Let $K$ be an ideal of a commutative ring $R$ with an identity $1$ and $x$ be a given element of a module $X$ over $R$. Prove that the subset
$$A = Kx = \{\alpha x \ |\ \alpha\in K \}$$
of $X$ is a submodule of $X$
Well, to begin, I remember that an ideal $K$ of a commutative ring $R$ is an additive subgroup of $R$ such that for $r\in R$ and $k\in K$ we have $rk = kr\in R$ (the ring is commutative).
Now, I need to prove that $A$ is a submodule of $X$, that is, prove that for $u,v\in A, u+v \ in A$ and for $\alpha\in R, u \in A$ we have $\alpha u \in A$. Well, elements of $A$ are of the form: $\alpha x$, so two of them would be
$$\alpha x, \beta x$$
where $\alpha, \beta \in K$
I think that I can just do
$$\alpha x + \beta x = (\alpha+\beta)x = \gamma x \in K$$
because $\alpha+\beta = \gamma \in K$ since an ideal is an additive subgroup, and therefore $\alpha+\beta$ is also an element of $K$
For verifying the second condition, I need to test that:
$$\beta (\alpha x) \in K$$
when $\alpha x \in K$
For that I would do:
$$\beta(\alpha x) =^1 (\beta\alpha)x$$
1 - since $\alpha x\in A$, we have that $\alpha\in K$, therefore $\alpha \in R$ and we can see this equality as this because $x$ is from a module over $R$. I don't know, however, what is $\alpha\beta$ since $K$ is an ideal (and therefore an additive group). Could somebody help me?
So you need to show that $\alpha\beta\in K$. But this is clear since $\beta\in K$ and $\alpha\in R$ and $K$ is an ideal in $R$, so by definition of the ideal, $\alpha\beta\in K$.