Is $\gcd(p_1^{n_1},\cdots,p_k^{n_k})=1$ if each $p_i$ is a prime number?

Question

Is $\gcd(p_1^{n_1},\dots,p_k^{n_k})=1$ if each $p_i$ is a prime number and $n_i\geq0$? I just really want to know if it is the case or not, I am not really intersted in a proof. Also, if this is the case, does it follow that $\operatorname{lcm}(p_1^{n_1},\dots,p_k^{n_k})=p_1^{n_1}\cdots p_k^{n_k}$?

Thanks in advance.

Answer 1

$\gcd(p_1^{n_1},\ldots, p_k^{n_k})=1$ if and only if $p_i\ne p_j$ for some $1\le i<j\le k$.

The second statement is slightly more complicated. If the $p_i$ are all pairwise distinct, then $\operatorname{lcm}(p_1^{n_1},\ldots, p_k^{n_k})=p_1^{n_1}\cdots p_k^{n_k}.$

If they are not distinct, then denote by $p_{i_1},\ldots, p_{i_r}$ the distinct primes amongst $\{p_1,\ldots, p_k\}$. Let $n_{i_\ell}=\max\{n_i: p_i=p_{i_\ell}\}.$ Then $$ \operatorname{lcm}(p_1^{n_1}\cdots p_k^{n_k})=\prod_{j=1}^r p_{i_r}^{n_{i_\ell}}.$$