F is a field, prove congruence defines an equivalence relation?

[kathrynag]

Let F be a field. Given p(x) in F[x], prove that congrence modulo p(x) defines an equivalence relation on F[x].

congruence modulo p(x) means a(x)=b(x) (mod p(x)) for a(x), b(x) in F[x].
Well, I know an equivalence relation on the integers means:
a~a
if a~b then b~a
if a~b and b~c, then a~c

First show a~a
a(x)=a(x)(mod p(x))
a(x) always is equivalent to itself

Next show b~a
Assume a~b
Then a(x)=b(x)(mod p(x)
We have p(x)|(a(x)-b(x)
Then a(x)-b(x)=s(x)p(x)
S0 a(x)-s(x)p(x)=b(x)
or b(x)-a(x)=-s(x)p(x)
then b~a

Next show a~c
Assume a~b and b~c
So p(x)|a(x)-b(x) and p(x)=b(x)-c(x)
So a(x)-b(x)=p(x)s(x) and b(x)-c(x)=p(x)r(x)
We can say b(x)=p(x)r(x)+c(x)
So we have a(x)-[p(x)r(x)+c(x)]=p(x)s(x)
a(x)-c(x)=p(x)s(x)+p(x)r(x)
a(x)-c(x)=p(x)[s(x)+r(x)]
So a~c

I was just wondering if someone could look this over?

 

[Nino MMP]

I understand the concept but I'm not sure this is a correct proof, or if there are any errors or areas that need improvement.Your proof looks good! The only suggestion I have is to make it more concise by using more algebraic manipulation to plug in values from one step to another. For example, when you show that b~a, you can combine the first two equations to get b(x) = a(x) - s(x)p(x). This way, you don't have to restate the same equation over and over again.