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I am currently studying integral domains and see that if $ab = ac$ and $a \neq 0$, then we have $b = c$. I was wondering whether the same is true for three nonzero subsets $I, J, K$ of an integral domain $R$, namely, if $I, J, K$ be three nonzero subsets of an integral domain $R$ such that none of them are equal with $R$ and contains $1$ and satisfies $IJ = IK$, then does this imply that $J = K$? (For nonempty subsets $A, B$ of $R$, we mean $AB = \{ab \mid a \in A \text{ and } b \in B\}$).

Also, if the answer is not affirmative, then what kind of domains satisfy this cancellative property about subsets?

Ratanjit
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  • Do you mean "does this imply that $J=K$"? Also, I assume that $R$ is an integral domain, but it worth making this explicit in your question. – Joe Mar 12 '23 at 19:24
  • Though there are trivial counterexamples, its worth noting that if every nonzero fin.gen. ideal is cancellable then the domain is a Prufer domain - see here for over $30$ characterizations of such, e.g. they are those domains satisfying the Chinese Remainder Theorem for ideals, or Gauss's Lemma for polynomial content ideals, or for ideals: $\rm\ A\cap (B + C) = A\cap B + A\cap C:,\ $ or $\rm\ (A + B)\ (A \cap B) = A\ B:,\ $ or $\rm\ A\supset B\ \Rightarrow\ A:|:B\ $ for fin. gen. $\rm:A:$ etc. – Bill Dubuque Mar 12 '23 at 19:32
  • There was a small typo in that question, so I edited my question, I am sorry about it. – Ratanjit Mar 12 '23 at 19:33

2 Answers2

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The real numbers are a field, and therefore an integral domain, and yet $$ (0,\infty)\cdot (1,\infty)=(0,\infty)\cdot (0,\infty)=(0,\infty) \, . $$ In fact, any integral domain $R$ which satisfies the desired property is isomorphic to $\mathbb Z/2\mathbb Z$. The proof is simple: since $$ \left(R\setminus\{0\}\right)\cdot \left(R\setminus \{0\}\right)=\left(R\setminus\{0\}\right)\cdot \{1\}=R\setminus\{0\} \, , $$ it follows that $R\setminus \{0\}=\{1\}$, so $R=\{0,1\}$. Since $1\neq0$ in an integral domain, $R$ has two elements, and any ring with two elements is isomorphic to $\mathbb Z/2\mathbb Z$. (Note that we also implicitly used the fact that $1\neq0$ to conclude that $R\setminus\{0\}$ contains $1$ when we were doing the above calculation.)

Even in $\mathbb Z/2\mathbb Z$, we have $\{0\}\cdot\{0\}=\{0\}\cdot \{1\}$, but I understand from your question that this is okay.

darij grinberg
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Joe
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We should definitely not expect this cancellation property of set multiplication. It's not even true in the real numbers: take for example $J=\{1\}$ and $K=\{-1,1\}$ and $I=[-7,7]$ (or any set that's symmetric with respect to the origin).

Greg Martin
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