How to prove that the intersection of ideals is an ideal

Question

Given $R$ is a ring, $X\subseteq H_i$ and $H_i$ is an ideal of $G$ for each $i=1,2,...,n$.
Prove that $H_1∩H_2∩...∩H_n$ is an ideal of $G$ and contains $X$.

That is a question I get from random source. I myself am not sure whether the question is right or not since $G$ is not defined yet, so I just assume that $G$ is the ring $R$ .

A literature states that:
Given non-empty set $S$, $S \subseteq R$, $S$ is an ideal if
$\forall s_1,s_2 \in S$, $s_1-s_2 \in S$, and
$(\forall r \in R) (\forall s \in S)$ $rs,sr \in S$.


Here is my try:
For $i=1,2,...,n$, $H_i$ is an ideal of $R$ which means that $H_i\subseteq R$ and thus also $H_1∩H_2∩...∩H_n \subseteq R$
To prove that $H_1∩H_2∩...∩H_n$ is an ideal of $R$, It should be shown that
$\forall a,b \in H_1∩H_2∩...∩H_n$, $a-b \in H_1∩H_2∩...∩H_n$, and
$(\forall r \in R) (\forall a \in H_1∩H_2∩...∩H_n)$ $ra,ar \in H_1∩H_2∩...∩H_n$.
What confused me is that I don't know what guarantee $a-b,ra,ar \in H_1∩H_2∩...∩H_n$, and $X$ is contained in $G$.
Can anyone explain? Thank you.

Answer 1

My comment above is already a hint. And here the definition along with a further hint are listed.
Definition:
A subset $I$ of a ring $R$ is called an ideal if it satisfies the following two conditions:
I. $$\forall a, b\in I, a-b\in I.$$
II. $$\forall a\in I, r\in R, ra\in I.$$

Notice that the first implies that $I$ is a subgroup under the addition in $R.$
Hint:
If an element $a$ belongs to every element $I_\lambda$ of a family $\{I_\lambda\}_{\lambda\in\Lambda}$ of ideals, then also $a\in \bigcap_{\lambda\in\Lambda} I_\lambda.$

Hope this helps.

Answer 2

Put $S=H_1\cap\cdots\cap H_n$.

To see that $S$ is an ideal, just verify that for any $a\in S$ and $r\in R$, we have (i) $ra\in S$ because $\forall i:a\in H_i$ by definition of $S$ and $\forall i:ra\in H_i$ by definition of ideal, hence $ra\in S$; we also have that (ii) $a,b\in S\Rightarrow a+b\in S$, since certainly $\forall i:a,b\in H_i$ by definition of $S$ and $\forall i:a+b\in H_i$ by definition of ideal, hence $a+b\in S$.