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I am studying a little bit of ideals and come up with the exercise to show that the ideal $\langle 3,x-1,y-2\rangle$ is not equal to $\langle 1\rangle$ in the polynomial ring $\mathbb{Z}[x,y]$. At first sight, it seems like the elements $3,x-1,y-2$ cannot generate every polynomial. However, I've couldn't been able to find a rigorous proof. Any help will be appreciate it :).

Thanks in advance.

Yeipi
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    $\Bbb Z[x,y]/\langle 3,x-1,y-2\rangle\cong\dots\ne{0}$. – Anne Bauval Jun 16 '24 at 21:41
  • Isomorphic to...? – Yeipi Jun 16 '24 at 21:53
  • Try to find out! To warm up, prove that $\Bbb Z[x]/(x-a)\cong\Bbb Z$. – Anne Bauval Jun 16 '24 at 21:54
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    Hint: evaluate $1 = 3f + (x-1)g + (y-2)h,$ at $,(x,y)=(1,2),\Rightarrow, 1 = 3f(1,2)\Rightarrow 3^{-1}\in \Bbb Z.\ $ This is a duplicate so please delete it when all is clear. $\ \ $ – Bill Dubuque Jun 16 '24 at 21:58
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    ...and then, that $\Bbb Z[x,y]/(x-a,y-b)\cong\Bbb Z$ (for any integers $a,b$). – Anne Bauval Jun 16 '24 at 22:02
  • Completing Anne Bauval's series of isomorphisms, $\mathbb{Z}[x,y]/\langle 3, x-1, y-2 \rangle \cong \mathbb{Z}/\langle 3 \rangle \not\cong {0}$. – Geoffrey Trang Jun 16 '24 at 22:23
  • Further (not needed), my prior comment shows an integer $,n\in I = (3,x!-!a,x!-!b)!\iff! \color{#c00}{3\mid n},,$ so the natural map of $,\Bbb Z,$ into $,R=\Bbb Z[x,y]/I,$ has kernel $,\color{#c00}{3\Bbb Z},,$ and it is onto by $!\bmod I!:\ f(x,y)\equiv f(a,b)\in\Bbb Z,,$ so $,R\cong \Bbb Z/\color{#c00}3,$ by the first isomorphism theorem. $\ \ $ – Bill Dubuque Jun 16 '24 at 22:41
  • As mentioned above, evaluating at $(x,y)=(1,2)$ shows it can't be $(1)$, same method as here in the dupe (and many other answers). $\ \ $ – Bill Dubuque May 17 '25 at 21:28

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Let $I=<3, x-1, y-2>$ in $\mathbb{Z}[x,y]$ and suppose $I=<1>$ , then $1\in I$ hence there exist $p,q,r\in \mathbb{Z}[x,y]$ such that

$$1= 3p(x,y)+(x-1)q(x,y)+(y-2)r(x,y)$$

and by substituting $(x,y)=(1,2)$ into the above equation we get

\begin{align} 1&= 3p(1,2)+(1-1)q(1,2)+(2-2)r(1,2)\\ &= 3p(x,y) \end{align}

now let $k:=p(1,2)\in\mathbb{Z}$ then $3k=1$ and $3^{-1}=k\in\mathbb{Z}$ and this is a contradiction.

user140776
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