# Why is it necessary to show that the set R is closed under multiplication in order to prove that (R,+,*) is a Ring.

I am currently reading John B Fraleigh's book A First Course in Abstract Algebra. The author in chapter 23, insists on the point that it is necessary for one to show that the R is closed under multiplication (along with the other Ring axioms) in order to completely show that is a Ring. However, in the definition of a Ring, I've come across only three axioms, which are as follows:

1. (R,+) must be an Abelian Group
• must be associative
2. For all a,b,c $\in$R the left distributive law and the right distributive law holds.

Now where is it mentioned that a*b must belong to R for all a,b $\in$ R. There are consequences of this fact, namely in showing that the set of all pure imaginary complex numbers ri for r $\in$ $\Re$ (with the usual addition and multiplication) is not a Ring. Since any two pure imaginary numbers are not closed under multiplication, the author writes that this set does not define a Ring.

• @user1952009 That's not correct - multiplication isn't mentioned in the group axioms. – Noah Schweber Jun 2 '16 at 3:34
• @NoahSchweber : the monoid axioms for rings, the group axioms for fields – reuns Jun 2 '16 at 3:34
• @user1952009 Note that the OP's axiomatization only says that multiplication is associative and satisfies distributivity - there is no mention of the monoid laws. So, that is not correct. – Noah Schweber Jun 2 '16 at 3:35
• @NoahSchweber : instead of saying "that is not correct" say that the usual definition of rings is that $\times$ is a monoid, right ? en.wikipedia.org/wiki/Ring_(mathematics)#Definition – reuns Jun 2 '16 at 3:36
• @user1952009 But the OP's question is about their given definition, so that's just not relevant (hence, not a correct answer to the OP's question). – Noah Schweber Jun 2 '16 at 3:36

## 1 Answer

In the three axiom definition, being closed under the operation is linguistically implicit, but generally explicitly ensured by defining the operations to be functions whose codomains are $R$. A ring must always be closed under addition and multiplication, but just like you lumped the left and right distributive laws into one axiom, many modern texts hide the closure axiom in the definitions. I've seen axiomatizations of rings with $11$ axioms, but in the end everyone has essentially the same definition.

• Note that we see this convention elsewhere - an $n$-ary operation on a set $X$ is almost always defined as a map $X^n\rightarrow X$, and multiplication, addition, etc. are binary operations. – Noah Schweber Jun 2 '16 at 3:38
• @Matt Samuel : There is no "disagreement" about the multiplicative identity, there's only different conventions as how we should name the structure "ring with identity" and "ring without identity", and these conventions always have a reason to exist ; the same story applies to the assumption of commutativity. An algebraic geometer would not call a ring a commutative unital ring since he/she always wants to work with commutative unital rings ; a non-commutative algebraist would be annoyed to assume that all his/her rings are commutative and write "non-commutative" all the time. – Patrick Da Silva Jun 2 '16 at 5:05
• Can the closure property be deduced from the three axioms I've mentioned in the question? If so, how? – model_checker Jun 2 '16 at 6:19
• @User no. It would be in the definition of the multiplication map. – Matt Samuel Jun 2 '16 at 10:03