# In every local ring exist nonzero nilpotent ideals?

I read that every local ring is clean so there exist clean rings with nonzero nilpotent ideals, i know that a local ring has a unique maximal ideal, but i don't know why do they say this implication.

thank you.

• In a local ring $R=R^\times\cup m$ where $m$ is the unique maximal ideal, $a$ or $a+1$ is a unit, thus it is clean. $k[x]/(x^2)$ is a local ring. $k[x]/(x^2-x)$ is clean but non local (its non-units are $rx,r-rx$ so one of $a,a-1,a-x,a-(1-x)$ is a unit ) Commented Jan 12, 2020 at 5:14

Because it is easy to produce local rings with nilpotent ideals, witnessing this statement. $$\mathbb Z/ 4\mathbb Z$$ is local with a nilpotent maximal ideal. Thus it is local, clean, and has such an ideal.
Based on the above, it sounds like you misunderstood the strategy of quantification. The claim only asserted “$$\exists$$ clean ring with nonzero nilpotent ideal” but you interpreted its solution as “$$\forall$$ local ring, local ring has nonzero nilpotent ideal, hence $$\exists$$ a clean ring like that.” Really it is already sufficient that “$$\exists$$ local ring such that it has nonzero nilpotent ideal, hence $$\exists$$ a clean ring like that.”