Intersection of any set of ideals is an ideal

Question

Prove that the intersection of any set of Ideals of a ring is an Ideal.

I'm looking for hints.

Let A, B both be Ideals of a ring R.

Suppose $I \equiv A\cap B$.

Since A and B are both Ideals of a ring R, A and B are both Subrings of a ring R. In particular, we have that $\left ( A,+ \right ),\left ( A\setminus \left \{ 0 \right \} ,\cdot \right ),\left ( B,+ \right ),\left ( B\setminus \left \{ 0 \right \},\cdot \right )$ are Abelian.

Now, Suppose $x_{1},x_{2} \in I$.

I'm not entirely sure how I can justify $x_{1}+\left ( -x_{2} \right ) \in I.$ Might be overthinking this but I might have to use the fact that I is the intersection.

Answer 1

Hints, as requested:

Let $I = \bigcap\limits_{j \in J} I_j$ be an intersection of ideals $I_j$ (where $J$ is an indexing set, finite or otherwise).

1. Show that for any $x, y \in I$, $x - y \in I$. Use the fact that $x, y \in I_j$, $\forall j \in J$.
2. Show that for any $r \in R$ and $x \in I$, $rx \in I$. Again, use the fact that $x \in I_j$, $\forall j \in J$. [If the ring is not commutative, this works only for left ideals, and the proof is similar for right ideals and two-sided ideals].

---

Solution (using above hints):

1. Since $x, y \in I_j$, and $I_j$ is an ideal, $x - y \in I_j$, $\forall j \in J$. Therefore, $x - y \in \bigcap\limits_{j \in J} I_j = I$.
2. Similarly, since $x \in I_j$, $rx \in I_j$, $\forall j \in J$. Therefore, $rx \in I$. The proof is similar for right ideals and two-sided ideals (or alternatively, a two-sided ideal is both a left ideal and a right ideal).

Note: If rings are defined to have $1$, it is enough to show $x + y \in I$, since $-y = (-1)y$.