Which of the following is/are true in case of rings? 
Which of the following is/are true. 
  
  
*
  
*$\mathbb{Z}[x]$ is a principal ideal domain.
  
*$\mathbb{Z}[x,y]/\langle y+1 \rangle$ is a unique factorization domain.
  
*If $R$ is a PID and $p$ is a non-zero prime ideal, then $R/p$ has finitely many prime ideals.
  
*If $R$ is a PID, then any subring of $R$ containing 1 is again a PID.
  

Option 1 is obviously false. Also for option 3, since $R$ is a PID, so $R/p$ is a field and field has only one prime ideal. So option 3 is true. But I have no clue how to solve option 2 and 4. Can anybody help me with these? Thanks.
 A: You can use (the falsity of) (1) to aid in (4). The ring $\mathbb{Q}[X]$ is a PID, but contains $\mathbb{Z}[X]$ as a subring containing $1$.
A: Since @BenWest has shown that option 4 is false, I will only talk about option 2 in my answer.
For option 2, consider the map $\phi \colon \mathbb{Z}[x,y] \to \mathbb{Z}[x]$ defined by $$\phi(f(x,y)) = f(x,-1).$$ $\phi$ is a ring homomorphism with $$\ker(\phi) = \{ f(x,y) \in \mathbb{Z}[x,y] : f(x,-1) = 0 \}.$$ We will show that the kernel is the principal ideal $(y+1)$. This will imply, from the first isomorphism theorem, that $\operatorname{im}(\phi) \cong \mathbb{Z}[x,y]/(y+1)$, which gives an explicit description of the quotient.
Now, let $f(x,y) \in \ker(\phi)$. View $f$ as lying in $\mathbb{Z}[y][x]$. So, $f(x,y) = a_0(y) + a_1(y) x + \dots + a_n(y) x^n$, for some $a_i(y) \in \mathbb{Z}[y]$. Therefore, $f(x,-1) = 0 \implies a_i(-1) = 0$ for all $i \implies (y+1) \mid a_i(y)$ for all $i \implies f(x,y) = (y+1) g(x,y)$ for some $g(x,y) \in \mathbb{Z}[x,y]$. Conversely, every element in $\mathbb{Z}[x,y]$ of the form $(y+1)k(x,y)$ lies in $\ker(\phi)$. Hence, $\ker(\phi) = (y+1)$.
Thus, $\mathbb{Z}[x,y]/(y+1) \cong \operatorname{im}(\phi) = \{ f(x,-1) : f(x,y) \in \mathbb{Z}[x,y] \}$. Obviously, $\operatorname{im}(\phi) \subseteq \mathbb{Z}[x]$. Conversely, if $f(x) \in \mathbb{Z}[x]$, then we define $g(x,y) := f(x) \in \mathbb{Z}[x,y]$, so $g(x,-1) = f(x) \in \operatorname{im}(\phi)$. Hence, $\mathbb{Z}[x] \subseteq \operatorname{im}(\phi)$.
Thus, $\operatorname{im}(\phi) = \mathbb{Z}[x]$, which is a UFD. Therefore, option 2 is true.
