Polynomials of Congruent Ring Elements are Congruent
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Theorem
Let $R$ be a commutative ring with unity.
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:
- $\ds \map F x \equiv \map F y \pmod I$
Proof
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:
- $\ds \sum_{k \mathop = 0}^r a_k x^k \equiv \sum_{k \mathop = 0}^r a_k y^k \pmod I$
From Left Cosets are Equal iff Product with Inverse in Subgroup:
- $\forall a, b \in R: a \equiv b \pmod I \iff a + I = b + I$
We have:
\(\ds \paren {\sum_{k \mathop = 0}^r a_k x^k} + I\) | \(=\) | \(\ds \sum_{k \mathop = 0}^r \paren {a_k x^k + I}\) | Quotient Ring Addition is Well-Defined | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^r \paren {\paren{a + I} \paren{x+ I}^k}\) | Quotient Ring Product is Well-Defined | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^r \paren {\paren{a + I} \paren{y+ I}^k}\) | Left Cosets are Equal iff Product with Inverse in Subgroup | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^r \paren {a_k y^k + I}\) | Quotient Ring Product is Well-Defined | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\sum_{k \mathop = 0}^r a_k r^k} + I\) | Quotient Ring Addition is Well-Defined | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{k \mathop = 0}^r a_k x^k\) | \(\equiv\) | \(\ds \sum_{k \mathop = 0}^r a_k y^k \pmod I\) | Left Cosets are Equal iff Product with Inverse in Subgroup |
$\blacksquare$