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I am reading section 12 (Flipping Moduli) of the paper "The polynomial $X^2+Y^4$ captures its primes" by Friedlander and Iwaniec.

At page 997, just below equation (12.7) we start estimating the following sum $$ V(f,g)=\sum_d f(d)\sum_{\substack{r_1s_2\equiv r_2s_1 (d)\\}} \alpha_{r_1,s_1}\bar{\alpha}_{r_2,s_2}\left(\frac{d}{r_1r_2}\right)g\left(\left|\frac{s_1}{r_1}-\frac{s_2}{r_2}\right|\right) $$ where $R\leq r\leq 2R$, $S\leq s\leq 2S$ and $f$ is supported on $\frac{1}{2}D\leq d\leq 3D$. They start by reducing the variables $r_1, r_2$ by the common divisor $c=(r_1,r_2)$ and removing the resulting condition $(c,d)=1$ (we know that both $r_1$ and $r_2$ are coprime to $d$, hence so is their gcd) by Mobius inversion, i.e. $$ 1_{\text{gcd}(c,d)=1}=\sum_{m|\text{gcd}(c,d)}\mu(m) $$ Using these tools we obtain equation (12.8) $$ V(f,g)=\sum_c\sum_{m\mid c}\mu(m)V_{c,m})(f,g) $$ where $V_{c,m}(f,g)$ is defined as $$ V_{c,m}(f,g)=\sum_d f(dm)\sum_{\substack{r_1s_2\equiv r_2s_1 (dm)\\ (r_1,r_2)=1}}\alpha_{cr_1,s_1}\bar{\alpha}_{cr_2,s_2}\left(\frac{d}{r_1r_2}\right)g\left(\left|\frac{s_1}{r_1}-\frac{s_2}{r_2}\right|\right) $$ They then proceed to obtain a bound for $V_{c,m}(f,g)$. After two pages they are able to prove the following bound (the inequality after equation (12.16)) \begin{equation} V_{c,m}(f,g)\ll\left\{(cm)^{-1/2}(DHRS)^{1/2}(2\log 2RS)^3 +\left[c^{-3/2}mD^{-1}(RS)^{3/2}+RS^{3/4}+SR^{3/4}\right](RS)^\epsilon\right\}\sum_{r,s}|\alpha_{cr,s}|^2 \end{equation} They then say that summing over this bound over $m$ and $c$ as in equation (12.8) yields $$ V(f,g)\leq \mathcal{H}(D,H,R,S)\sum_r\sum_s\tau(r)|\alpha_{r,s}|^2 $$ where $\mathcal{H}(D,H,R,S)$ satisfies the bound \begin{multline} \mathcal{H}(D,H,R,S)\ll (DHRS)^{1/2}(\log 2RS)^4\\ +\left[D^{-1}(RS)^{3/2}+RS^{3/4}+SR^{3/4}\right](RS)^\epsilon \end{multline} I am not able to understand how summing over $m$ and $c$ yields this bound. In particular what is the range over which $c$ varies? I think it would suffices to prove the following bounds $$ \sum_c\sum_{m\mid c}\mu(m)(cm)^{-1/2}\ll \log 2R $$ and $$ \sum_c\sum_{m\mid c}\mu(m)(c)^{-3/2}m\ll \log 2R $$ I am able to show, using the multiplicativity of $\mu$ that $$ \sum_{m\mid c}\mu(m)m^{-1/2}\leq 1\qquad\text{ and }\qquad\sum_{m\mid c}\mu(m)m\leq c $$ but I think that this estimation is too crude. In particular using such bound both equations reduce to showing that $$ \sum_c c^{-1/2}\ll \log 2R $$ but doesn't look to be correct. Overall I don't know if what I have done so far is correct and I should I proceed now. Thank you!

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  • $\begingroup$ Thank you for asking your question in such helpful detail! It's even slightly worse than you posit, because the $\ll$ bound for $V_{cm}(f,g)$ means that $\mu(m)$ must be replaced by $|\mu(m)|$ (hence effectively by $1$) when summing over $m$. Still, that part of the sum should be covered by the $\tau(r)$ in the eventual bound. $\endgroup$
    – Greg Martin
    Commented Dec 18, 2018 at 18:39
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    $\begingroup$ However, I see what you mean about the sums over $c$. Since $r_1$ and $r_2$ are close to $R$, their gcd $c$ should take values from $1$ up to $R$ish, meaning that $\sum_c c^{-1/2}$ looks to be about $R^{1/2}$ ... and a quick look through that section of the paper doesn't find anything to contradict this. So, I'm not being very helpful here, other than to say that I too don't see how this step goes yet. $\endgroup$
    – Greg Martin
    Commented Dec 18, 2018 at 18:40
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    $\begingroup$ At the end of the day, you have to bound $$\sum_{c|r}\sum_{m|c} \Bigl({1\over\sqrt{cm}}+{m\over c^{3/2}}\Bigr)$$ in terms of $r$, and while I'm not sure why they just don't say (e.g) this is $\tau_3(r)$ via lazily turning the summand into $1$, the asserted bound of $\tau_2(r)\log R$ seems correct. $\endgroup$
    – literature-searcher
    Commented Dec 18, 2018 at 19:43

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