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I know that the existence of unity implies non existence of zero divisor.

Is the converse true? (because I have prooved the latter - its quite simple).

If not; how do I go about this?

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    $\begingroup$ "existence of unity implies non existence of zero divisor." Not true at all. The ring $\Bbb Z_4$ has both unity $1$ and a zero divisor $2$. And $2\Bbb Z$ has neither. So there is no relation (at least not that simple) between the existence of unity and the existence of zero divisors. $\endgroup$
    – Arthur
    Apr 10, 2019 at 16:27
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    $\begingroup$ What definition of $\,\Bbb Z[x]\,$ are you using where this isn't immediate? $\endgroup$ Apr 10, 2019 at 16:32

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The constant polynomial $1=1\cdot x^0$ is a unit. The ring is certainly an integral domain, being a subring of the PID $\Bbb Q[x]$. Also, $R[x]$ is an integral domain if and only if $R$ is an integral domain, see this duplicate:

Prove that $R[x]$ is an integral domain if and only if $R$ is an integral domain.

Since $\Bbb Z$ is an integral domain, also $\Bbb Z[x]$ is. Also the units of $\Bbb Z[x]$ are the units of $\Bbb Z$. For this, see here:

What are the units of Z[x]?

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