corollaryroutine

Enlarging exceptional primes changes only constant

So enlarging SS to S′S' only rescales the height bound, same cc
discards extra primes without changing the exponent
Corollary·hei:enlarge-exceptions·§2.2 Contact primes and ordinary integer witnesses

Statement (paper)

If S′⊃SS'\supset S is any fixed finite set of primes, then h(s)≤H(∏q∈S′∖Sq)cMS′(s)c(s∈P1(Q)∖C). \hgt(s)\le H\left(\prod_{q\in S'\setminus S}q\right)^c M_{S'}(s)^c \qquad(s\in\mathbf P^1(\Q)\setminus\mathcal C). In particular enlarging the exceptional set changes the constant but leaves the exponent unchanged.

In plain English

For any finite set of primes S′⊃SS'\supset S, every s∈P1(Q)∖Cs\in\mathbf P^1(\Q)\setminus\mathcal C satisfies h(s)≤H(∏q∈S′∖Sq)cMS′(s)c\hgt(s)\le H\left(\prod_{q\in S'\setminus S}q\right)^c M_{S'}(s)^c, with the same exponent cc as in the original height bound. So enlarging the exceptional set only changes the constant.

Why it's here

This follows from the height bound in hei-height, since passing from SS to S′S' removes the primes in S′∖SS'\setminus S from the contact product MS(s)M_S(s) and so changes it by at most the product of those primes. It lets later arguments discard finitely many extra primes, such as those arising from a particular root tuple, without affecting the exponent in the uniform height bound for ring elements.
Classified routine:
This is a direct consequence of the height bound hei-height: moving from S to S' removes only the primes in S'\S from the contact radical M_S(s), so M_S(s) changes by at most their product and the bound holds with the same exponent c and a larger constant.

Sources

PaperOpen
Section
§2.2 Contact primes and ordinary integer witnesses
Page
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Label
hei:enlarge-exceptions
Line
sections/02-reduction.tex:270
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