Solve for $p(x-1) = x+1$ in $\mathbb C[x]/(x^3-x^2)$ (or give software to help)

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I'm wondering if there exists any reasonable software for doing computations with (polynomial) rings, similar to GAP for groups. This question seems to have asked for something similar, but it went unanswered.

The particular problem that I need solving, if anyone can do it independently of the above request, is figuring out which polynomials $p$ makes the equation $p(x-1) = x+1$ true in $R = \mathbb C[x]/(x^3-x^2)$.

I believe $p$ exists by the following reasoning: The only prime ideals of $R$ are $(x)$ and $(x-1)$ (in particular, these are the prime ideals of $\mathbb C[x]$ containing $x^3 - x^2$). We have that $(x+1)(x-1)x^2 = x^4 - x^2 = 0$, so $x+1$ is a zero-divisor, so it isn't a unit, so it's in one of the prime ideals. It's not in $(x)$ since it's clear that there are no polynomials with non-zero constant terms in this ideal. Thus $x+1 \in (x-1)$.

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I don't think it is possible. The element $x+1$ is a unit, where $x-1$ is not (you showed that $x-1$ is a prime element which cannot be a unit). To show $(1+x)p = 1$ mod $(x^2-x^3)$, take $p = 1 - x + \frac12x^2$. We have $$ (1+x)(1 - x + \frac12x^2) = 1 - x + \frac12x^2 + x - x^2 + \frac12x^3 = 1 $$ mod $(x^2-x^3)$ since $\frac12x^3 = \frac12x^2$ in $R$.

In your calculation $(x+1)(x-1)x^2 = x^4 - x^2 = 0$, notice that $(x-1)x^2 = x^3-x^2$ is already zero in $R$.