how to formally prove that $R[x]/ \langle f(x)\rangle$ is isomorphic to $R[α]$

Question

Let $R$ be a ring, let $\alpha$ be a root of the polynomial $f(x)\in R[x]$. How can I formally prove that the kernel of the evaluation map $\phi:R[x]\rightarrow R[\alpha]$ defined as $f(x)\mapsto f(\alpha)$ is $\langle f(x)\rangle$?

EDIT: the original statement is incorrect. But now how to formally prove that $R[x]/\langle f(x)\rangle$ is isomorphic to $R[α]$?

EDIT: In Artin algebra page 339, we denote the ring we get by adjointing a toot of $f$ by $R$, if the 2 rings are not isomorphic, why are we able to do that? Could someone please explain? Thanks in advance！

Answer 1

The idea that comes across here is that when we're modding out by the ideal generated by $f(x)$, the class $\overline{x}$ of $x$ in the quotient ring satisfies $f(\overline{x})=0$ because $f(\overline{x})=\overline{f(x)}=\overline{0}$. So if you define $\alpha=\overline{x}$, then $R[\alpha]$ is exactly isomorphic to $R[x]/(f(x))$, because the minimal polynomial for $\alpha$ is exactly $f$ by definition. So you may use the first isomorphism theorem with the map $x\rightarrow\alpha$, and the kernel will be exactly multiples of $f$.

Answer 2

The screenshot of that proposition is just a definition. For an abstract ring $R$, the notion $R[\alpha]$ does make any sense a priori. This is why he defines the notion $R[\alpha]$ as $R[\alpha] := R[x]/(f)$. $\alpha$ is just a symbol, namely a symbol for the residue class of $x$ in $R[x]/(f)$. The definition is however motivated from the following fact:

Let $R$ be a normal domain and $\alpha$ an element of the algebraic closure of the fraction field $K$ of $R$, which is integral over $R$. Then $R[\alpha] \cong R[x]/(f)$, where $f$ is the unique monic minimal polynomial of $\alpha$ over $K$, which happens to have coefficients in $R$.

Note that in this case, the notion $R[\alpha]$ makes sense a priori, because it ist just defined as the intersection of all subrings of $\overline K$, that contain $R$ and $\alpha$.