Question

66. Let R be a commutative ring with identity. An ideal I of R is irreducible if it cannot be expressed as the intersection of two ideals of R neither of which is contained in the other. the following. (a) If P is a prime ideal then P is irreducible. (b) If z is a non-zero element of R, then there is an ideal I, maximal with respect to the property that r gI, and I is irreducible. (c) If every irreducible ideal of R is a prime ideal, then 0 is the only nilpotent element of R.

Answer 1

a) We will prove the contrapositive.

Suppose is an ideal that is not irreducible. then there are ideals $P_1,P_2\subseteq R$ such that $P_j\nsubseteq P_i$ if $i\neq j$, and $P_1\cap P_2=P$. Since $P_j\nsubseteq P_i$ if $i\neq j$, there are elements $x_1\in P_1\setminus P_2$ and $x_2\in P_2\setminus P_1$ . Note that $x_1x_2\in P_1\cap P_2=P$ . Since $x_1\in P_1\setminus P_2$ and $P_1\cap P_2=P$, we know $x_1\notin P$. Similarly,
2 P
. Thus, we have found elements $x_1,x_2\in R$ such that $x_1,x_2\notin P$ but $x_1x_2\in P$. Hence, is not prime.

Thus, if is an ideal that is not irreducible then is not prime. Equivalently, if is prime, then is irreducible.

b) Let $x\in R$ be a non-zero element. Let ${\cal I}_x$ be the set of all ideals $I\subseteq R$ such that $x\notin I$. Note that ${\cal I}_x$ is not empty since the zero ideal is contained in it. Partially order ${\cal I}_x$ by inclusion. Then every chain of ideals $I_1\subseteq I_2\subseteq \cdots$ in ${\cal I}_x$ has an upper bound in ${\cal I}_x$, namely,

$I=\bigcup_{k=1}^\infty I_k$

because it is an ideal and does not contain . By Zorn's lemma, ${\cal I}_x$ has a maximal element. Call it $I_x$. This is an ideal in ${\cal I}_x$, so $x\notin I_x$. We claim that $I_x$ is irreducible. Note that proving this claim will prove the statement in question.

Proof of claim: If possible, suppose that there are ideals $P,Q\subseteq R$, neither is contained in the other, such that $I_x=P\cap Q$. Since $x\notin I$, one of them, say , must be such that it does not contain . Thus, $P\in {\cal I}_x$ . Since $I_x$ is maximal in ${\cal I}_x$, and $P\in {\cal I}_x$ , we have $P\subseteq I_x$. Since $P\subseteq I_x$ and $I_x=P\cap Q$, we conclude . But then $P= I_x=P\cap Q~\Rightarrow ~P\subseteq Q$ , a contradiction. Thus, there are no ideals $P,Q\subseteq R$, such that $I_x=P\cap Q$, and neither is contained in the other. Hence, $I_x$ is irreducible.

c) Let $x\neq 0$ be an element in . Let $I_x\subseteq R$ be an ideal as in b). Since every irreducible ideal is prime, $I_x\subseteq R$ is prime. Thus, there is a prime ideal $I_x\subseteq R$ such that $x\notin I_x$. Since every nilpotent element is contained in the intersection of all prime ideals, we conclude that is not nilpotent. Thus, the only nilpotent element is .