# 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 $x\not\in C$, $y\not\in B$. 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. $\square$

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).

Let $I$, $P_{1},\ldots,P_{n}$ be ideals in $R$ such that each $P_{i}$ is prime. If $I\subseteq P_{1}\cup\cdots\cup P_{n}$, then there exists $i\in\{1,\ldots,n\}$ 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 $I\not\subseteq P_{1}\cup\cdots\cup P_{n}$. For $i\in\{1,\ldots,n\}$ define

 $\overline{P_{i}}=P_{1}\cup\cdots\cup P_{i-1}\cup P_{i+1}\cup\cdots\cup P_{n}.$

By our assumption (and induction hypothesis) $I\not\subseteq\overline{P_{i}}$ for any $i\in\{1,\ldots,n\}$. Thus for any $i$ there is $x_{i}\in I$ such that $x_{i}\not\in\overline{P_{i}}$.

Now for any $i\in\{1,\ldots,n\}$ define $\overline{x_{i}}=x_{1}\cdots x_{i-1}x_{i+1}\cdots x_{n}\in I$. Then we have

 $\overline{x_{1}}+\cdots+\overline{x_{n}}\in I$

and thus there is $j\in\{1,\ldots,n\}$ such that $\overline{x_{1}}+\cdots+\overline{x_{n}}\in P_{j}$. Since $\overline{x_{i}}\in P_{j}$ for any $i\neq j$, then we have that

 $\overline{x_{j}}\in P_{j}.$

But $P_{j}$ is prime, so there is $k\neq j$ such that $x_{k}\in P_{j}\subseteq\overline{P_{k}}$. Contradiction. $\square$

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 $I\not\subseteq A_{i}$ for any $i=1,2,3$. $\square$

Title ideals contained in a union of ideals IdealsContainedInAUnionOfIdeals 2013-03-22 19:03:55 2013-03-22 19:03:55 joking (16130) joking (16130) 4 joking (16130) Theorem msc 13A15 IdealIncludedInUnionOfPrimeIdeals