# Non Noetherian ring with only one prime ideal

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?

• 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. – rschwieb Jan 30 at 14:06
• Why don't post the edit as an answer? – 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.
• 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 $I$? – ZFR Jan 30 at 1:12
• @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 $I$. – Badam Baplan Jan 30 at 6:32