abc=a+b+c in ring

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Source: RNMO SHL 2025, grade 12

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#1 h Apr 6, 2025, 5:55 PM • 1 Y

Y by PikaPika999

In which finite rings can we find three (not necessarily distinct) nonzero elements so that their sum equals their product?

David-Andrei Anghel

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#2 h Apr 6, 2025, 9:22 PM • 2 Y

Y by PikaPika999, RobertRogo

Answer: Anything but $\mathbb{F}_3.$

If $\mathrm{char} A \not\in \{2,3\}$ then we can choose $1,2,3.$

If $\mathrm{char} A = 2$ then we can choose $1,1,1.$

Now let $\mathrm{char} A = 3$ and consider the following cases:

Case 1. $A$ is not a field

Subcase 1.1. There is some nonzero nilpotent element $a \in A.$

We can choose $a^{k-1}, a^{k-1}, a^{k-1},$ where $k$ is the nilpotence order of $a.$

Subcase 1.2. There are no nonzero nilpotent elements in $A.$

By a standard result, there are nonzero zero divisors in $A.$ (Proof) Let $a,b \in A \setminus \{0\}$ such that $ab=0.$ Then we can choose $a,b,-(a+b).$ Indeed, the element $-(a+b)$ is nonzero, because otherwise we have $b=-a,$ so $a^2=0,$ which contradicts the assumption of this subcase.

Case 2. $A$ is a field.

It is easy to see that $\mathbb{F}_3$ does not work. To prove that any $A \neq \mathbb{F}_3$ works, it is enough to find $a,b \in A$ such that $a+b+1=ab,$ which is equivalent to $(a-1)(b-1)=2.$ Let $\alpha \in A \setminus \{0,1,2\}.$ Then we can choose $a=1+ \alpha$ and $b=1+2\alpha^{-1}.$ Clearly $a \neq 0$ and $b \neq 0$ (the former is equivalent to $\alpha=2$ and the latter is equivalent to $\alpha=1$ ).

This post has been edited 3 times. Last edited by Filipjack, Apr 7, 2025, 6:15 PM

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#4 h Apr 8, 2025, 9:31 AM

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Alternative approach for proving that $A$ is a field in the above Subcase 1.2:

Assume that $A$ has a nontrivial idempotent element, denoted as $e$ . Taking $x=-1, y=e, z=1-e$ yields $x+y+z=0=xyz$
Now it's a known result (due to Mr. Marian Andronache) that if $A$ is a finite ring such that $N(A)=\{0\}$ and $\text{Id}(A)=\{0,1\}$ then $A$ is a field. Hint for the proof

This post has been edited 2 times. Last edited by RobertRogo, Apr 8, 2025, 9:32 AM

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