$M\subseteq p\implies (M)\subseteq p$ where $p$ is a prime ideal, $(M)$ the ideal generated by $M$

Question

I'm stuck on this result needed for a larger problem:

Let $M\subseteq R$, $p$ be a prime ideal of $R$ then: $$M\subseteq p\implies (M)\subseteq p$$ where $(M)$ denotes the ideal generated by $M$. I attempted to show a contradiction using ideal properties, but I'm not getting anywhere.

How can the inclusion be derived?

Answer 1

Well as @TokenToucan pointed out in the comments this is true for any ideal $I$. If $M \subset R $ which is contained in an ideal $I$ of $R$, then by definition $(M)$ is the intersection of all the ideals of $R$ which contain $M$. Now the ideal $I$ contains $M$ by hypothesis, hence $(M) \subset I$.