What is the dimension of the ring $R=\mathbb{Q}[x]/((x+1)^2)$ as a vector space over $\mathbb{Q}$?
My try:
Let $I=((x+1)^2)$. Using division algorithm we can say $R=\{(a+bx)+I~|~a, b \in \mathbb{Q} \}\cong \mathbb{Q} \times \mathbb{Q}$.
Hence $\dim_\mathbb{Q} R=2.$
Is this a right approach? Please mention if I've made any mistake. Does this method works for polynomial of degree $n$, I mean, can we say:
If $f(x) \in \mathbb{Q}[x]$ is of degree $n$, and $R:=\mathbb{Q}[x]/(f(x))$, then $\dim_\mathbb{Q} R=n.$