Polynomials of Congruent Ring Elements are Congruent

Theorem

Let $I$ be an ideal of $R$.

Let $x, y \in R$.

Let:

$x \equiv y \pmod I$

where the notation indicates congruence modulo $I$.

Let $\map F X \in R \sqbrk X$ be a polynomial in one variable over $R$.

Then:

$\ds \map F x \equiv \map F y \pmod I$

Let $\map F X = \ds \sum_{k \mathop = 0}^r a_k X^k$ where $X$ is the indeterminate and $a_0, a_1, \ldots, a_r \in R$.

It has to be shown:

$\ds \sum_{k \mathop = 0}^r a_k x^k \equiv \sum_{k \mathop = 0}^r a_k y^k \pmod I$

$\forall a, b \in R: a \equiv b \pmod I \iff a + I = b + I$

We have:

$\ds \paren {\sum_{k \mathop = 0}^r a_k x^k} + I$

$=$

$\ds \sum_{k \mathop = 0}^r \paren {a_k x^k + I}$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^r \paren {\paren{a + I} \paren{x+ I}^k}$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^r \paren {\paren{a + I} \paren{y+ I}^k}$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^r \paren {a_k y^k + I}$

$\ds $

$=$

$\ds \paren {\sum_{k \mathop = 0}^r a_k r^k} + I$

$\ds \leadsto \ \ $

$\ds \sum_{k \mathop = 0}^r a_k x^k$

$\equiv$

$\ds \sum_{k \mathop = 0}^r a_k y^k \pmod I$

$\blacksquare$