Prime Spectra and Radicals

The displayed equivalence is exactly the defining membership criterion in [L1].

Taking $I=R$ in step 1.1 gives $\sqrt{R}=R$, because every element already has its first power in $R$. If $R$ is the zero ring, then $(0)=R$, so the same observation gives $\sqrt{(0)}=R$.

Steps 1.1 and 2.1 prove the claim and record the unit-ideal and zero-ring boundaries explicitly.

If $a\in I$, then $a^1\in I$, so $a\in \sqrt{I}$ by [L1]. Thus $I\subseteq \sqrt{I}$. If $r\in R$ and $x\in \sqrt{I}$, choose $n\ge 1$ with $x^n\in I$. Then $(rx{)}^n=r^n{x}^n\in I$, so $rx\in \sqrt{I}$.

Let $x,y\in \sqrt{I}$. Choose $m,n\ge 1$ with $x^m\in I$ and $y^n\in I$, and set $N=m+n$. By [L2], every term of $(x+y{)}^N$ has the form $\left(\genfrac{}{}{0px}{}{N}{k}\right)x^k{y}^{N-k}$. For each $k$, either $k\ge m$ or $N-k\ge n$; otherwise $k\le m-1$ and $N-k\le n-1$, which would force $N\le m+n-2$. Hence each term lies in $I$, so $(x+y{)}^N\in I$ and $x+y\in \sqrt{I}$.

Steps 1.1 and 1.2 show that $\sqrt{I}$ is an ideal containing $I$. If $I\subseteq J$ and $x\in \sqrt{I}$, any power of $x$ lying in $I$ also lies in $J$, so $x\in \sqrt{J}$. Thus radical is order-preserving.

If $x\in \sqrt{\sqrt{I}}$, choose $m\ge 1$ with $x^m\in \sqrt{I}$, and then choose $n\ge 1$ with $x^{mn}\in I$. By [L1], this means $x\in \sqrt{I}$. Together with step 2.1 applied to $I\subseteq \sqrt{I}$, this proves $\sqrt{\sqrt{I}}=\sqrt{I}$.

The radical construction therefore sends ideals to radical ideals, contains the original ideal, and is order-preserving and idempotent.

Since $I_1\cap \cdots \cap I_n\subseteq I_j$ for every $j$, [L1] gives $\sqrt{I_1\cap \cdots \cap I_n}\subseteq \sqrt{I_j}$ for every $j$, hence $\sqrt{I_1\cap \cdots \cap I_n}\subseteq \sqrt{I_1}\cap \cdots \cap \sqrt{I_n}$.

Conversely, let $x\in \sqrt{I_1}\cap \cdots \cap \sqrt{I_n}$. For each $j$, choose $m_j\ge 1$ with $x^{m_j}\in I_j$, and set $N=m_1+\cdots +m_n$. Then $x^N=x^{N-m_j}{x}^{m_j}\in I_j$ for every $j$, so $x^N\in I_1\cap \cdots \cap I_n$. Hence $x\in \sqrt{I_1\cap \cdots \cap I_n}$.

The two inclusions prove the stated equality.

Every generator $ij$ of $IJ$ lies in both $I$ and $J$, so $IJ\subseteq I\cap J$. Therefore $\sqrt{IJ}\subseteq \sqrt{I\cap J}=\sqrt{I}\cap \sqrt{J}$ by [L1].

If $x\in \sqrt{I}\cap \sqrt{J}$, choose $m,n\ge 1$ with $x^m\in I$ and $x^n\in J$. Then $x^{m+n}=x^m{x}^n\in IJ$, so $x\in \sqrt{IJ}$.

Step 1.1 gives $\sqrt{IJ}\subseteq \sqrt{I}\cap \sqrt{J}$, step 1.2 gives the reverse inclusion, and [L1] identifies that common ideal with $\sqrt{I\cap J}$.

Let $x+I\in R\mathrm{/}I$. By [L2], $x+I\in \sqrt{J\mathrm{/}I}$ exactly when $(x+I{)}^n=x^n+I\in J\mathrm{/}I$ for some $n\ge 1$, and that happens exactly when $x^n\in J$. Applying [L2] again shows that this is equivalent to $x\in \sqrt{J}$, so $x+I\in \sqrt{J\mathrm{/}I}$ exactly when $x+I\in \sqrt{J}\mathrm{/}I$.

Step 1.1 proves $\sqrt{J\mathrm{/}I}=\sqrt{J}\mathrm{/}I$. Consequently, $J\mathrm{/}I$ is radical exactly when $J\mathrm{/}I=\sqrt{J\mathrm{/}I}$, exactly when $J\mathrm{/}I=\sqrt{J}\mathrm{/}I$, and exactly when $J=\sqrt{J}$.

The quotient radical is therefore exactly the quotient of the radical, and radical ideals correspond across the quotient map.

If $a\mathrm{/}s\in S^{-1}\text{\hspace{0.17em}⁣}\sqrt{I}$, choose $n\ge 1$ with $a^n\in I$. Then $(a\mathrm{/}s{)}^n=a^n\mathrm{/}{s}^n\in S^{-1}I$, so $a\mathrm{/}s\in \sqrt{S^{-1}I}$ by [L1]. This proves $S^{-1}\text{\hspace{0.17em}⁣}\sqrt{I}\subseteq \sqrt{S^{-1}I}$.

Conversely, let $r\mathrm{/}s\in \sqrt{S^{-1}I}$. Choose $n\ge 1$ with $r^n\mathrm{/}{s}^n\in S^{-1}I$, and then choose $a\in I$ and $u\in S$ with $r^n\mathrm{/}{s}^n=a\mathrm{/}u$. By [L2], some $t\in S$ satisfies $t(u{r}^n-a{s}^n)=0$. Hence $(tu)r^n=ta{s}^n\in I$, so $((tu)r{)}^n=(tu{)}^{n-1}((tu)r^n)\in I$. Thus $(tu)r\in \sqrt{I}$ by [L1], and $r\mathrm{/}s=((tu)r)\mathrm{/}((tu)s)$ lies in $S^{-1}\text{\hspace{0.17em}⁣}\sqrt{I}$.

Steps 1.1 and 1.2 prove the equality $S^{-1}\text{\hspace{0.17em}⁣}\sqrt{I}=\sqrt{S^{-1}I}$.

Let $x+N\in R\mathrm{/}N$ be nilpotent. Then $(x+N{)}^m=N$ for some $m\ge 1$, so $x^m\in N$. By [L1], some power of $x^m$ is zero, hence some power of $x$ is zero, so $x\in N$. Therefore $x+N=0+N$, and the only nilpotent element of $R\mathrm{/}N$ is zero. Thus $R\mathrm{/}N$ is reduced by [L1].

Let $\phi :R\to A$ with $A$ reduced. If $x\in N$, then $x^m=0$ for some $m\ge 1$, so $\phi (x{)}^m=0$. Reducedness of $A$ forces $\phi (x)=0$, so $N\subseteq \ker \phi$. Therefore $\bar{\phi }(x+N):=\phi (x)$ is well-defined, and it is unique because $\pi$ is surjective.

The quotient by the nilradical is reduced and is universal among maps from $R$ to reduced rings.

Let $\mathrm{\Sigma }$ be the set of ideals $J⊴R$ with $I\subseteq J$ and $J\cap S=\mathrm{\varnothing }$, ordered by inclusion. The ideal $I$ lies in $\mathrm{\Sigma }$, so $\mathrm{\Sigma }\mathrm{\ne }\mathrm{\varnothing }$.

If $\mathcal{C}\subseteq \mathrm{\Sigma }$ is a chain, then $J=\bigcup \mathcal{C}$ is an ideal containing $I$. Moreover $J\cap S=\mathrm{\varnothing }$, because if $s\in J\cap S$ then $s\in C$ for some $C\in \mathcal{C}$, contradicting $C\cap S=\mathrm{\varnothing }$. Thus every chain in $\mathrm{\Sigma }$ has an upper bound.

Zorn's lemma yields a maximal member $\mathfrak{p}$ of $\mathrm{\Sigma }$. Because $1\in S$ by [L2] and $\mathfrak{p}\cap S=\mathrm{\varnothing }$, the ideal $\mathfrak{p}$ is proper.

Suppose $ab\in \mathfrak{p}$ while $a\mathrm{\notin }\mathfrak{p}$ and $b\mathrm{\notin }\mathfrak{p}$. By maximality of $\mathfrak{p}$ in $\mathrm{\Sigma }$, the larger ideals $\mathfrak{p}+(a)$ and $\mathfrak{p}+(b)$ must meet $S$. Choose $\nu =m+ra\in S\cap (\mathfrak{p}+(a))$ and $\omega =n+tb\in S\cap (\mathfrak{p}+(b))$ with $m,n\in \mathfrak{p}$ and $r,t\in R$. Then $\nu \omega =mn+mtb+nra+rtab\in \mathfrak{p}$, because $m,n,ab\in \mathfrak{p}$. But $\nu \omega \in S$ by [L2], contradicting $\mathfrak{p}\cap S=\mathrm{\varnothing }$. Therefore $\mathfrak{p}$ is prime by [L1].

The ideal $\mathfrak{p}$ is prime, contains $I$, and is disjoint from $S$, exactly as required.

The set $S=\{1,f,f^2,\dots \}$ is multiplicative, and the hypothesis says exactly that $I\cap S=\mathrm{\varnothing }$.

Applying [L1] to the ideal $I$ and the multiplicative set $S$ yields a prime ideal $\mathfrak{p}$ with $I\subseteq \mathfrak{p}$ and $\mathfrak{p}\cap S=\mathrm{\varnothing }$. In particular $f\mathrm{\notin }\mathfrak{p}$.

This is the required separating prime.

Let $x\in \sqrt{I}$. Choose $n\ge 1$ with $x^n\in I$. Since $I\subseteq \mathfrak{p}$, one has $x^n\in \mathfrak{p}$.

Repeatedly applying primality from [L2] to the factorization $x^n=x\cdot x^{n-1}$ shows that $x\in \mathfrak{p}$: if $x\mathrm{\notin }\mathfrak{p}$, then $x^{n-1}\in \mathfrak{p}$; repeating the same argument eventually forces $x\in \mathfrak{p}$ after all.

Every element of $\sqrt{I}$ lies in $\mathfrak{p}$, so $\sqrt{I}\subseteq \mathfrak{p}$.

Since $f\mathrm{\notin }\sqrt{I}$, [L1] says that $f^n\mathrm{\notin }I$ for every integer $n\ge 1$.

Applying [L2] to step 1.1 yields a prime ideal $\mathfrak{p}$ with $I\subseteq \mathfrak{p}$ and $f\mathrm{\notin }\mathfrak{p}$.

This prime separates $f$ from the radical of $I$.

Let $x\in \sqrt{I}$. By [L1], every prime ideal containing $I$ also contains $x$. Therefore $x$ belongs to the displayed intersection.

Let $x\mathrm{\notin }\sqrt{I}$. By [L2], there is a prime ideal $\mathfrak{p}$ containing $I$ with $x\mathrm{\notin }\mathfrak{p}$. Hence $x$ does not belong to the displayed intersection.

Steps 1.1 and 1.2 prove that an element belongs to $\sqrt{I}$ exactly when it belongs to every prime ideal containing $I$, which is the claimed equality.

By [L1], $\mathrm{Nil}(R)=\sqrt{(0)}$.

Applying [L2] to the zero ideal gives $\sqrt{(0)}={\bigcap }_{\mathfrak{p}\in \mathrm{Spec}R}\mathfrak{p}$. Combining this with step 1.1 yields the claimed formula for the nilradical.

Therefore the nilradical is exactly the intersection of all prime ideals of $R$.

If $R$ is reduced, then $\mathrm{Nil}(R)=(0)$ by [L1]. Applying [L2] yields $(0)={\bigcap }_{\mathfrak{p}\in \mathrm{Spec}R}\mathfrak{p}$.

Conversely, if $(0)={\bigcap }_{\mathfrak{p}\in \mathrm{Spec}R}\mathfrak{p}$, then [L2] shows that $\mathrm{Nil}(R)=(0)$. Now [L1] gives that $R$ is reduced.

Steps 1.1 and 1.2 prove the equivalence.

Let $\mathfrak{p}\in V(J)$. By [L1], this means $J\subseteq \mathfrak{p}$. Since $I\subseteq J$, one also has $I\subseteq \mathfrak{p}$.

The containment $I\subseteq \mathfrak{p}$ says exactly that $\mathfrak{p}\in V(I)$ by [L1]. Therefore every element of $V(J)$ lies in $V(I)$.

Hence $V(J)\subseteq V(I)$.

If $\mathfrak{p}\in V\text{\hspace{0.17em}⁣}\left({\sum }_{\lambda }{I}_{\lambda }\right)$, then ${\sum }_{\lambda }{I}_{\lambda }\subseteq \mathfrak{p}$. Since every $I_{\lambda }$ is contained in that sum, one has $I_{\lambda }\subseteq \mathfrak{p}$ for every $\lambda$, so $\mathfrak{p}\in V(I_{\lambda })$ for all $\lambda$.

Conversely, if $\mathfrak{p}\in V(I_{\lambda })$ for every $\lambda$, then each $I_{\lambda }$ lies in $\mathfrak{p}$. Because $\mathfrak{p}$ is an ideal, it contains every finite sum of elements coming from the family, hence it contains ${\sum }_{\lambda }{I}_{\lambda }$. Therefore $\mathfrak{p}\in V\text{\hspace{0.17em}⁣}\left({\sum }_{\lambda }{I}_{\lambda }\right)$.

Steps 1.1 and 1.2 prove the displayed equality.

If $\mathfrak{p}\in V(I_j)$ for some $j$, then $I_j\subseteq \mathfrak{p}$. Every element of $I_1\cdots I_n$ lies in $I_j$, so $I_1\cdots I_n\subseteq \mathfrak{p}$, and therefore $\mathfrak{p}\in V(I_1\cdots I_n)$. This proves $V(I_1)\cup \cdots \cup V(I_n)\subseteq V(I_1\cdots I_n)$.

Conversely, let $\mathfrak{p}\in V(I_1\cdots I_n)$ and suppose that no $I_j$ is contained in $\mathfrak{p}$. For each $j$, choose $a_j\in I_j\setminus \mathfrak{p}$. Then $a_1\cdots a_n\in I_1\cdots I_n\subseteq \mathfrak{p}$, so repeated use of [L2] forces some $a_j\in \mathfrak{p}$, a contradiction. Hence $I_j\subseteq \mathfrak{p}$ for some $j$, and $\mathfrak{p}\in V(I_j)$.

The two inclusions prove $V(I_1\cdots I_n)=V(I_1)\cup \cdots \cup V(I_n)$.

Every prime ideal contains $0$, so $V((0))=\mathrm{Spec}(R)$. No prime ideal equals the whole ring, so no prime ideal contains $R$; hence $V(R)=\mathrm{\varnothing }$.

The arbitrary-sum identity is exactly [L2], and the finite-product identity is exactly [L3].

Steps 1.1 and 1.2 supply the four listed vanishing-set identities.

If $V(I)=V(J)$, then the two ideals are contained in exactly the same prime ideals. Applying [L1] to both ideals shows that $\sqrt{I}$ and $\sqrt{J}$ are intersections over the same family of prime ideals, hence $\sqrt{I}=\sqrt{J}$.

Conversely, suppose $\sqrt{I}=\sqrt{J}$. If $\mathfrak{p}\in V(I)$, then $\mathfrak{p}$ contains $I$, so [L1] gives $\sqrt{I}\subseteq \mathfrak{p}$. Therefore $\sqrt{J}\subseteq \mathfrak{p}$, hence $J\subseteq \sqrt{J}\subseteq \mathfrak{p}$ by [L2], and $\mathfrak{p}\in V(J)$. The same argument with $I$ and $J$ reversed shows $V(J)\subseteq V(I)$.

Steps 1.1 and 1.2 prove that vanishing sets agree exactly when radicals agree.

Every prime ideal contains $0$, so no prime lies in $D(0)$. Every prime ideal is proper, so it does not contain $1$; hence every prime lies in $D(1)$. Therefore $D(0)=\mathrm{\varnothing }$ and $D(1)=\mathrm{Spec}(R)$.

Let $\mathfrak{p}\in \mathrm{Spec}(R)$. Then $\mathfrak{p}\in D(fg)$ exactly when $fg\mathrm{\notin }\mathfrak{p}$. Because $\mathfrak{p}$ is prime, this is equivalent to saying that neither $f$ nor $g$ lies in $\mathfrak{p}$, that is, $\mathfrak{p}\in D(f)\cap D(g)$. The same prime-ideal property shows that $f^n\in \mathfrak{p}$ exactly when $f\in \mathfrak{p}$, so $D(f^n)=D(f)$.

Steps 1.1 and 1.2 prove the stated identities for principal distinguished subsets.

Let $J=(f_1,\dots ,f_n)$. If $f\in \sqrt{J}$ and $\mathfrak{p}\in D(f)$, then $f\mathrm{\notin }\mathfrak{p}$. A prime ideal containing $J$ would contain every element of $\sqrt{J}$ and therefore would contain $f$, so $\mathfrak{p}$ cannot contain $J$. Hence at least one $f_i$ is omitted by $\mathfrak{p}$, which means $\mathfrak{p}\in D(f_i)$ for some $i$. Thus $D(f)\subseteq D(f_1)\cup \cdots \cup D(f_n)$.

Conversely, assume $D(f)\subseteq D(f_1)\cup \cdots \cup D(f_n)$. If $\mathfrak{p}$ contains $J$ and omitted $f$, then $\mathfrak{p}$ would lie in the left-hand side and hence in some $D(f_i)$, contradicting $f_i\in J\subseteq \mathfrak{p}$. Therefore every prime ideal containing $J$ also contains $f$, so $V(J)=V(J+(f))$. By [L1], the radicals of $J$ and $J+(f)$ agree; since $f\in J+(f)$, this forces $f\in \sqrt{J}$.

Steps 1.1 and 1.2 prove the equivalence. The final sentence is just the definition of membership in a radical ideal.