Okay, let's run some radical translations of shit I've already written:
\[
\sqrt{\mathcal{I}_{p^k}} = \sigma(k)\\
\]
for any prime \(p\). And we can write a straight bound:
\[
\sqrt{\mathcal{I}_{p^kq^u}} \le \sigma(k) \sigma(u)\\
\]
While additionally we have:
\[
|\sqrt{\mathcal{I}_{ab}}| \le |\sqrt{\mathcal{I}_{a}}||\sqrt{\mathcal{I}_{b}}|
\]
so long as \((a,b)=1\), that \(a\) and \(b\) are co prime..... I think the word for this is "sub" multiplicative, or some bs like that...
Fuck, Vittorio-- I think you are a couple steps away from a hard result........
I should add that I have ultimately proven the sub radical formula:
\[
|\sqrt{\mathcal{I}_{ab}}| \le |\sqrt{\mathcal{I}_{a}}||\sqrt{\mathcal{I}_{b}}|
\]
by just noting that the left hand adds a bunch of Iverson \([a=b]\), and the right hand ignores that. Where, all we do is add a couple \(1)'s on the right... nothing more.
Once you add this, you have created a weird ring effect, and a better zeta function. We can write \(|\sqrt{\mathcal{I}_m}| = q(m)\). And this is not multiplicative, but \(q(mn) \le q(m)q(n)\) when \((m,n)=\) are coprime......
\[
\sqrt{\mathcal{I}_{p^k}} = \sigma(k)\\
\]
for any prime \(p\). And we can write a straight bound:
\[
\sqrt{\mathcal{I}_{p^kq^u}} \le \sigma(k) \sigma(u)\\
\]
While additionally we have:
\[
|\sqrt{\mathcal{I}_{ab}}| \le |\sqrt{\mathcal{I}_{a}}||\sqrt{\mathcal{I}_{b}}|
\]
so long as \((a,b)=1\), that \(a\) and \(b\) are co prime..... I think the word for this is "sub" multiplicative, or some bs like that...
Fuck, Vittorio-- I think you are a couple steps away from a hard result........
I should add that I have ultimately proven the sub radical formula:
\[
|\sqrt{\mathcal{I}_{ab}}| \le |\sqrt{\mathcal{I}_{a}}||\sqrt{\mathcal{I}_{b}}|
\]
by just noting that the left hand adds a bunch of Iverson \([a=b]\), and the right hand ignores that. Where, all we do is add a couple \(1)'s on the right... nothing more.
Once you add this, you have created a weird ring effect, and a better zeta function. We can write \(|\sqrt{\mathcal{I}_m}| = q(m)\). And this is not multiplicative, but \(q(mn) \le q(m)q(n)\) when \((m,n)=\) are coprime......
