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

Decide which of the following are subrings of $\mathbb{Q}$.

1. The set of all rational numbers with odd denominators (when written in lowest terms).
2. The set of all rational numbers with even denominators (when written in lowest terms).
3. The set of all nonnegative rational numbers.
4. The set of squares of rational numbers.
5. The set of all rational numbers with odd numerators (when written in lowest terms).
6. The set of all rational numbers with even numerators (when written in lowest terms).

1. In this previous exercise, we showed that this set is a subgroup of $\mathbb{Q}$. Moreover, if $a/b$ and $c/d$ are written in lowest terms and $b$ and $d$ are odd, then when $ac/bd$ is written in lowest terms, its denominator must divide $bd$ and thus is odd. So this subset of $\mathbb{Q}$ is closed under multiplication and is indeed a subring.
2. In this previous exercise, we showed that this set is not a subgroup of $\mathbb{Q}$. Thus it is not a subring of $\mathbb{Q}$.
3. This subset is not a group, since only 0 has an additive inverse. Thus it is not a subgroup of $\mathbb{Q}$, and thus not a subring.
4. Note that $1 = 1^2$ is in this set, but that $1+1 = 2$ is not the square of a rational. To see this, suppose otherwise that $(a/b)^2 = 2$; then $a^2 = 2b^2$. By the fundamental theorem of arithmetic, the number of times 2 divides $a^2$ and $2b^2$ must be the same. However, 2 divides $a^2$ and even number of times, while it divides $2b^2$ an odd number of times, a contradiction. Thus this subset is not a subring, as it is not closed under addition.
5. This set is not closed under addition since $1/3$ is in lowest terms, but $2/3$ is in lowest terms and has an even numerator. Thus this subset is not a subring.
6. First we show that this subset $A_6$ is a subgroup. If $a/b$ and $c/d$ are in lowest terms and $a$ and $c$ are even, then $(a/b) + (-c/d) = (ad-bc)/bd$. Now $b$ and $d$ must be odd, so that when reduced to lowest terms, the numerator of $(ad-bc)/bd$ is even since 2 divides $a$ and $b$. Since $A_6$ contains $2 = 2/1$, $A_6 \leq \mathbb{Q}$ is a subgroup. Now if $a/b$ and $c/d$ are in $A_6$ and in lowest terms, then $b$ and $d$ are odd, so that when expressed in lowest terms, 2 must divide $ac$. Thus $A_6$ is closed under multiplication, and is a subring of $\mathbb{Q}$. Now suppose $A_6$ has an identity element $u/v$. Then for all $a/b$, we have $ua/vb = a/b$, so that $uab = vab$. Thus $u = v$. Since $u$ must be even and $v$ odd, we have a contradiction. So $A_6$ is a ring, but not a unital ring.

• Bobby Brown  On October 31, 2010 at 5:48 pm

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.

• nbloomf  On October 31, 2010 at 7:47 pm

Oops- you’re right.

Thanks!

• yang lu  On April 6, 2011 at 11:03 pm

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.

• nbloomf  On April 6, 2011 at 11:17 pm

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

• yang lu  On April 6, 2011 at 11:39 pm

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?

• nbloomf  On April 7, 2011 at 1:24 pm

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.

• James  On April 14, 2011 at 11:50 pm

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.

• Will  On November 4, 2011 at 4:48 am

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.

• Danny  On November 25, 2011 at 12:07 pm

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?

• Danny  On November 25, 2011 at 12:12 pm

Sorry, I realize my mistake now 🙂

• nbloomf  On November 25, 2011 at 10:53 pm

No worries. 🙂