Tensor product of modules: $\Bbb Z[x]/\langle f(x) \rangle \otimes_{\Bbb Z} \Bbb Z/p\Bbb Z \cong (\Bbb Z/p\Bbb Z)[x]/\langle f(x) \rangle$

This is a question about tensor product of modules.

How to show that $$\Bbb Z[x]/\langle f(x) \rangle \otimes_{\Bbb Z} \Bbb Z/p\Bbb Z \cong (\Bbb Z/p\Bbb Z)[x]/\langle f(x) \rangle$$ for any prime $p$ and irreducible polynomial $f(x)\in\Bbb Z[x]$?

Attempt: I start with the map $$\phi:\Bbb Z/p\Bbb Z[x] \to \Bbb Z[x]/\langle f(x) \rangle \otimes_{\Bbb Z} \Bbb Z/p\Bbb Z$$ defined by $$\phi(a_0+a_1x+\cdots+a_nx^n)=1\otimes a_0+x\otimes a_1+\cdots+x^n\otimes a_n.$$ It is easy to show that $\phi$ is a well-defined surjective module homomorphism, so it suffices to show that $$\ker\phi=\langle f(x) \rangle \subset\Bbb{Z}/p\Bbb{Z}[x].$$ But this is where I am stuck. Suppose $$\phi(a_0+\cdots+a_nx^n)=1\otimes a_0+\cdots+x^n\otimes a_n=(a_0+\cdots+a_nx^n)\otimes 1=0.$$ I am tempted to say that this implies that $a_0+\cdots+a_nx^n=0\in\Bbb{Z}[x]/\langle f(x)\rangle$, but I am not sure how to justify this.

Recall the following isomorphism for an $R$-Module $M$ -
$R/I \otimes_R M \simeq M/IM$
Let $M = \mathbb{Z}[X]/\left(f\right)$$\implies M \otimes_\mathbb{Z} \mathbb{Z}/p\mathbb{Z} \simeq M/pM$
$pM = p \mathbb{Z}[X]/\left(f\right) = \big(p,f\big)/(f) \in \mathbb{Z}[X]/(f)$