what is the dimension of the following quotient ring as a $\mathbb{C}$-vector space ?
$$\mathbb{C}\left[x,\frac{1}{x-x^2}\right]/\left(x-x^3\right)$$
How to find a basis(consisted of polynomials) of this vector space?
2026-03-27 23:48:48.1774655328
about the dimension of a quotient ring as a $\mathbb{C}$-vector space
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Notice that $(x-x^3) = \bigl((x-x^2)(x+1)\bigr) = (x+1)$ as ideals in $\mathbb C[x, \frac1{x-x^2}]$. Therefore, we have an isomorphism $$ \mathbb C\left[x,\frac1{x-x^2}\right]/(x-x^3) = \mathbb C\left[x,\frac1{x-x^2}\right]/(x+1) \cong \mathbb C $$ induced by $x\mapsto -1$ (and hence $\frac1{x-x^2}\mapsto -\frac12$). The dimension of your ring is therefore 1 and a basis is given by 1.