Polynomial quotient rings

Question

I have a quotient ring $R=\Bbb Z[t]/(1-t)^3$.

It is asked to show that $\overline{2t^3-2}=\overline{6t^2-6t}$ and $\overline{1-4t^3}=\overline{4t-6t^2-t^4}$.

I feel like I am missing some important idea, this is how I proceeded.

$(1-t)^3=(1-3t+3t^2-t^3)$ is the neutral element here, this should be an ideal of $R$.

so I am tempted to write $\overline{2t^3-2}=\overline{2t^3-2+2*(1-3t+3t^2-t^3)}=\overline{6t^2-6t+1}$ which produces the wrong result.

I fail to see where I went wrong.

$\Bbb Z[t]=\{a+bt|a,b\in\Bbb Z\}$ so I am thinking of the equivalent classes in $R=\Bbb Z[t]/(1-t)^3$ as the residue classes if I divided by $(1-t)^3$.

Is it because I am thinking of $(1-t)^3=(1-3t+3t^2-t^3)$ as $0$, but thinking of it as identity element makes no sense for me.

Answer 1

The first one is true, writing

$$(t-1)^3=0\mod (t-1)^3\iff t^3-3t^2+3t-1=0\mod (t-1)^3$$ $$\iff 3t^2-3t=t^3-1\mod (t-1)^3$$

But then we see that $2(3t^2-3t)=6t^2-6t= 2(t^3-1)=2t^2-2\mod (t-1)^3$.

For the other one, use $a=b\mod I\iff a-b\in I$, so that

$$(1-4t^3)-(4t-6t^2-t^3)=1-4t+6t^2-3t^3$$

however, for this to be $=0\mod (t-1)^3$ we need $3t^3-6t^2+4t-1=p(t)(t-1)^3$. But then necessarily $\deg p(t)=0$ and indeed $p(t)=3$ by considering just the first coefficient, again impossible.

Answer 2

$\begin{eqnarray}{\bf Hint}\ \ &&g(t)\equiv h(t)\ \ {\rm in}\ \ \Bbb Z[t]\ {\rm mod}\ (t\!-\!1)^3\\ \iff && (t\!-\!1)^3\!\mid g(t)\!-\!h(t) =: f(t)\ \ {\rm in}\ \ \Bbb Z[t]\\ \iff && 0 = f(1) = f'(1) = f''(1)\ \ \text{by Taylor series expansion at}\,\ t = 1 \end{eqnarray}$

Example $ $ your $\,f = 2t^3\!-\!6t^2\!+6t\!-\!2,\ f'\! = 6t^2\!-\!12t\!+\!6,\ f''\! = 12t\!-\!2\ $ so we easily compute that $\, f(1) = f'(1)=f''(1) = $ coefficient sum $= 0\,$ for all three polynomials.