Give an example of non Noetherian ring with only one prime ideal.

In one of the topics I have read that $k[x_1,x_2,x_3,\dots]/(x_ix_j)$ will be an example. I know that the ring $k[x_1,x_2,x_3,\dots]$ is NOT Noetherian ring. But can anyone show in detail why the above example will work?

I am studying commutative algebra for a short time. So the detailed explanation would be very useful.

I would be very grateful for your help!

And please do not duplicate this question since others topics has only answer no explanation.

EDIT: I will provide detailed explanation of rschwieb's answer in order to clarify things for myself.

Let $\mathbb{k}$ is field and consider the ring $\mathbb{k}[x_1,x_2,\dots] $ and ideal generated by all products $x_ix_j$ for $1\leq i\leq j<\infty$ and call this ideal $J$. Consider the following quotient-ring $R:=\mathbb{k}[x_1,x_2,\dots]/J$. Consider and ideal $I$ in $R$, where $I=(x_1+J,x_2+J,\dots)$. Note that you've written $(x_1,x_2,\dots)$ but I guess that it's incorrect since element in $R$ has form $f+J$, right?

Also note that an ideal $I$ is nilpotent because $I^2$ is zero $R$, i.e. $I^2=J$. Ideal $I$ is maximal in $R$ because the quotient-ring $R/I$ is field because any nonzero element in $R/I$ has form $(c+J)+I$, where $c\in \mathbb{k}$ and $c\neq 0$. But it definitely has inverse since $\mathbb{k}$ is field.

Thus, an ideal $I$ in $R$ is nilpotent and maximal. Hence $R$ has unique prime ideal which is $I$.

Is my reasoning correct?

  • $\begingroup$ Note that you've written $(x_1,x_2,\dots)$ but I guess that it's incorrect . No, I wrote "$(x_1,x_2,\dots)$ in the quotient" meaning $(x_1,x_2,\dots)/I$, which is the same thing as what you said, modulo the chosen notation. One should probably include the comment Badam Baplan made to make the final conclusion clear. $\endgroup$ – rschwieb Jan 30 at 14:06
  • $\begingroup$ Why don't post the edit as an answer? $\endgroup$ – user26857 Jan 30 at 16:26

I think you mean the $i,j$ in the denominator range over all indices.

In that case, the ideal $(x_1,x_2\ldots)$ in the quotient $k[x_1,x_2\ldots]/I$ where $I$ is generated by the pairwise products of indeterminates is both maximal and nilpotent, so it is the unique prime ideal.

  • $\begingroup$ Could you give more details, please? I really cannot understand what ideal is prime here? And in what ring do we consider our ideal? I would be very grateful for detailed and good answer. $\endgroup$ – ZFR Jan 29 at 22:25
  • $\begingroup$ @ZFR I wrote more explicitly about the ideal and ring. You should be able to confirm now. $\endgroup$ – rschwieb Jan 29 at 22:30
  • $\begingroup$ How to prove that if $R$ commutative ring and an ideal $I$ is nilpotent and maximal then there is only one prime ideal in $R$, namely $I$? $\endgroup$ – ZFR Jan 30 at 1:12
  • $\begingroup$ I guess that you are using this fact here, right? $\endgroup$ – ZFR Jan 30 at 1:20
  • 1
    $\begingroup$ @ZFR If $I$ is nilpotent and maximal then we have $I^n = 0$ for some $n$. If $P$ is any prime of $R$ then we thus have $I^n \subseteq P$. Since $P$ is prime we have $I \subseteq P$, hence $I = P$ by maximality of $I$. $\endgroup$ – Badam Baplan Jan 30 at 6:32

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