Let $R$ be a commutative reduced ring with identity with the property that if $I$ and $J$ are two ideals of $R $ such that if $I+J$ is not contained in any minimal prime ideal, then there exist ideals $I'$ and $J'$ of $R$ such that $I'J'=0$ and the ideals $I+I'$ and $J+J'$ are not contained in any minimal prime ideal.

Is there any characterization for such a ring? Or

Is there a reduced ring that dose not have this property?

Note that clearly Noetherian rings have this property.

  • $\begingroup$ Related: is there a (necessarily non-Noetherian) commutative ring (with 1) with the property that some minimal prime is contained in the union of all the others? $\endgroup$ – Kevin Buzzard Jan 16 '17 at 19:50
  • $\begingroup$ @KevinBuzzard. Here is an example. Begin with the polynomial ring $S=k[x_1,x_2,x_3,\dots]$ in countably many variables. Let $E$ be the ideal generated by $x_ax_b$ for all $1\leq a < b$. Let $R$ be $S/E$. For every prime $\mathfrak{p}$ of $S$ that contains $E$, for every $a = 1 ,2 ,\dots$, if $x_a\not\in \mathfrak{p}$, then for every $b\neq a$, $x_b\in \mathfrak{p}$. Thus, the minimal primes over $E$ are of the form $\mathfrak{p}_a= \langle x_b | b\neq a \rangle$. For every $f\in \mathfrak{p}_a$, since $f$ has only finitely many terms, there exists $c>a$ such that $f\in \mathfrak{p}_c$. $\endgroup$ – Jason Starr Jan 17 '17 at 9:01
  • $\begingroup$ By the way, the ring $R$ in my previous comment does have the property in the OP's question. An ideal $K$ containing $E$ is contained in no $\mathfrak{p}_a$ precisely if, for every $a$, there exists $f\in E$ with a nonzero term $c x_a^n$ for $c\in k^\times$ and $n\geq 0$. If $K=I+J$ has this property, then let $I'$, resp. $J'$, be generated by those $x_a$ such that $I$, resp. $J$, is contained in $\mathfrak{p}_a$. For every $a$, if $x_a\in I'$, then $x_a\not\in J'$. Thus $I'J' \subset E$. Yet $I+I'$ and $J+J'$ are contained in no $\mathfrak{p}_a$. $\endgroup$ – Jason Starr Jan 17 '17 at 9:12
  • $\begingroup$ Typo correction: "there exists $f\in E$" --> "there exists $f\in K$". $\endgroup$ – Jason Starr Jan 17 '17 at 9:54

I am posting as an "answer" the generalization of the example in my comment above. The ring $R$ has your property if the ring $R$ has the following property: for every minimal prime ideal $\mathfrak{p}$, there exists $x_{\mathfrak{p}}\in R\setminus \mathfrak{p}$ that annihilates $\mathfrak{p}$. In this case, let $I'$, resp. $J$', be the ideal generated by all those elements $x_{\mathfrak{p}}$ for minimal primes $\mathfrak{p}$ that do not contain $I$, resp. $J$.

By hypothesis, no minimal prime $\mathfrak{p}$ contains both $I$ and $J$. Thus, for every generator $x_{\mathfrak{p}}$ of $I'$ and for every general $x_{\mathfrak{q}}$ of $J'$, $\mathfrak{p}$ does not equal $\mathfrak{q}$. Thus, there exists $y_{\mathfrak{p}}\in \mathfrak{p}\setminus \mathfrak{p}\cap \mathfrak{q}$. Since $x_{\mathfrak{p}}y_{\mathfrak{p}} = 0$, and since $y_{\mathfrak{p}}$ is not in $\mathfrak{q}$, necessarily $x_\mathfrak{p}$ is in $\mathfrak{q}$. Thus, since $x_{\mathfrak{q}}$ annihilates $\mathfrak{q}$, $x_{\mathfrak{p}}\cdot x_{\mathfrak{q}}$ equals $0$. Therefore $I'J'$ equals $\{0\}$. Yet for every minimal prime $\mathfrak{p}$, if $\mathfrak{p}$ contains $I$, resp. $J$, then $\mathfrak{p}$ does not contain $I'$, resp. $J'$. Therefore $I+I'$ and $J+J'$ are contained in no minimal primes.

Please note, for every ring $R$, for every minimal prime $\mathfrak{p}$, for every minimal prime $\mathfrak{q}\neq \mathfrak{p}$, there does exist $y_{\mathfrak{q}}\in \mathfrak{q}\setminus \mathfrak{p}$. If $R$ is Noetherian, then there are only finitely many such minimal primes $\mathfrak{q}$ different from $\mathfrak{p}$, and the product $x_{\mathfrak{p}}$ of the finitely many elements $y_{\mathfrak{q}}$ is an element that is not in $\mathfrak{p}$, yet it is in every other minimal prime. Thus $x_{\mathfrak{p}}\cdot \mathfrak{p}$ is contained in every minimal prime. Assuming that the ring $R$ is also reduced, $x_{\mathfrak{p}}\cdot \mathfrak{p}$ equals $\{0\}$. Thus the hypothesis above holds for reduced, Noetherian rings.

Edit. My partial answer above does not account for every example. Let $X$ be an infinite set, let $(k_i)_{i\in X}$ be an indexed collection of fields, and let $R$ be the product of fields $\prod_{i\in X} k_i$. For every element $a\in R$, there is an associated subset of $X$, $Z(a) = \{i\in X | a_i = 0\}$. For every proper ideal $\mathfrak{A}\subset R$, the subset of the power set of $X$, $Z(\mathfrak{A}) = \{Z(a) | a\in \mathfrak{A}\}$, is a filter of $X$. Conversely, for every filter $\mathcal{F}$ of $X$, the set $I(\mathcal{F}) = \{a\in R | Z(a)\in \mathcal{F}\}$ is an ideal of $R$. These are order-reversing, inverse bijections. The prime ideals are all maximal ideals, and they correspond to ultrafilters. For a filter $\mathcal{F}$, if there exists nonzero $a\in R$ such that $a\cdot I(\mathcal{F})$ equals $\{0\}$, then for every $i\in Z(a)$, $\mathcal{F}$ is contained in the principal ultrafilter associated to $i$. Thus, for every non-principal ultrafilter $\mathcal{F}$, there exists no such $a$. Thus the property I identify above fails for $R$.

On the other hand, for filters $\mathcal{F}$ and $\mathcal{G}$, and ideals $I=I(\mathcal{F})$, $J=I(\mathcal{G})$, $I+J$ equals $I(\langle \mathcal{F},\mathcal{G}\rangle)$, where $\langle \mathcal{F},\mathcal{G}\rangle$ is the (pseudo)filter of all sets $C$ that contain some set $A\cap B$ for $A\in \mathcal{F}$ and $B\in \mathcal{G}$ (this is a true filter if and only if it does not contain the empty set). Since the minimal prime ideals are the maximal ideals, $I+J$ is contained in no minimal prime ideal if and only if $I+J$ equals the whole ring. This is equivalent to the condition that there exists $A\in \mathcal{F}$, $B\in \mathcal{G}$ such that $A\cap B$ is the empty set. In this case, define $I'$, resp. $J'$, to be the ideal generated by the idempotent $a$, resp. $b$, with $Z(a)=X\setminus A$, resp. $Z(b)=X\setminus B$. Since $ab$ equals $0$, $I'J'$ equals $\{0\}$. Thus, the ring $R=\prod_{i\in X} k_i$ satisfies your property even though it does not satisfy my property.


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