Give an example of an extension $B/A$ of rings, with $B$ an integral domain and a nonzero prime ideal $\mathfrak{p}$ of B such that $\mathfrak{p} \cap A=(0).$
I don't know where to begin with this.. I know that $\mathfrak{p}$ is a prime ideal if $A\setminus \mathfrak{p}$ is closed under multiplication, ie if $ab\in \mathfrak{p} \implies a\in \mathfrak{p}$ or $b \in \mathfrak{p}$. Since this is pretty much the only information given, I'm not sure how to work with this. I am also not sure if $(0)$ represents something other than simply $\{0\}$?
Does this mean that the ideal $\mathfrak{p}$ would need to be generated by some element of $B \setminus A?$
We have been working with completely abstract terms, so giving an example is a challenge here for me.
What about $\mathbb R[x]$ (polynomials in one variable with real coefficients) as an extension of $\mathbb R$. Then $(x)$ is a prime ideal that contains no constants, so $(x)\cap\mathbb R=(0)$.