This question has two parts
- I came out with this conjecture in the title: In a ring with no nilpotent elements (0) is not decomposable, and I think I have a proof of it:
Since the ring has no nilpotent elements we now that $ \sqrt{0}=0$ so $ 0$ has no minimal prime ideals belonging to 0. The minimal prime ideals belonging to zero are precisely the minimal prime ideals of the ring. Since every ring has minimal prime ideals $0 $ cannot be decomposable.
Is this proof right?
- I want to prove that in the ring $C(X)$ the set of continuous real valued functions of $f$ on $X$, a Compact Hausdorff space, 0 is not decomposable. A proof of this can be seen here. Using my previous "proof" I came up with this alternative proof:
the maximal ideals are of the form $M_c=\{f:f(c)=0\}$. Since the set of nilpotent elements is a subset of the Jacobson Radical = $\bigcap_{c\in \mathbb{R}}M_c =0$ we have that 0 is not decomposable.
I think that my proof is not ok since I'm not using the Hausdorff hypothesis which I think is important. The compactness is used in the claim "the maximal ideals are of the form $ M_c$" can some one point out what is wrong?
Thank you all in advance
Note: When I post questions of the type "what is wrong here" is so common to get answers of the type "your proof is wrong since the conclusion is wrong" which for me is equivalent to say "your proof is wrong because it is wrong". The kind of answer I'm looking for is "your proof is wrong because in this specific part of your proof you are wrong" so please avoid the first kind of answers.
Note 2
An ideal $I$ is decomposable if there are $Q_1,⋯,Q_m$ primary ideals such that I=⋂Qi. The $\sqrt{Q_i}$ are said to belong to I. Note that it is not necessarily true that $\sqrt{Q_i} \subseteq I $ nevertheless we still say it belongs to $ I$.
The first sentence of your proof is already false; in an integral domain we have $\sqrt{0}=0$, but $0$ is a minimal prime ideal belonging to $0$.