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Problem: Find the zero divisors and the units of the quotient ring $\mathbb Z[X]/\langle X^3 \rangle$.

If $a \in \mathbb Z[X]/ \langle X^3 \rangle$ is a zero divisor, then there is $b \neq 0_I$ such that $ab=0_I$. I think that the elements $a=X+ \langle X^3 \rangle$ and $b=X^2+ \langle X^3 \rangle$ are zero divisors because we have:

$$ab=XX^2+ \langle X^3 \rangle =X^3+ \langle X^3 \rangle = \langle X^3 \rangle.$$

I couldn't think of any other divisors so I suspect these two are the only ones. Am I correct? If that is the case, how could I show these are the only zero divisors?

As for the units I don't know what to do. Any suggestions would be appreciated.

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Suppose $ab = 0$ in $\mathbb{Z}[X]/(X^3)$. Then $X^3 \mid ab$ in $\mathbb{Z}[X]$. If $X \nmid a$, then $X^3 \mid b \implies b = 0$ in $\mathbb{Z}[X]/(X^3)$. Thus every zerodivisor is a multiple of $X$, i.e. is of the form $cX$.

If $a = a_0 + a_1X + a_2X^2$, $b = b_0 + b_1X + b_2X^2$, then $ab$ has constant term $a_0b_0$, which equals $1$ iff $a_0, b_0 = \pm 1$. Thus every unit has constant term $\pm 1$, i.e. is of the form $\pm 1 + cX$.

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    $\begingroup$ To see that $\pm 1 + cX$ is indeed a unit, note that $X$ is nilpotent, and unit $+$ nilpotent $=$ unit $\endgroup$ – zcn Sep 14 '14 at 6:26
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About zero divisors, you're wrong. What about $2X + 3X^2$, for example?

Concerning the units, why don't you write $f(x) = a + bx + cx^2$, and $g(x) = A + Bx + Cx^2$, and see what has to happen in order for the product $f(x)g(x)$ to reduce to $1$ after you mod out by the ideal generated by $X^3$.

You can actually try doing the same sort of thing to find the zero divisors.

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You are asking for which $a \in \mathbb{Z}[X]$ there exists $b \in \mathbb{Z}[X]$ such that $X^3 | ab$ but not $X^3 | b$.

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