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Decide whether a given set of rationals is a subring


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.5

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).


Solution:

(1) In Exercise 1.1.6, 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 Exercise 1.1.6, 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.


Linearity

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