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$f,g\in P$ is a set of polynomials with coefficients in R. $fTg$ if $f-g=c$ for $\exists c\in R$. Show T is an equivalence relation on P

I'm assuming we can show it's symmetric because if f-g=c then g-f=-c, and since c is in R then -c is in R as well.

Then f-f=0 which is in R, but I don't know how to show it's transitive

user8714896
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  • Note $ f,T, g\iff H(f) = H(g)\ $ where $H(p) = p-p(0),,$ so it's a special case of this general result on equivalence kernels (whose proof is short and obvious). – Bill Dubuque Nov 21 '19 at 07:26

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$f-g=c$ and $g-h=d$ implies $f-h =c+d$ by addition.

Kavi Rama Murthy
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