## Archive for April, 2008

### operations on ideals

4 April, 2008

Given a commutative ring R, one can do different operations with ideals. If you are familiar with operations on subgroups in groups, a number of these thing will look very familiar.

The easiest operation is ideal intersection, that we have already encountered: given two ideals $I,J\subseteq R$, their intersection $I\cap J$ is an ideal itself. Indeed, the intersection of subgroups (we talk about additive subgroups of the additive group of R) is a subgroup, and for any $x\in I\cap J$ and $r\in R$ one has
$rx\in I$ and $rx\in J$, implying $rx\in I\cap J$, as required from $I\cap J$ to be an ideal.
Similarly, we can see that the intersection of an arbitrary number of ideals is an ideal.

Note that the union of two ideals need not be an ideal: e.g. $(2)\cup (3)\subseteq\mathbb{Z}$ is not an ideal, as it is already not a subgroup of the additive group of R, as can be seen by taking $3 \in (3)$, $-2 \in (2)$ and computing $3+(-2)=1\not\in (2)\cup (3)$.

The rest of the operations are of more algebraic flavour. The first is the ideal sum $I+J$ of two ideals $I,J\subseteq R$. By definition, $I+J=\{x+y\mid x\in I, y\in J\}$. This is indeed an ideal: it is a subgroup of the additive group of R, and also $r(x+y)=rx+ry\in I+J$ for any $r\in R$. As well, note that $I+J$ is the minimal w.r.t. inclusion ideal of R that contains $I\cup J$.
One can generalise this to the sum $\sum_{\alpha\in A}I_\alpha$
of an arbitrary collection $A$ of ideals: take finite
(in sense that they have only finitely many nonzero summands) sums $\sum_{x\alpha\in I_\alpha x_\alpha}$ for
the set of elements of $\sum_{\alpha\in A}I_\alpha$. Again, we note that $\sum_{\alpha\in A}I_\alpha$ is the
minimal ideal containing $\cup_{\alpha\in A}I_\alpha$.

The second operation is ideal product $I\cdot J$ of two ideals $I,J\subseteq R$. By definition, $IJ=(xy\mid x\in I, y\in J)$, that is, the ideal generated by the products $xy$, where $x\in I, y\in J$. This is an ideal, just by definition. Note that $IJ\subseteq I\cap J$.
Moreover, $IJ=I\cap J$ iff $I$ and $J$ are coprime, i.e.
$I+J=(1).$ Indeed, $(I+J)(I\cap J)=I(I\cap J)+J(I\cap J)\subseteq IJ.$ It is easy to check that
this product is associative, and we can drop brackets:
$I\cdot(J\cdot K)=(I\cdot J)\cdot K=I\cdot J\cdot K.$
A particular important kind of ideal product is the $k$-th power $I^k$ of an ideal $I.$
By definition $I^k=I\cdot I\cdot I\cdot\dots\cdot I$, where the product is taken $k$ times.

The third operation is in some sense the inverse of multipication, it is ideal quotient
$(I:J)=\{x\in R\mid xJ\subseteq I\}$. In particular $(0:J)$ is called the annihilator of $J.$
You are invited to check the following properties of the ideal quotient: $(I:I)=R$, $(I:R)=I.$