The Wikipedia article about Elliott-Halberstam (EH for short) conjecture says that the so-called Bombieri-Vinogradov theorem, which is a weaker form of EH conjecture, is in some sense an averaged form of the Generalized Riemann Hypothesis (that is, the analogue of RH for Dirichlet L-functions). So, would the full EH conjecture follow from GRH?
Thanks in advance.

Edit March 29th 2021: denoting by $\Theta_{EH}:=\sup\{\theta\vert EH(\theta)\}$, EH conjecture is equivalent to $\Theta_{EH}=1$. The proven lower bound for it has increased from $1/2$ (Bombieri-Vinogradov) to a greater value. On the other hand, the proven upper bound for the de Bruijn-Newman constant $\Lambda$ has decreased from $1/2$ to $0.2$ (Platt and Trudgian), while the proof of Newman's conjecture by Rodgers and Tao implies $\Lambda\geq 0$, and RH is equivalent to this constant vanishing. So that, it seems that $\Theta_{EH}$ and $\Lambda$ are "dual", the duality relation being the involution $s\mapsto 1-s$. Is it thus likely that $\theta<1-\Lambda\Longrightarrow EH(\theta)$?


The Elliott-Halberstam conjecture is not known to follow from GRH. Even the weak version of EH (which is with $Q=x^{1/2+\epsilon}$ for any fixed $\epsilon>0$) does not follow from GRH. On the other hand, it is known that the Elliott-Halberstam conjecture almost implies the twin primes conjecture, i.e., it implies that there are infinitely many pairs of primes at distance $≤ 16$ (now $\le 12$, see Sylvain's comment). Furthermore, the Bombieri-Vinogradov theorem is indeed an amazingly strong unconditional replacement for the GRH bound (and has as natural strengthening the EH conjecture).

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    $\begingroup$ EH even implies that $16$ can be replaced by $12$, as shown by James Maynard. $\endgroup$ – Sylvain JULIEN Dec 29 '13 at 19:17
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    $\begingroup$ I think a big problem in preventing GRH from showing EH is the possible non-uniformity in implicit constants in the error terms. Perhaps a uniform version of GRH would suffice. $\endgroup$ – Stanley Yao Xiao Dec 30 '13 at 0:59

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