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.

• Hint: $\mathbb{Q}[x,y]$ is a UFD, and ... – user10354138 May 4 at 14:27

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.

• Could you explain why $1\notin (x+1,y)$? – Bonrey May 4 at 14:50
• For any ideal $I$ of a ring $R$ if $1\in I$ then since $RI \subseteq I$ it means that $R\subseteq I$, and so $I=R$. – Theo C. May 4 at 14:54
• Ok. I think I got this. Thank you! – Bonrey May 4 at 15:08

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.

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.

Hint:

What can you say about the degree of $$f$$ in $$x$$? in $y? •$f$must be a constant? – Bonrey May 4 at 14:44 • Yes. Is it possible, from$x+1$and$y$? – Bernard May 4 at 14:49 • I guess only if$f\ne0$– Bonrey May 4 at 14:52 • You can hav$0$, but no othe constant, considering again the degrees in$x$and in$y$. – Bernard May 4 at 14:54 • If$f=0$then we have$\begin{cases}x+1=0*g_1 \\ y=0*g_2\end{cases} \Rightarrow \begin{cases}x=-1 \\ y = 0\end{cases}$. But the statement should work for any$x$and$y\$. Am I wrong? – Bonrey May 4 at 14:59