Nilradical of absolutely flat ring Suppose $A$ is an absolutely flat ring (i.e. every $A$-module is flat). Is it true that nilradical of $A$ is trivial, i.e. $\mathfrak{N}(A)=\{0\}$? 
I believe the answer is yes. Here is my attempted proof:

We use the following characterization of absolutely flat rings (Chapter
  2, Exercise 27 in Atiyah & Macdonald). $A$ is absolutely flat ring $\Leftrightarrow$
  $(x)=(x^2)$ for each $x\in A$. Now, suppose $x^{n}=0$ for
  $n\in\mathbb{N}$, where $n$ is minimal. If $x\neq 0$, we necessarily
  have $n\ge 2$. Since $(x)=(x^2)$, we have $x=ax^2$ for some $a\in A$. Thus,
  $x^{n-1}=ax^{n}=0$, contradicting minimality of $n$. Hence, $x=0$ is
  the only nilpotent element of $A$.

Is my proof correct?
I have done Chapter 2, Exercise 28 in Atiyah & Macdonald, which lists important properties of absolutely flat rings, but this simple property was not mentioned there.
 A: That looks ok.
Another way is to note (or learn now) that a ring has no nonzero nilpotents if it has no nonzero elements with square zero (exercise!) then you can say "suppose a is nonzero but $a^2=0$. Then there exists an x such that $a=a^2x=0$ a contradiction."
Another path, if you know that the Jacobson radical of an absolutely flat ring is zero, is notice that the intersection of all prime ideals (equal to the nil radical) is contained in the intersection of maximal ideals (the Jacobson radical) and hence both are zero.

Bonus noncommutative info
Not all Von Neumann regular rings lack nilpotents (any matrix ring $M_n(F)$ over a field with n>1 has nilpotents). The ones that don't have any nilpotents are called strongly regular rings, and they are exactly the VNR rings whose idempotents are all central.
A: Here is another bit of information that you can get out of absolutely flat rings that is very useful. 


If $A$ is absolutely flat then for any multiplicative subset $S \subseteq A$, we have $S^{-1}A$ being absolutely flat.


Thus in particular for any maximal ideal $\mathfrak{m} \subseteq A$, we have $A_{\mathfrak{m}}$ being absolutely flat. However a local ring that is absolutely flat is a field and so this means that $\operatorname{Jac}(A)_{\mathfrak{m}} \subseteq A_{\mathfrak{m}}$ is zero for all maximal ideals $\mathfrak{m}$. The Jacobson radical is an ideal in the localization because $\operatorname{Jac}(A) \subseteq \mathfrak{m}$ for any maximal ideal of $A$. Since being zero is a local property this means that $\operatorname{Jac}(A) = 0$. The same proof mutadis mutants also works to show that the nilradical of an absolutely flat ring is zero.


Exercise: Proof the result mentioned above. Hint: You will not believe it, but for any two $S^{-1}A$ - modules $M$ and $N$, the modules $M \otimes_A N $ and $M \otimes_{S^{-1}A} N$ are isomorphic as $S^{-1}A$ - modules!! Now why is $M \otimes_A N$ a $S^{-1}A$ - module?
Proof the a local ring that is absolutely flat is a field: Let $A$ be a local ring with maximal ideal $\mathfrak{m}$. Choose any $x \in \mathfrak{m}$. Then we have $\mathfrak{m}(x) = (x)$. Indeed, $\mathfrak{m}(x) \subseteq (x)$ and the other inclusion follows from $(x) = (x^2)$. By the Nakayama Lemma it follows $x = 0$ and since this was any element of $\mathfrak{m}$, we have $\mathfrak{m} = 0$. In other words, $A$ is a field.


