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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.

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

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  • $\begingroup$ Could you explain why $1\notin (x+1,y)$? $\endgroup$ – Bonrey May 4 at 14:50
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    $\begingroup$ 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 $I=R$. $\endgroup$ – Theo C. May 4 at 14:54
  • $\begingroup$ Ok. I think I got this. Thank you! $\endgroup$ – Bonrey May 4 at 15:08
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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.

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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.

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Hint:

What can you say about the degree of $f$ in $x$? in $y?

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  • $\begingroup$ $f$ must be a constant? $\endgroup$ – Bonrey May 4 at 14:44
  • $\begingroup$ Yes. Is it possible, from $x+1$ and $y$? $\endgroup$ – Bernard May 4 at 14:49
  • $\begingroup$ I guess only if $f\ne0$ $\endgroup$ – Bonrey May 4 at 14:52
  • $\begingroup$ You can hav $0$, but no othe constant, considering again the degrees in $x$ and in $y$. $\endgroup$ – Bernard May 4 at 14:54
  • $\begingroup$ 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? $\endgroup$ – Bonrey May 4 at 14:59

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