localization of quotient of a polynomial ring

Question

Let $A=k[x]$ where $k$ is a field, $f=x(x-1)$, $g=x(x^2+1)$ and $S=\{x^n; n\geq 1\}$. Is it correct that $S^{-1}(A/(f))\cong k$ and $S^{-1}(A/(g))\cong \mathbb C$ (where $k=\mathbb R$ in the last case)?

Answer 1

$S^{-1} (k[x]/x(x-1)) \simeq (S^{-1} k[x])/x(x-1)$

since $x$ is now invertible $(S^{-1} k[x])/x(x-1) \simeq (S^{-1} k[x])/(x-1) \simeq k$

$S^{-1} (\mathbb{R}[x]/x(x^2+1)) \simeq \mathbb{R}[x]/(x^2+1) \simeq \mathbb{C}$

by the same reasons.

Another approach: $S^{-1} (k[x]/(x*f)) = S^{-1} (k[x]/x \oplus k[x]/f) = S^{-1} k[x]/x \oplus S^{-1} k[x]/f$ and $S^{-1} k[x]/x = 0$ since $x$ is now invertible.

Answer 2

Yes. To prove it, you can use the Chinese remainder theorem: $$k[x]/\bigl(x(x-1)\bigr)\simeq k[x]/(x)\times k[x]/(x-1), $$ $$\text{so }\quad S^{-1}\Bigl(k[x]/\bigl(x(x-1)\bigr)\Bigr)\simeq S^{-1}\bigl(k[x]/(x)\bigr)\times S^{-1}\bigl(k[x]/(x-1)\bigr)\simeq \{0\}\times S^{-1}k \simeq \{0\}\times k. $$ Similarly for the second case.