If $I\subset \mathbb{Z}[X]$ prime ideal, then $I\cap \mathbb{Z}$ prime ideal in $\mathbb{Z}$

Question

Let $I\subset \mathbb{Z}[X]$ be a prime ideal. Show that $I\cap \mathbb{Z}$ is a prime ideal in $\mathbb{Z}$.

I think i have to start with the evaluation homomorphism \begin{align*} \phi: \mathbb{Z}[X]&\rightarrow \mathbb{Z},\\ f(X)&\mapsto f(a). \end{align*}

But I'm not sure how to use this and if this is the right approach. Can anyone help me with this? Thanks!

Answer 1

If $a,b\in\mathbb{Z}$ such that $ab\in \mathbb{Z}\cap I\subset I$, so $a\in I$ or $b\in I$ wich is equivalent to $a\in I\cap\mathbb{Z} $ or $b\in I\cap \mathbb{Z}.$

Answer 2

You could consider the homomorphism $\phi : \mathbb{Z} \longrightarrow \mathbb{Z}[X]/I$ given by $\phi(n) = n + I$. Its kernel is $\mathbb{Z} \cap I$, and is a prime ideal since $\mathbb{Z}[X]/I$ is an integral domain.

Answer 3

Suppose $\;rs\in I\cap\Bbb Z\;$ , then

$$\begin{cases}r\in I\;\;\text{or}\;\;s\in I\\{}\\\text{Both}\;\;r,s\in\Bbb Z\end{cases}\implies r\in I\cap\Bbb Z\;\;\text{or}\;\;s\in I\cap\Bbb Z$$