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Lifting a direct summand of a free module
After your edit, the answwer is trivially yes: Split the map $R_I\rightarrow S_I$ with a map $f_I:S_I\rightarrow R_I$. Lift $f$ arbitrarily to a map $f:S\rightarrow R$. Then the composition $R\rightarrow S\rightarrow R$ is the identity mod $I$, hence an isomorphism. 
Oct 14 
comment 
Lifting a direct summand of a free module
Multiplication by $1+i$ becomes a split injection mod $I$. 
Oct 14 
answered  Lifting a direct summand of a free module 
Oct 6 
comment 
polynomials with roots on the unit circle
Even in the quadratic case, $P$ can be anything of the form $Ax^2+Bx+A$, with $2A<B<2A$, which doesn't translate into any sort of congruence condition mod primes (other than the trivial condition that two coefficients that are equal remain equal after reduction). So I'm unclear on what you're looking for. 
Oct 5 
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Why do roots of polynomials tend to have absolute value close to 1?
The other answers are quite enlightening, but it was the rephrasing in this answer that made me go "Oh! Of course!". If you pull all the coefficients of an $m$degree polynomial from a reasonably tight probablility distribution, then the coefficients will all be pretty close to each other, so the polynomial will be close to a factor of $x^{m+1}1$. Of course it's natural to think about how to make this more precise, but to me, this is the insight that answers the original "Why should I expect this?" question. 
Oct 3 
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Why do roots of polynomials tend to have absolute value close to 1?
@WłodzimierzHolsztyński: Well, yes, but when the roots are nonreal, they "travel" in the degenerate sense that they don't actually travel at all. 
Oct 2 
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Why do roots of polynomials tend to have absolute value close to 1?
For polynomials of the form $x^2+bx+c$, the absolute values of the nonreal roots depend only on $c$, so that as $b$ varies, we get lots of roots all on a circle. Is this the same phenomenon writ small, and does it shed any light? 
Oct 1 
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Surjectivity of the algebraic Kfunctor
@MatthiasWendt: I don't think the localization sequence is relevant here; the OP is asking about the map $K_q(R)\rightarrow K_q(S)$, not about a map $K_q(S)\rightarrow K_q(R)$. 
Oct 1 
awarded  Organizer 
Oct 1 
revised 
If matrices describe simplices, what do matrix operations describe?
deleted "simplicialstuff" tag, which i understand to apply to simplicial sets and their generalizations. 
Sep 23 
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is $x_{n}\ll \overline{x}_{n}^{2}$?
What does $\ll$ mean? 
Sep 8 
awarded  Nice Answer 
Aug 17 
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A group allowing exactly 7 group topologies
"This is why they say 7". Who is they? 
Aug 4 
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Universal quantifier in Russell's Theory of descriptions  Who is the UNIVERSE?
PS. In ZFC, I can prove: "For all $A$, $A\subset A$." Are you worried that there is no set of all sets? 
Aug 4 
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Universal quantifier in Russell's Theory of descriptions  Who is the UNIVERSE?
Andreas beat me to this with his answer, but it's models, not theories, that have "universes of discourse". 
Aug 4 
answered  About Weil's proof of “Weil conjectures for curves and abelian varieties” 
Jul 21 
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Is there a truly general voting impossibility theorem that applies to real elections?
You can address both problems (I) and (II) if two separate polities  say Texas and California  both adopt this system, with Texas voters making their payments to Californians and California voters making their payments to Texans. Then none of the democracy fees are wasted (which addresses problem II) and the people who put less monetary value on getting their preferred election outcomes will receive, on average, cash payments that they value more highly than they value those outcomes (at least assuming that the average payments are the same in both polities). 
Jul 21 
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commutative algebra, diagonal morphism
@VivekShende: But we don't need $ker(A[x]\rightarrow B)$ to be finitely generated; we just need a certain quotient of it to be finitely generated, no? 
Jul 20 
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Derivatives of infinite order
The axiom of choice??? 
Jul 12 
awarded  Nice Answer 