Let $k$ be a field. Then $k[x]/(x^2)$ has only one prime ideal.

It is clear that $(X)$ is the maximal ideal thus prime where $X$ is the class elements $x$. How do I know there aren't other prime ideals? I tried to define homomorphism $\phi':k[x]/(x^2)\to k$ by evaluation but it seems there is no other way to define the map well defined other than evaluating at 0 due to $\phi'([f])=\phi'([g])$ not equal for different representatives.

  • $\begingroup$ Accepted solution at linked question explains this very well. $\endgroup$
    – rschwieb
    Feb 15, 2017 at 4:57

1 Answer 1


Suppose $\mathfrak{p}$ is a prime ideal. Then $x\cdot x = 0\in \mathfrak{p}$, so $x\in \mathfrak{p}$, and $(x)\subseteq \mathfrak{p}$. Since, as you note, $(x)$ is maximal, $(x) = \mathfrak{p}$.


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