## Decide whether a given set of rationals is a subring

Decide which of the following are subrings of .

- The set of all rational numbers with odd denominators (when written in lowest terms).
- The set of all rational numbers with even denominators (when written in lowest terms).
- The set of all nonnegative rational numbers.
- The set of squares of rational numbers.
- The set of all rational numbers with odd numerators (when written in lowest terms).
- The set of all rational numbers with even numerators (when written in lowest terms).

- In this previous exercise, we showed that this set is a subgroup of . Moreover, if and are written in lowest terms and and are odd, then when is written in lowest terms, its denominator must divide and thus is odd. So this subset of is closed under multiplication and is indeed a subring.
- In this previous exercise, we showed that this set is not a subgroup of . Thus it is not a subring of .
- This subset is not a group, since only 0 has an additive inverse. Thus it is not a subgroup of , and thus not a subring.
- Note that is in this set, but that is not the square of a rational. To see this, suppose otherwise that ; then . By the fundamental theorem of arithmetic, the number of times 2 divides and must be the same. However, 2 divides and even number of times, while it divides an odd number of times, a contradiction. Thus this subset is not a subring, as it is not closed under addition.
- This set is not closed under addition since is in lowest terms, but is in lowest terms and has an even numerator. Thus this subset is not a subring.
- First we show that this subset is a subgroup. If and are in lowest terms and and are even, then . Now and must be odd, so that when reduced to lowest terms, the numerator of is even since 2 divides and . Since contains , is a subgroup. Now if and are in and in lowest terms, then and are odd, so that when expressed in lowest terms, 2 must divide . Thus is closed under multiplication, and is a subring of . Now suppose has an identity element . Then for all , we have , so that . Thus . Since must be even and odd, we have a contradiction. So is a ring, but not a unital ring.

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## Comments

In 4, the candidate subring is not closed under addition or subtraction. 16-9=7, which is not a square of a rational. Q is only a group under addition.

Oops- you’re right.

Thanks!

for number 1, S = the set of all rational numbers with odd denominators is not a subgroup of Q because take 2/9 (its in S) but 9/2 is not in S so S is not a subgroup of Q. Thus S is not a subring of Q.

You are correct that is not a multiplicative subgroup, but it is an additive subgroup. Ring elements need not have multiplicative inverses.

Sorry, I am kind of lost. I guess I am not understanding the definition of the subring: A subring of the ring R is a subgroup of R that is closed under multiplication.

Do you think you can further explain?

Sure.

A ring is a set with two binary operators, “plus” and “times”, one unary operator “minus”, and one nullary operator (i.e. an element) “zero”, which satisfy some axioms.

A subring of a ring is a subset which is closed under the operators.

So it is true that a subring is an (additive) subgroup which is closed under multiplication, though strictly speaking this follows from the definition.

Now a ring might have other operations defined on it- for instance, in the rationals, every nonzero element has a multiplicative inverse. Subrings do not need to be closed under “bonus operators” like this.

It seems like the major confusion here is that the rationals are a group under addition and (excluding zero) under multiplication as well. In the context of rings, the group structure we care about is the additive structure.

It’s worth noting that the answer to number 6 depends on your definition of “ring.” Many authors will say that a ring must contain 1, in which case #6 is not a subring.

Dummit and Foote omit the requirement of a multiplicative identity in their axioms for a ring. So, in view of the fact that this is an exercise from their text, the correct answer is that #6 is a subring.

for number 1, if we take 1/3 * 1/3 = 1/6 which is not in the set. Doesn`t that make it not a subring?

Sorry, I realize my mistake now 🙂

No worries. 🙂