How do I show $(f_1(x), f_2(x), ..., x-a) = (f_1(a), ..., f_r(a), x-a)$ where $a \in R$, $R$ is a commutative ring, $f_i(x) \in R[x]$

Question

Let $R$ be a commutative ring, $a \in R$, and $\forall i = 1, ...,r \ \ f_i(x) \in R[x]$.

Prove the equality of ideals

$(f_1(x), ..., f_r(x), x-a ) = (f_1(a), ...f_r(a), x-a)$.

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We want to show that $$\forall g_1(x), ..., g_{r+1}(x) \in R[x] \ \ \exists y_1(x), ..., y_{r+1}(x) \in R[x]: \ \ g_1(x)f_1(x) + ... + g_r(x)f_r(x) + g_{r+1}(x)(x-a) = y_1(x)f_1(a) + ... + y_r(x)f_r(a) + y_{r+1}(x)(x-a)$$ and vice versa.

Answer 1

$(f_i(x)-f_i(a))(a)=0$. Write $f_i(x)-f_i(a)=q(x)(x-a)+c$ where $c\in R$, you have $f_i(a)-f_i(a)=q(a)(a-a)+c=0$, implies that $c=0$, thus $f_i(x)\in (f_1(a),...,f_r(a),x-a)$.

On the other hand, $f_i(x)=q(x)(x-a)+c$ this implies that $c=f_i(a)=f_i(x)-q(x)(x-a)$ thus $f_i(a)\in (f_1(x),...,f_r(x),x-a)$.

Answer 2

as the coefficient dominant of $x-a$ is 1 ivertible in R the euclidien division is ok for any polynomial by $x-a$. so $f_i(x)=g(x)+b $ for some constante $b$ the equality with x=a come $f(a)=b$ and then as reauired the ideal $(f_1(x),...,f_r(x),(x-a))$ is the ideal $(f_1(a),...,f_r(x),(x-a))$