Product of ideals ($IJ$)

The product $IJ$ consists of finite sums of products $xy$ with $x\in I$ and $y\in J$.

For an ideal $I$ in a commutative ring $R$, put $I^0=R$ and let $I^n$ for $n\ge 1$ be the product of ideals with $n$ factors equal to $I$. It consists of finite sums of products of $n$ elements of $I$. These ideals form a descending chain $R\supseteq I\supseteq I^2\supseteq \cdots$.

Every positive power has the same radical of an ideal as $I$:One inclusion follows from $I^n\subseteq I$. Conversely, if $x^m\in I$, then $x^{mn}\in I^n$, giving the other inclusion. Different ideals can therefore define the same radical ideal, even when their successive powers are all distinct.

An ideal $I$ is idempotent when $I^2=I$. Then $I^j=I$ for every $j\ge 1$, so a nonzero idempotent ideal gives a nonzero intersection of all its powers. A semigroup algebra with arbitrarily divisible positive exponents supplies examples in a non-Noetherian integral domain. If $I$ is finitely generated, the determinant trick applied to $I=I^2$ produces $r\in I$ such that $(1+r)I=0$.

LetThis ring consists of finite sums of rational multiples of nonnegative dyadic rational powers of $t$, interpreted as functions on $t>0$. Every pair of elements lies in one polynomial ring $\mathbb{Q}[u_n]$, so $R$ is an integral domain. The constant term defines a surjective ring homomorphism $R\to \mathbb{Q}$. Its kernel of a ring homomorphism $\mathfrak{m}$ is nonzero and is a maximal ideal, by the maximal ideal quotient criterion.

Every $f\in \mathfrak{m}$ has the form $u_{n}g(u_n)$ for some $n$ and some polynomial $g$. The identityexpresses it as a product of two elements of $\mathfrak{m}$. Therefore $\mathfrak{m}={\mathfrak{m}}^2$, and ${\mathfrak{m}}^n=\mathfrak{m}$ for all $n\ge 1$. This shows that strict descent of powers of an ideal can fail in an integral domain, even for a nonzero maximal ideal. By the determinant trick, such an ideal cannot be finitely generated, so this ring is not a Noetherian ring.