Question about modules and ideals Definitions:
*An ideal $I$ in a unitary ring $A$ is a set $I \subset A$ which satisfies the following properties:

*

*$0\in I$ ,

*If $a, b \in I$, then $a + b \in I$,

*If $a \in A$ and $b \in I$ then $ab,ba \in I$
*Let $A$ be a unit ring. A left $A-$module is a triple $(M, +, ·)$ such that $M$ is a set, $ + : M \times M \rightarrow M$ is an internal operation on $M$ and $·$ is what is called an external operation on $M$ with domain of operators on $A$, which simply means that $· : A \times M \rightarrow M$. Furthermore the following properties must be met:

*

*$(r + s) + t = r + (s + t)$ for all $r, s, t \in M$.

*$r + s = s + r$ for all $r, s \in M$.

*There exists an element $0\in M$ such that $r + 0 = r$ for all $r \in M$.

*For every $r\in M$ there exists an element $−r\in M$ such that $r + (−r) = 0$.

*$a(r + s) = ar + as$ for all $a \in A$ and all $r, s \in M$.

*$(a + b)r = ar + br$ for all $a, b \in A$ and all $r\in M$.

*$a(br) = (ab)r$ for all $a, b \in A$ and all $r\in M$.

*$1r = r$ for all $r \in M$.

*Let $A$ be a unitary ring and $M$ be an $A-$module. We will say that a
module $N$ is a submodule of $M$ if $N\subset M$ and the operations of $N$ are the
same as those of $M$.
Question:
If $A,B$ are units ring.Suppose that $A$ is a $B$-module. Then if $I$ is an ideal of $A$. Then $I$ is an $B$-submodule of the $B$-module $A$?
 A: If $I$ is an ideal of $A$, then $(I,+)$ is a subgroup of $(A,+)$, with $a \cdot b, b \cdot a \in I$ for all $a \in A$, $b \in I$.
If we take $c \in B$, we have that $c\cdot a = (c \cdot 1) \cdot a \in I$, because $c \cdot 1 \in A$, and $I$ is an ideal.
Then $I$ is closed by the action of $B$, and thus $I$ is a $B-$submodule of $A$
A: Your question is not clear.  You are regarding $A$ as both a $B$-module and a unitary ring, but you have not specified what (if any) conditions you impose on how the structures interact.  For example do you demand
$$C1:\qquad a_1+_{\tiny \rm ring}a_2=a_1+_{\tiny B{\rm -module}}a_2,$$
for all $a_1,a_2\in A$?
I would guess you do, but if you take the question literally, this need not be the case (so there is no relationship between ideals in $A$ and submodules in $A$).
Further do you demand $$C2:\qquad b(a_1a_2)=(ba_1)a_2,$$
for all $b\in B$ and $a_1,a_2\in A$?
One previous answer assumes you do, whilst the other previous answer assumes you do not.  I can see why both could be natural assumptions.
The nice answer by @ZAF (+1) shows that if you assume $C1, C2$ then ideals in $A$ are also submodules.  Note the converse need not hold.  Let $B=\mathbb{Z}$ and $A=\mathbb{Z}[x]$.  Then $A$ is naturally a module over $B$, satisfying both $C1,C2$.  Then the degree $0$ elements in $A$ (e.g. the integers) are a $B$-submodule.  However they are not an ideal.
However if you assume $C1$ but not $C2$, then ideals need not be submodules.
Note in your condition (3) you define ideals as $2$-sided ideals, so you cannot use the example given by @azif00.
Instead let $A=\mathbb{Z}\times \mathbb{Z}$ and let $B=\mathbb{Z}[x]$, where $$x(n,m)=(m,n),$$ for all $n,m\in \mathbb{Z}$.
Then the set $I=\{(n,0)|\,\, n\in \mathbb{Z}\}$ is a ($2$-sided) ideal in $A$.  However it is not a submodule, as $x(1,0)=(0,1)\notin I$.
A: No. Given a unitary ring $A$, consider the ring $B := A^{\rm op} \times A$, where $A^{\rm op}$ is the opposite ring of $A$. For $(x,y) \in B$ and $a \in A$ define $(x,y)a := yax$. This makes $A$ a $B$-module, and a $B$-submodule of $A$ is exactly the same as a two-sided ideal of $A$. Hence, if $I$ is a left- (or right-) ideal of $A$ which is not a two-sided ideal, $I$ cannot be a $B$-submodule of $A$.
