[proofplan]
We prove the two implications directly from the definition of an $R$-submodule. If $N$ is a submodule, then scalar multiples of elements of $N$ remain in $N$, and the sum of two such scalar multiples remains in $N$. Conversely, the assumed closure under all two-term $R$-linear combinations recovers closure under addition, additive inverses, and scalar multiplication by choosing the coefficients $1_R$, $-1_R$, and $0_R$ appropriately.
[/proofplan]
[step:Use submodule closure to obtain every two-term linear combination]
Assume that $N$ is an $R$-submodule of $M$. Let $a,b \in R$ and let $x,y \in N$. Since $N$ is closed under the scalar action of $R$, both $ax$ and $by$ belong to $N$. Since $N$ is closed under the addition inherited from $M$, their sum also belongs to $N$:
\begin{align*}
ax + by \in N.
\end{align*}
Thus every two-term $R$-linear combination of elements of $N$ lies in $N$.
[/step]
[step:Recover the submodule operations from the linear combination hypothesis]
Assume conversely that for every $a,b \in R$ and every $x,y \in N$,
\begin{align*}
ax + by \in N.
\end{align*}
Because $N$ is nonempty, choose an element $y_0 \in N$.
First, let $x,y \in N$. Taking $a = 1_R$ and $b = 1_R$ in the hypothesis gives
\begin{align*}
1_R x + 1_R y \in N.
\end{align*}
Since $M$ is a unital left $R$-module, $1_R x = x$ and $1_R y = y$, hence
\begin{align*}
x+y \in N.
\end{align*}
Therefore $N$ is closed under addition.
Next, let $x \in N$. Taking $a = -1_R$, $b = 0_R$, and $y = y_0$ in the hypothesis gives
\begin{align*}
(-1_R)x + 0_R y_0 \in N.
\end{align*}
The module axioms give $0_R y_0 = 0_M$ and $(-1_R)x = -x$, so
\begin{align*}
-x \in N.
\end{align*}
Therefore $N$ is closed under additive inverses.
Finally, let $r \in R$ and $x \in N$. Taking $a = r$, $b = 0_R$, and $y = y_0$ in the hypothesis gives
\begin{align*}
rx + 0_R y_0 \in N.
\end{align*}
Since $0_R y_0 = 0_M$, this says
\begin{align*}
rx \in N.
\end{align*}
Thus $N$ is closed under scalar multiplication by elements of $R$.
[guided]
The converse is the part where the two-term linear combination condition must be unpacked carefully. We assume that for every $a,b \in R$ and every $x,y \in N$,
\begin{align*}
ax + by \in N.
\end{align*}
The subset $N$ is nonempty, so we may fix one element $y_0 \in N$. This fixed element is used whenever we want to turn the two-variable condition into a one-variable closure statement.
To prove closure under addition, take arbitrary $x,y \in N$. We choose the coefficients $a = 1_R$ and $b = 1_R$. The hypothesis gives
\begin{align*}
1_R x + 1_R y \in N.
\end{align*}
Because $M$ is a unital left $R$-module, the identity element $1_R$ acts as the identity on $M$. Hence $1_R x = x$ and $1_R y = y$, so the displayed element is exactly $x+y$. Therefore
\begin{align*}
x+y \in N.
\end{align*}
This proves that $N$ is closed under addition.
To prove closure under additive inverses, take arbitrary $x \in N$. We must produce $-x$ as a permitted linear combination of elements of $N$. Since the condition expects two elements of $N$, we use the fixed element $y_0 \in N$ for the second input and choose coefficients $a = -1_R$ and $b = 0_R$. The hypothesis gives
\begin{align*}
(-1_R)x + 0_R y_0 \in N.
\end{align*}
The module axioms give $0_R y_0 = 0_M$, and the scalar action of the additive inverse $-1_R$ gives $(-1_R)x = -x$. Thus the displayed element is $-x$, and therefore
\begin{align*}
-x \in N.
\end{align*}
This proves that $N$ is closed under additive inverses.
To prove closure under scalar multiplication, take arbitrary $r \in R$ and $x \in N$. Again we use the fixed element $y_0 \in N$ for the second input, this time with coefficients $a = r$ and $b = 0_R$. The hypothesis gives
\begin{align*}
rx + 0_R y_0 \in N.
\end{align*}
Since $0_R y_0 = 0_M$, this reduces to
\begin{align*}
rx \in N.
\end{align*}
Thus $N$ is closed under scalar multiplication by every element of $R$.
[/guided]
[/step]
[step:Conclude that the subset is exactly a submodule]
From the previous step, $N$ is nonempty, closed under addition, closed under additive inverses, and closed under scalar multiplication by elements of $R$. Therefore $N$ is an additive subgroup of the underlying abelian group of $M$ and is stable under the $R$-action. By the definition of an $R$-submodule, $N \le M$. This proves the converse implication and completes the proof.
[/step]
