$\mathbb{Z}/n\mathbb{Z}[T]$ and zero divisors

I read the following in wiki, but I can't understand what is meant by "divisor" there.

Notice that $(\mathbb{Z}/2\mathbb{Z})[T]/(T^{2}+1)$ is not a field since it admits a zero divisor $(T+1)^2=T^2+1=0$ (since we work in $\mathbb{Z}/2\mathbb{Z}$ where $2=0$)

I understand that if $a^2 \bmod p = 0$, then $p$ then isn't prime, but what did they mean about "divisor"?

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If your question is what do they mean by zero divisor, then an element $a$ is a zero divisor (note that this is one term), if there is some $b$ such that $ab=0$. i.e. $a$ divides zero. So the ring described has such an element, and hence is not a field. – Joe Tait Aug 22 '12 at 14:36
@Joe The definition requires $\,b\ne 0\,$ (else every element would be a zero-divisor by taking $\,b = 0\,).\ \$ – Bill Dubuque Aug 22 '12 at 15:14
indeed, thank you for pointing that out – Joe Tait Aug 22 '12 at 15:26
@tomasz I didn't mention that because some authors do employ the convention that $\,0\,$ is a zero-divisor (in contexts where this proves convenient). Further, they may assume that the reader will deduce the appropriate convention from the context. – Bill Dubuque Aug 22 '12 at 15:46
A first thing you need to be aware of is that "zero divisor" is an indivisible [sic] term, and that it is not the same thing as "divisor of $0$". If one allows the general term "divisor" to be applied to $0$ at all, then every element $a$ is a divisor of $0$, since $0$ is a multiple $0a$ of $a$; this does not mean that all elements are zero divisors. – Marc van Leeuwen Aug 22 '12 at 16:10

An nonzero element $a$ of a ring is a zero divisor if there is another nonzero element $b$ such that $ab=0$. In particular, a zero divisor can't have an inverse since otherwise $a^{-1}ab=a^{-1}*0$ hence $b=0$, a contradiction.
That being the case, in your example $a=b=T+1$ is the zero divisor hence is noninvertible and so your ring is not a field.