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I thought of the ring: $\mathbb{R}[x,y]/(xy,x-2,y-2)$ so then you have that $[xy]=[0]$, $[x]=[2]\neq[0]$ and $[y]=[2]\neq[0]$. But then you also have that $[xy]=[x]\cdot[y]=[2]\cdot[2]=[4]$, so $[4]=[0]$. Idk if you can do this, it looks weird.

If anyone has an other example, feel free to say it :).

Kasper Van Gaever
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  • You've got the zero ring, both cosets $[x], [y]$ become zero in the quotient. A simpler example, $\Bbb R[x,y]/(xy)$, would do just fine. – Amateur_Algebraist Oct 30 '23 at 21:11
  • @Amateur_Algebraist but then you dont have that $x,y \neq0$ – Kasper Van Gaever Oct 30 '23 at 21:15
  • Why? The equality is meant as (cosets of) formal polynomials. It does not consider the possible vanishing of these polynomials at all. – Amateur_Algebraist Oct 30 '23 at 21:17
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    $, \bar x = 0\in R[x,y]/(xy)\iff x\in (xy)\iff x = xyf\overset{\large x,y,=,1,0}\Longrightarrow 1 = 0\ \ $ – Bill Dubuque Oct 30 '23 at 21:19
  • @BillDubuque why can you just choose $x$ and $y$ in your last implication arrow – Kasper Van Gaever Oct 30 '23 at 21:27
  • We evaluated the the polynomial equality at $,x,y = 1,0,,$ i.e. $f(x,y) = g(x,y)\Rightarrow f(1,0) = g(1,0).,$ If you are familiar with evaluation homomorphisms and/or the universal mapping property of polynomial rings then you can express it in that language. Compare to this proof that polynomial rings are not fields. If you know such universal properties of polynomial & quotient rings then it is instructive to use them to show that $,\Bbb Z[x,y]/(xy)$ is the universal ring with a zero-divisor. – Bill Dubuque Oct 30 '23 at 21:37
  • Such universal examples can prove extremely powerful, e.g. see this proof of Sylvester's determinant identity, which works in a universal determinant ring, and exploits that is is a domain, so we can cancel before evaluating (to remove an "apparent singularity" when the cancelled determinant is zero). Beware that even some grad students and professors have had problems understanding how it avoids division by zero (analytic bias obstructs algebraic thought). – Bill Dubuque Oct 30 '23 at 21:51
  • If you study a bit of algebraic geometry down the road, you can see that roughly, the reason $\langle xy, x-2, y-2 \rangle$ is the unit ideal is that the system of equations $xy=0, x-2=0, y-2=0$ has no simultaneous solutions. (Though that's possibly a bit of a lie - algebraic geometry would work better if the example were $\mathbb{C}[x,y] / \langle xy, x-2, y-2 \rangle$.) – Daniel Schepler Oct 30 '23 at 23:09

2 Answers2

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I don't think the replies are helpful at your level of understanding. In the ring you have constructed, both $[x]$ and $[y]$ are $[0]$. So it is not a good example. I'll show that $[x] = [0].$ To see this, remark that $x(y-2) = xy - 2x.$ So, $x = \frac{1}{2} \cdot xy - \frac{1}{2}x \cdot (y-2)$ which implies that $x \in (xy, x-2, y-2)$ is in the ideal so $[x] = [0].$ The equality $[y] = [0]$ is proved similarly.

Indeed as one commentor noted, $\mathbb{R}[x,y]/(xy)$ should work. We trivially have $[xy] = [x][y] = [0]$. But $[x]$ and $[y]$ are not $[0].$ @Bill Dubuque gave an argument why this is the case. He assumed that $[x] = [0]$ and found a contradiction $1 = 0$. So $[x] \neq [0]$ and similar for $[y]$.

Neckverse Herdman
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Seems like a simpler example would be $\Bbb R[t]/(t^3)$, where $x=t, y=t^2$.

Robert Shore
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  • Yes, of course we can specialize $,y,$ in the universal example $,\Bbb R[x,y]/(xy),$ to any nonconstant $,f(x)\ [\neq x]\ \ $ – Bill Dubuque Oct 30 '23 at 23:02