Given a polynomial ring in 4 variables then.... MULTIPLE CHOICE CORRECT ANSWER
Consider the ideal $I := (ux, uy, vx, uv)$ in the polynomial ring $\mathbb{Q}[u, v, x, y]$, where
$u, v, x, y$ are indeterminates. Choose the correct statement(s) from below:
$i):-$ Every prime ideal containing $I$ contains the ideal $(x, y)$;
$ii):-$ Every prime ideal containing $I$ contains the ideal $(x, y)$ or the ideal $(u, 
v)$;
$iii):-$ Every maximal ideal containing $I$ contains the ideal $(u, v)$;
$iv):-$ Every maximal ideal containing $I$ contains the ideal $(u, v, x, y)$.
ATTEMPT
Let P be the Prime-Ideal containing  $I := (ux, uy, vx, uv)$ $I \subseteq P$ $\implies ux,vx,vy \in P$ 
If $x\notin I$ then since $ux,vx \in P$ we have $u,v \in P \implies (u,v) \in P$
on the other hand
If $u\notin I$ then since $uy,ux \in P$ we have $x,y \in P \implies (x,y) \in P$
Is my partial solution correct?
I had done with Options i) and ii).
What about iii),iv)?
 A: Your question is incorrect because all the four options are false.
(i) is false because $I\subseteq \langle u,v\rangle$.
(ii) is false because $I\subseteq \langle u,x\rangle$.
For (iii) and (iv)....
Notice that $\langle u-1,v,x,y\rangle$ is a maximal ideal. For a proof see Magdiragdag's answer. 
Notice also that $I\subseteq\langle u-1,v,x,y\rangle$.
(iii) is false because $\langle u,v\rangle\not\subset\langle u-1,v,x,y\rangle$.
(iv) is false as $\langle u,v,x,y\rangle\not\subset\langle u-1,v,x,y\rangle$.
To prove $\langle u,v\rangle$ is prime..
Consider the evaluation map $f:\mathbb{Q}[u,v,x,y]\rightarrow \mathbb{Q}[z_1,z_2]$ defined as
$$f(u)=f(v)=0 \text{ and } f(x)=z_1 \text{ and } f(y)=z_2$$
Then $$\frac{\mathbb{Q}[u,v,x,y]}{\langle u,v\rangle}\cong \mathbb{Q}[z_1,z_2]$$
To prove that $\ker (f)=\langle u,v\rangle$ you may want to look at Magdiragdag's answer.
To prove $\langle u-1,v,x,y\rangle$ is maximal...
Consider the evaluation map $f:\mathbb{Q}[u,v,x,y]\rightarrow \mathbb{Q}[u,v,x,y]$
$$f(u)=1, \text{ } f(v)=v, \text{ }  f(x)=x,  \text{ }  f(y)=y$$
Then $\ker(f)=\langle u-1,v,x,y\rangle$. (For explanation see Magdiragdag's answer.)
Then we it follows
$$\frac{\mathbb{Q}[u,v,x,y]}{\langle u-1,v,x,y\rangle}\cong \mathbb{Q}$$
