ideals contained in a union of ideals

Assume that $R$ is a commutative ring.

Lemma. Let $A$, $B$, $C$ be ideals in $R$ such that $A\subseteq B\cup C$. Then $A\subseteq B$ or $A\subseteq C$.

Proof. Assume that this is not true. Then there are $x,y\in A$ such that $x\in B$, $y\in C$ and $x\notin C$, $y\notin B$. Obviously $x+y\in A\subseteq B\cup C$ and without loss of generality we may assume that $x+y\in B$. Then $y=(x+y)-x\in B$. Contradiction. $\mathrm{\square }$

Remark. This lemma is also true if we exchange ring with a group and ideals with subgroups (because we didn’t use multiplication and commutativity of addition in proof).

Proposition. Let $I$, $P_1,\mathrm{\dots },P_n$ be ideals in $R$ such that each $P_i$ is prime. If $I\subseteq P_1\cup \mathrm{\cdots }\cup P_n$, then there exists $i\in \{1,\mathrm{\dots },n\}$ such that $I\subseteq P_i$.

Proof. We will use the induction on $n$. For $n=2$ our lemma applies. Let $n>2$. Assume that $I⊈P_1\cup \mathrm{\cdots }\cup P_n$. For $i\in \{1,\mathrm{\dots },n\}$ define

$$\overline{P_i}=P_1\cup \mathrm{\cdots }\cup P_{i-1}\cup P_{i+1}\cup \mathrm{\cdots }\cup P_n.$$

By our assumption (and induction hypothesis) $I⊈\overline{P_i}$ for any $i\in \{1,\mathrm{\dots },n\}$. Thus for any $i$ there is $x_i\in I$ such that $x_i\notin \overline{P_i}$.

Now for any $i\in \{1,\mathrm{\dots },n\}$ define $\overline{x_i}=x_1\mathrm{\cdots }{x}_{i-1}{x}_{i+1}\mathrm{\cdots }{x}_n\in I$. Then we have

$$\overline{x_1}+\mathrm{\cdots }+\overline{x_n}\in I$$

and thus there is $j\in \{1,\mathrm{\dots },n\}$ such that $\overline{x_1}+\mathrm{\cdots }+\overline{x_n}\in P_j$. Since $\overline{x_i}\in P_j$ for any $i\ne j$, then we have that

$$\overline{x_j}\in P_j.$$

But $P_j$ is prime, so there is $k\ne j$ such that $x_k\in P_j\subseteq \overline{P_k}$. Contradiction. $\mathrm{\square }$

Counterexample. We will show, that if $P_i$’s are not prime, then the thesis no longer hold, even when $n=3$. Consider the ring of polynomials in two variables over a simple field of order $2$, i.e. ${\mathbb{Z}}_2[X,Y]$. Let $R={\mathbb{Z}}_2[X,Y]/(X^2,XY,Y^2)$. For $W(X,Y)\in {\mathbb{Z}}_2[X,Y]$ we shall write $\overline{W(X,Y)}=W(X,Y)+(X^2,XY,Y^2)\in R$. Then it is easy to see, that

$$R=\{\overline{0},\overline{1},\overline{X},\overline{Y},\overline{X}+\overline{Y},\overline{X}+\overline{1},\overline{Y}+\overline{1},\overline{X}+\overline{Y}+\overline{1}\}.$$

Let

$$I=\{\overline{0},\overline{X},\overline{Y},\overline{X}+\overline{Y}\};$$

$$A_1=\{\overline{0},\overline{X}\};$$

$$A_2=\{\overline{0},\overline{Y}\};$$

$$A_3=\{\overline{0},\overline{X}+\overline{Y}\}.$$

It can be easily checked, that $I,A_1,A_2,A_3$ are all ideals and $I\subseteq A_1\cup A_2\cup A_3$ but obviously $I⊈A_i$ for any $i=1,2,3$. $\mathrm{\square }$