How to prove that a ring ideal is not principal? I need to prove that an ideal $(x+1, y)$ in the ring $\mathbb{Q}[x,y]$ is not principal.
I already tried to prove the statement by contradiction supposing that $(x+1,\ y)$ is principal. So, here is what I've done so far:
$$(x+1,\ y)=\{f_1*(x+1)+f_2*y\bigm|f_1,f_2\in R=\mathbb{Q[x,y]}\}$$
$(x+1,\ y)$ is principal, hence $(x+1,\ y)=(f): f\in R$
$$f_1=1,\ f_2=0 \Rightarrow x+1\in(f)$$
$$f_1=0,\ f_2=1 \Rightarrow y\in(f)$$
Therefore, $\begin{cases}x+1\ \vdots\ f \\ y\ \vdots\ f\end{cases}\Rightarrow
\begin{cases}x+1=f*g_1\\y=f*g_2\end{cases}$ where $g_1$ and $g_2$ are some functions.  
But I don't understand how to end the solution. I'd be very grateful if anyone could explain to me what to do next.
 A: If $(x+1,y)=\bigl(f(x,y)\bigr))$, for some $f(x,y)\in\mathbb Q[x,y]$, then $f(x,y)\mid y$. And $f[x,y]$ cannot be a constant non-zero polynomial, because then we would have $\bigl(f(x,y)\bigr))=\mathbb Q[x,y]$. But the only non constant polynomials that divide $y$ are those of the form $ky$, with $k\in\mathbb Q\setminus\{0\}$. Since no such polynomial divides $x+1$, a contradiction is reached.
A: The Lemma with $\rm \,D=\Bbb Q[x],\ c = x\!+\!1\,$ yields $\,(x\!+\!1,y)\,$ principal $\,\Rightarrow\,\overbrace{(x\!+\!1)g(x)=1\,\Rightarrow\, 0 = 1}^{\large {\rm eval\ at}\ x\ =\ -1}$
Lemma $ $ Let $\rm \,D\,$ be a domain, $\rm \,\color{#c00}{0\ne }c\in D.\,$ If $\rm \ (c,y) = (f)\ $ in $\rm \,D[y]\,$ then $\rm \,c\,$ is a unit in $\rm D$.
Proof $\rm\ \ f\ \in\ (c,y)\, \Rightarrow\, f\ =\ c\, g_1 + y\,h_1.\, $ Eval at $\rm\: y = 0\ \Rightarrow\ f(0)\ =\ cg_1(0)\ =\ cd,\:$ $\rm \,d\in  D$
Hence $\rm\ \ \color{#c00}{0\neq} c\ \in\ (f)\ \Rightarrow\ c\ =\ f\, g\ \color{#c00}\Rightarrow\ deg\ f\, =\, 0\, \Rightarrow\  f\, =\, f(0)\, =\, cd,\, $ by $\,\rm D\,$  a domain.
Therefore $\rm\,\ \ \ y\ \in\ (f)\ \Rightarrow\, y\ =\ f\, h\ =\ cdh.\,\ $ Eval at $\rm\ y = 1\ \Rightarrow\ 1\: =\ cd\,h(1)\, \Rightarrow\, c\,$  a unit.
A: Well, if $y\in\langle f\rangle$, then $f\mid y$ and so $f=1$ or $\alpha y$ for some rational number $\alpha\ne0$. Moreover, $f\mid x+1$ and this is only possible for $f=1$. But $1\not\in\langle x+1,y\rangle$ and so the ideal cannot be principle.
A: Hint:
What can you say about the degree of $f$ in $x$? in $y?
