ideal must be proper, so no prime ideals exist. Therefore $\mathrm{Spec}(R)=\mathrm{\varnothing }$.

Since every subset named in [L2] is defined as a subset of $\mathrm{Spec}(R)$, step 1.1 forces all of them to be empty.

Hence the zero ring has empty prime spectrum, and all the displayed $V$ and $D$ boundary sets are empty as well.

The ideals $(0)$, $(x)$, $(y)$, and $(x+y)$ are prime in $A$, and each is contained in $\mathfrak{m}$ because their generators lie in $(x,y)$.

By [L1], each of these four primes determines a distinct prime ideal of $A_{\mathfrak{m}}$, and every prime of $A_{\mathfrak{m}}$ comes from some prime of $A$ contained in $\mathfrak{m}$.

This records a concrete finite family of primes inside the prime localization.

Every element of $(\bar{x},\bar{y})$ is a $k$-linear combination of positive-degree residue classes, and every such monomial is nilpotent because powers eventually hit one of the relations $x^3=0$, $x^{2}y=0$, or $y^4=0$. Thus $(\bar{x},\bar{y})$ is the nilradical.

Any monomial of total degree $5$ in $\bar{x}$ and $\bar{y}$ either has $\bar{y}$-exponent at least $4$ or $\bar{x}$-exponent at least $2$ together with a positive $\bar{y}$-exponent, or else $\bar{x}$-exponent at least $3$; each case is zero in $A$. Hence $(\bar{x},\bar{y}{)}^5=(0)$. The theorem [L1] guarantees that some common exponent must exist; this computation shows that $5$ works in this example.

Therefore the nilradical of this Noetherian quotient has a concrete common nilpotence exponent.

Let $\mathfrak{n}=({\bar{x}}_1,{\bar{x}}_2,\dots )$. Each generator ${\bar{x}}_i$ is nilpotent, and every element of $\mathfrak{n}$ involves only finitely many generators, so in this commutative ring it is a finite sum of commuting nilpotents and hence nilpotent. Therefore $\mathfrak{n}\subseteq \mathrm{Nil}(A)$. Conversely, the quotient $A\mathrm{/}\mathfrak{n}\cong k$ is reduced, so every nilpotent element of $A$ lies in $\mathfrak{n}$. Thus $\mathrm{Nil}(A)=\mathfrak{n}$.

For every integer $N\ge 1$, the element ${\bar{x}}_{N+1}^N$ lies in ${\mathfrak{n}}^N$. It is nonzero in $A$ because only the $(N+1)$-st power of $x_{N+1}$ is killed by the defining relations. Hence ${\mathfrak{n}}^N\mathrm{\ne }0$ for every $N$, so $\mathfrak{n}$ is not nilpotent.

This shows that the Noetherian hypothesis in the nilradical-nilpotence theorem is essential.