Operations on Ideals

Proposition 1. Let $I$ and $J$ be ideals of a ring $R$. Then:

1. $IJ \subseteq I$ and $IJ \subseteq J$.

2. $IJ \subseteq I \cap J \subseteq I+J$.

Proof: We have

$$
IJ=\left\{\sum_{i=1}^n x_i y_i \mid x_i\in I,\ y_i\in J,\ n\in \mathbb{N}^*\right\}.
$$

Therefore, for any finite sum

$$
\sum_{i=1}^n x_i y_i \in IJ,
$$

since $J$ is an ideal of $R$, by the absorption property of ideals we have

$$
x_i y_i \in J
$$

for every $i$, and hence

$$
\sum_{i=1}^n x_i y_i \in J.
$$

Thus

$$
IJ\subseteq J.
$$

Similarly,

$$
IJ\subseteq I.
$$

This proves part (1).

Part (2) follows immediately from part (1), since

$$
IJ\subseteq I \cap J.
$$

Moreover, for every $x\in I\cap J$, we have

$$
x=x+0\in I+J.
$$

Hence

$$
I\cap J\subseteq I+J.
$$

Therefore,

$$
IJ\subseteq I\cap J\subseteq I+J.
$$

Proposition 2. Let $I$, $J$, and $K$ be ideals of a ring $R$. Then:

1. Commutativity of addition:
   $$
   I+J=J+I.
   $$

2. Associativity of addition:
   $$
   I+(J+K)=(I+J)+K.
   $$

3. Associativity of multiplication:
   $$
   (IJ)K=I(JK).
   $$

4. Left distributive law:
   $$
   I(J+K)=IJ+IK.
   $$

5. Right distributive law:
   $$
   (J+K)I=JI+KI.
   $$

Proof: Parts (1) and (2) are obvious.

We now prove part (3). Since

$$
IJ=\left\{\sum_{i=1}^n x_i y_i \mid x_i\in I,\ y_i\in J,\ n\in \mathbb{N}^*\right\},
$$

every element of $(IJ)K$ is a finite sum of the form

$$
\sum_{j=1}^m \left(\sum_{i=1}^n x_{ij}y_{ij}\right) z_j
=
\sum_{j=1}^m \sum_{i=1}^n x_{ij}y_{ij}z_j,
$$

where $x_{ij}\in I$, $y_{ij}\in J$, and $z_j\in K$.

Now each term satisfies

$$
x_{ij}y_{ij}z_j=x_{ij}(y_{ij}z_j)\in I(JK).
$$

Since $I(JK)$ is closed under addition, it follows that

$$
\sum_{j=1}^m \sum_{i=1}^n x_{ij}y_{ij}z_j\in I(JK).
$$

Hence

$$
(IJ)K\subseteq I(JK).
$$

Similarly,

$$
I(JK)\subseteq (IJ)K.
$$

Therefore,

$$
(IJ)K=I(JK).
$$

We next prove part (4). On the one hand, since

$$
J\subseteq J+K,
$$

we have

$$
IJ\subseteq I(J+K).
$$

Similarly,

$$
IK\subseteq I(J+K).
$$

Since $I(J+K)$ is an ideal and hence closed under addition, we obtain

$$
IJ+IK\subseteq I(J+K).
$$

On the other hand, every element of $I(J+K)$ is a finite sum of the form

$$
\sum_t x_t (y_t+z_t),
$$

where $x_t\in I$, $y_t\in J$, and $z_t\in K$.

For each term, we have

$$
x_t(y_t+z_t)=x_ty_t+x_tz_t\in IJ+IK.
$$

Since $IJ+IK$ is also closed under addition, it follows that

$$
\sum_t x_t(y_t+z_t)\in IJ+IK.
$$

Hence

$$
I(J+K)\subseteq IJ+IK.
$$

Therefore,

$$
I(J+K)=IJ+IK.
$$

Part (5) can be proved in exactly the same way as part (4).

The cover image of this article was taken in Irkutsk Oblast, Russia.