= Solution
{c}
\b[False: existence holds, but uniqueness fails.] First, the <Lasker–Noether theorem> supplies existence; here is an argument from the preceding part. If some proper <ideal> in a <Noetherian ring> were not a finite intersection of <irreducible ideals>, the <ascending chain condition> would let us choose such an <ideal> $I$ maximal under inclusion. It is not irreducible, so $I=J\cap K$ with both $J$ and $K$ strictly larger than $I$. By maximality, both are finite intersections of <irreducible ideals>, and therefore so is $I$, a contradiction. Part (d) makes each factor a <primary ideal>. The whole <ring> is represented by the empty intersection if that convention is needed.
For a counterexample to uniqueness, work in $R=k[x,y]$ over a <field> $k$ and put $I=(x^2,xy)$. The <Hilbert basis theorem> makes $R$ a <Noetherian ring>. There are two distinct <minimal primary decompositions>:
$$
\boxed{(x^2,xy)=(x)\cap(x^2,y)=(x)\cap(x^2,y-x).}
$$
The <ideal> $(x)$ is a <prime ideal>, hence a <primary ideal>. Each of the other factors has <quotient ring> isomorphic to $k[x]/(x^2)$, by substituting respectively $y=0$ and $y=x$. As in part (b), each is $(x,y)$-primary.
To check both intersections, let $Q_c=(x^2,y-cx)$, for $c=0,1$. Substitution identifies $R/Q_c$ with $k[x]/(x^2)$. The class of $xg(x,y)$ is zero exactly when $g(0,0)=0$, equivalently when $g\in(x,y)$. Therefore $(Q_c:x)=(x,y)$ and
$$
(x)\cap Q_c=x(Q_c:x)=x(x,y)=I.
$$
The two $Q_c$ are different, since $y$ belongs to $Q_0$ but has nonzero class $x$ modulo $Q_1$. Both decompositions are irredundant: $x\notin Q_c$, while $y-cx\notin(x)$. Applying the <radical of an ideal> operation to the two components in either decomposition gives $(x)$ and $(x,y)$, respectively.
The <second uniqueness theorem for primary decomposition> explains the distinction: the component at an <isolated prime of a primary decomposition> is unique, but <embedded primary components> need not be. The counterexample varies an embedded component.
Back to article page
