# Prove that the ring $\mathbb Z$ [x] (ring of all polynomials with integer coefficients) has unity.

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).

• "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. Apr 10, 2019 at 16:27
• What definition of $\,\Bbb Z[x]\,$ are you using where this isn't immediate? Apr 10, 2019 at 16:32
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: