Exercise III.4.11 (Aluffi): Let $R$ be a commutative ring, $a \in R$ and $f_1(x),...,f_r(x) \in R[x]$.
- Prove the equality of ideals \begin{equation} (f_1(x),...,f_r(x),x-a) = (f_1(a),...,f_r(a),x-a). \end{equation}
- Prove the useful substitution trick \begin{equation} \frac{R[x]}{(f_1(x),...,f_r(x),x-a)} \cong \frac{R}{(f_1(a),...,f_r(a))}. \end{equation}
So here is my current logic, I'm having trouble formulating my ideas/want to make sure I'm going down the correct path.
For the first equality it is clear that \begin{equation} (f_1(x),...,f_r(x), x-a) = (f_1(x)) + \dots (f_r(x)) + (x-a) \end{equation} and \begin{equation} (f_1(a),...,f_r(a),x-a) = (f_1(a)) + \dots (f_r(a)) + (x-a). \end{equation} Thus, my "modding out" by the ideal $(x-a)$ it would follow that equality holds correct? As all that is left are the polynomials generated by $f_1(x),...,f_r(x)$ where $f(a) \neq 0$. Is this logic correct?
Second, assuming the first equality holds. Recall that \begin{equation} \frac{R[x]}{(x-a)} \cong R. \end{equation} Thus I feel like the substitution trick should follow quickly after consideration of this fact as well as the first equality. However, I'm a little lost with regards to formulating a concrete proof.
Any hints/help will be greatly appreciated!
Here are some hints:
For the first equality, you're on the right track! Recall the correspondence theorem for Rings (page 142 of Aluffi). In particular, for ideals $I$ and $J$ containing $\mathfrak{a}$, we have $I=J$ if and only if $I / \mathfrak{a} = J / \mathfrak{a}$ in $R/\mathfrak{a}$.
Now if we consider $I = \langle f_i(x) \rangle$, $J = \langle f_i(a) \rangle$, and $\mathfrak{a} = \langle x-a \rangle$, we will have
$$ \langle f_i(x), x-a \rangle = \langle f_i(a), x-a \rangle \iff I + \mathfrak{a} = J + \mathfrak{a} \iff I/\mathfrak{a} = J/\mathfrak{a} $$
Of course, as you have noted, it is clear that $I/\mathfrak{a} = J/\mathfrak{a}$ since $x=a$ in the quotient ring, so $f_i(x) = f_i(a)$.
For the second equality, you are also on the right track. Notice:
$$ R[x]/(I + \mathfrak{a}) \cong (R[x]/\mathfrak{a})/(I/\mathfrak{a}) \cong (R[x]/\mathfrak{a})/(J/\mathfrak{a}) \cong R/(J/\mathfrak{a}) $$
Of course, it is easily checked that $J/\mathfrak{a} = \langle f_i(a) \rangle$.
Feel free to comment if there's anything you want me to clarify!
I hope this helps ^_^