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?
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.