$\frac{R[x]}{(f_1(x),\ldots,f_r(x),x - a)} \cong \frac{R}{(f_1(a),\ldots,f_r(a))}$ I am solving the following problem from Aluffi Algebra. Chapter 0.

Let $R$ be a commutative ring, $a \in R$, and $f_1(x),\ldots, f_r(x) \in R[x]$. Prove the following equality of ideals

*

*$(f_1(x),\ldots, f_r(x),x - a) = (f_1(a),\ldots,f_r(a),x - a).$

*Prove the following substitution trick $\frac{R[x]}{(f_1(x),\ldots,f_r(x),x - a)} \cong \frac{R}{(f_1(a),\ldots,f_r(a))}$.


I was able to prove (1). Clearly $(f_1(a),\ldots,f_r(a),x - a) \subset (f_1(x),\ldots, f_r(x),x - a)$. Conversely let $h(x) \in (f_1(x),\ldots,f_r(x),x - a)$.
It follows that
$h = s_1(x)f_1(x) + \ldots + s_r(x)f_r(x) + g(x)(x - a) = (x - a)u_1(x) + f_1(a) + (x - a)u_2(x) + f_2(a) + \ldots + (x - a)u_r(x) + f_r(a) + g(x)(x - a) \in (f_1(a), \ldots, f_r(a), x - a)$.
I am having few issues with problem 2. I think I have the idea.
$$\frac{R[x]}{(f_1(x),\ldots,f_r(x),x - a)} = \frac{R[x]}{(f_1(a),\ldots,f_r(a), x - a)} = \frac{R[x] / (x - a)}{(f_1(a),\ldots,f_r(a))} = \frac{R}{(f_1(a),\ldots,f_r(a))}$$
Why is it true that  $\frac{R[x]}{(f_1(a),\ldots,f_r(a), x - a)} = \frac{R[x] / (x - a)}{(f_1(a),\ldots,f_r(a))}$ can someone explain this ?
 A: There are a few ways to see this. Since you're reading Aluffi, you can verify it by checking that the two rings have the same universal property. We'll do this by unwinding the universal properties of quotients and polynomial rings that go into building them up.
In particular, a map $\frac{R[x]}{(f_1(a), \ldots, f_r(a), x-a)} \to S$ is the same thing as a map $R \to S$, plus a ~ bonus element ~ $s \in S$  (which is where we'll send $x$), so that the elements $f_k(a)$ and $x-a$ get sent to $0$ (do you see why?).
But a map $\frac{R[x] / (x-a)}{(f_1(a), \ldots, f_2(a)} \to S$ is a map $R[x]/(x-a) \to S$ so that the elements $f_k(a)$ get sent to $0$. This leads us to ask "what's a map $R[x] / (x-a) \to S$"? And we find it's a map from $R \to S$, plus a ~ bonus element ~ where we send $x$, so that $x-a$ gets sent to $0$.
Can you massage these descriptions until they look the same? Then, since these two rings have the same maps out of them, can you conclude they must be isomorphic? (Hint: yoneda).

Edit:
To clarify the yoneda hint, recall the functor $R \mapsto \text{Hom}(R,-)$ is a full and faithful embedding.
This means that isomorphisms of presheaves $\text{Hom}(R_1,-) \cong \text{Hom}(R_2,-)$ must come from isomorphisms of rings $R_1 \cong R_2$. This is useful because oftentimes it's easier to show two presheaves are isomorphic than to directly check the rings are.
But how do we show $\text{Hom}(R_1, - ) \cong \text{Hom}(R_2, -)$? There are lots of high powered tools for doing this, but for this problem we can use the regular old definition:

$\text{Hom}(R_1, -) \cong \text{Hom}(R_2,-)$ if and only if for every ring $S$ we have a bijection of sets $\text{Hom}(R_1,S) \cong \text{Hom}(R_2,S)$ which is natural in $S$.

So to show $R_1 \cong R_2$, it suffices to show $\text{Hom}(R_1,-) \cong \text{Hom}(R_2,-)$. But to do that it suffices to show that for every $S$ there is a (natural) bijection between maps $R_1 \to S$ and maps $R_2 \to S$.
"Naturality" has a technical definition, which I'm sure you know from reading Aluffi. But on a practical level it means that your argument doesn't care about $S$.
(There's an analogy here with programming language theory, where natural transformations agree with "polymorphic functions", but that's a discussion for another time).
So for us, we showed a bijection between maps $R_1 \to S$ and maps $R_2 \to S$. Moreover, our argument worked the same for every $S$ -- we didn't have to do any casework. Said another way, we showed $\text{Hom}(R_1, S) \cong \text{Hom}(R_2, S)$ naturally in $S$. But this is the definition of $\text{Hom}(R_1, -) \cong \text{Hom}(R_2, -)$ as functors! Then, applying yoneda, we conclude that $R_1 \cong R_2$ as rings.

I hope this helps ^_^
