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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.

Prism
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2 Answers2

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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.

rschwieb
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    Thanks for this great answer. (+1) To prove the exercise you mention, I would argue by proving the contrapositive: Assume $x^{n}=0$ for $x\neq 0$, where $n$ is minimal. Then $(x^{n-1})^2=x^{2n-2}=0$, because $2n-2\ge n$, as $n\ge 2$. Since $x^{n-1}\neq 0$ (using minimality of $n$), we have obtained a non-zero element whose square is $0$. And this isn't much different from the proof I have :( Is there slicker way of proving the Exercise? :) – Prism Jun 28 '13 at 01:01
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    The proof using the fact Nilradical is inside Jacobson radical is also pretty cool! (Though I will have to learn why Jacobson radical of absolutely flat ring is zero) – Prism Jun 28 '13 at 01:10
  • @prism yes, the exercise is no slicker than your proof, but I feel like breaking it out that way leaves you with a second bit of information to use in other circumstances ! – rschwieb Jun 28 '13 at 01:34
  • @rschwieb You can prove it by passing to the localization. –  Jun 28 '13 at 01:38
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    It is a cool fact, nonetheless. :) – Prism Jun 28 '13 at 01:38
  • Dear @BenjaLim : for those of us not fluent in thinking with localizations, it seems overcomplicated. The way I would do it is to point out that every nonzero ideal of a VNR ring contains a nonzero idempotents, but the Jacobson radical can't contain such a thing (since if x is in J, then 1-x is a unit, but if xx=x, 1-x is an idempotents too, whence x=0) but this is just a comfort thing! I will be sure to figure out the localization way sometime :) – rschwieb Jun 28 '13 at 01:59
  • Looks like my tablet has a habit of pluralizing idempotent. – rschwieb Jun 28 '13 at 02:02
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    @rschwieb: I would finish it like this: If $x$ is idempotent in J, then $1-x=1-x^2=(1-x)(1+x)$. Since $1-x$ is unit, multiplying both sides by $(1-x)^{-1}$, we get $1=1+x$, so $x=0$. Anyways, I like purely ring-theoretic arguments too! – Prism Jun 28 '13 at 02:08
  • @rschwieb If you are stuck with the localization argument, you can look at my answer below. –  Jun 28 '13 at 03:05
  • @Prism I added a proof in my answer below that a local ring that is absolutely flat is a field. –  Jun 28 '13 at 03:10
  • @Prism Another advantage to the approach I took is that it works for noncommutative rings too, where localization is sometimes harder, sometimes impossible. – rschwieb Jun 28 '13 at 13:08
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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?

<p><strong>Proof the a local ring that is absolutely flat is a field:</strong> 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.</p>
  • That is a sweet proof :) (+1) "Being zero" is a local property. This idea (Proposition 3.8 in A & M) seems to be used heavily! – Prism Jun 28 '13 at 01:36
  • @Prism Yes certainly. I have used that result many times over and over again. –  Jun 28 '13 at 01:37
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    @BenjaLim Nakayama seems like a bit of a sledgehammer here! A local ring has only 0 and 1 as idempotents. But each principal ideal is generated by an idempotent in an absolutely flat ring. Hence, the ring has no nontrivial ideals and is a field. (With minor rewording, works for noncommutative rings too.) – rschwieb Jun 28 '13 at 13:01
  • Dear @BenjaLim I hope not to come off as critical: really I do covet the ability to reason with localizations and other commutative algebra techinques, and I'm glad to have your demonstrations at hand. Maybe when I've digested them I will regard them as elementary as the techniques I suggested. I hope someone else gets something out of my techniques too. – rschwieb Jun 28 '13 at 13:06
  • Dear @BenjaLim In the sense of this essay I feel like a fox, knowing a lot of little things, and not yet knowing the "one big thing" (localization) that the hedgehog knows. I understand it is a sharp tool in commutative algebra :) – rschwieb Jun 28 '13 at 13:34
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    @rschwieb No worries! Localization and the Nakayama Lemma are bread and butter in commutative algebra. I wish I knew more techniques myself in noncommutative algebra! Unfortunately my background has been such that placed it was a heavy emphasis on commutative algebra. Regards, –  Jun 29 '13 at 16:34
  • To prove an absolutely flat local ring is a field, you can't apply Nakayama's lemma to m, since it may not be finitely generated. – Mathstudent Nov 11 '23 at 01:27