Question

Let be a prime and let $\mathbb{Q}_{p}$ be the set of rational numbers whose denominator (when written in lowest terms) is not divisible by .

i) Show, with the usual operations of addition and multiplication, that $\mathbb{Q}_{p}$ is a subring of $\mathbb{Q}$.

ii) Show that is a subring of $\mathbb{Q}$.

iii) Is $\mathbb{Q}_{p}$ a field? Explain.

iv) What is $\mathbb{Q}_{p}\cap A_{p,}$ where $A_{p}$ is the set of all fractions with denominator a power of

Answer 1

p be a prime number. So be the set of all rational numbers whose denominatos (when written in lowest terms] is not divisible by P. Plb where a is the lowest 1 terms = { 08:PXb, ged(a, b)=i} - (1) clearly spco het a az E Be then pfb, ; PX bz so e f (6, 62) and pfairbe) And a az ele Hence Sp be a subring of B. Since 2 and 5 both prime, by ② B and ass both subring of a Again, we know that intersection of two subrings is a subring. Hence and is a subring of ..

( p is a prime number , then 2XP, 3XP, .......(P-1)fp. :. god( P-t, P) = 4 mes. How since pt (e-12 because (R6<p) Let m be the inverse multiplication . of L with respect to en med here denominator of m is p, which divisible by is P. 1 . P-1 & Sp that is mese Since each Sp, then element of Sp is not Sp has no inverse a field. in An is the set of all а рел. оf P. fractions with denominaton -- Ap = : a€ 2, «211 Let XEB, A Ap . xE8p and xe Ap : :: x= wthere ae 2,621

Now 'xe Sp that is p., 021, DEZ By constraction of Sp, azo be the only possibility :: 0 Ap= {o}
0 be the identity element of the group (Qp ,+).