Exercise 1.3.13 (If $(a,b) = p$, what are the possible values of $(a^2,b)$?)

Proof. We denote ${\nu }_p(a)$ the largest exponent $\alpha$ such that $p^{\alpha }\mid a$.

Since $a\wedge b=p$, ${\nu }_q(a)={\nu }_q(b)=0$ if $q$ prime, $q\ne p$ (thus ${\nu }_q(a^k)={\nu }_q(b^k)=0$, and

(a)

• Case 1. ${\nu }_p(a^2)=2$ and ${\nu }_p(b)\ge 1$, thus $a\wedge b=p$ or $a\wedge b=p^2$.
• Case 2. ${\nu }_p(a^2)\ge 2$ and ${\nu }_p(b)=1$, thus $a\wedge b=p$.

The possible values of $a^2\wedge b$ are $p,p^2$.

(b)

• Case 1. ${\nu }_p(a^3)=3$ and ${\nu }_p(b)\ge 1$, thus $a^3\wedge b=p,p^2$ or $p^3$.
• Case 2. ${\nu }_p(a^3)\ge 3$ and ${\nu }_p(b)=1$, thus $a^3\wedge b=p$.

The possible values of $a^3\wedge b$ are $p,p^2,p^3$.

(c)

• Case 1. ${\nu }_p(a^2)=2$ and ${\nu }_p(b^3)\ge 3$, thus $a^2\wedge b^3=p^2$.
• Case 2. ${\nu }_p(a^2)\ge 2$ and ${\nu }_p(b^3)=3$, thus $a^2\wedge b^3=p^2$ or $p^3$.

The possible values of $a^2\wedge b^3$ are $p^2,p^3$.