1. ## Congruence uncertainty

Suppose f(x)and g(x)∈Z_7 [x]. Suppose also that f(x)≡g(x)(mod x). What can we say about f(0)and g(0)?

With this problem i am jus not sure if i am over thinking it because we have done congruences with ideals and in F[x].. would fall under just dealing with F[x]

2. Originally Posted by empressA88
Suppose f(x)and g(x)∈Z_7 [x]. Suppose also that f(x)≡g(x)(mod x). What can we say about f(0)and g(0)?

With this problem i am jus not sure if i am over thinking it because we have done congruences with ideals and in F[x].. would fall under just dealing with F[x]
f and g are 7 th degree polynomials in in x. Saying that they are congruent "mod x" f- g is a multiple x which, in turn, means f- g has no constant term.

3. is that f - g, or are you say f and g?

4. Originally Posted by HallsofIvy
f and g are 7 th degree polynomials in in x. Saying that they are congruent "mod x" f- g is a multiple x which, in turn, means f- g has no constant term.
Doesn't $f,g\in\mathbb{Z}_7[x]$ mean $f$ and $g$ are polynomials with coefficients in $\mathbb{Z}_7$?

If so then $f\equiv g\bmod{x} \implies x\mid f-g\implies$ the constant term of $f-g\equiv0\bmod{7}$.

But this constant term is equal to $f(0)-g(0)$. Therefore $f(0)\equiv g(0) \bmod{7}$.