$\begin{align}&(x\! -\! a,f_1(x),\ldots,f_r(x))\\ =\ &(x\! -\! a,f_1(a),\ldots,f_r(a))\end{align}\ $ [Ideal evaluation = mod reduction]

Question

$\begin{align}&(x\! -\! a,f_1(x),\ldots,f_r(x))\\[.1em] =\ &(x\! -\! a,f_1(a),\ldots,f_r(a))\end{align}\ $ [Ideal evaluation = mod reduction]

I am solving Aluffi chapter 0. I am completely stuck on this question.

Let R be a commutative ring, $a \in R$, and $f_1(x),\ldots, f_r(x) \in R[x]$.

Prove the equality of ideals

1. $(f_1(x),\ldots,f_r(x),x - a) = (f_1(a),\ldots,f_r(a),x - a)$
2. 
Prove the useful substitution trick $\frac{R[x]}{(f_1(x),\ldots,f_r(x), x - a)} \cong \frac{R}{(f_1(a),\ldots,f_r(a))}$

Answer 1

Hint:

You can divide each polynomial by $x-a$, obtaining $$f_k(x)=q_k(x)(x-a)+f_k(a),\quad k=1,\dots,r.$$

Answer 2

Hint: as in Euclidean algorithm for gcds, ideals are preserved by: mod out a generator by another

$$x\!-\!a,\, f \in I \iff x\!-\!a,\!\!\!\!\! \overbrace{f\,-\, q\, (x-a)}^{\textstyle f(a) = f\bmod x\!-a}\!\!\!\!\!\in I $$

where we used Polynomial Remainder Theorem, i.e. $\,f(a) = f\bmod x\!-\!a\,$ .

Remark $ $ More generally ideals are prserved by any unimodular transformations of generators ("change of basis")