Questions about a sentence concerning rings and monoids

Question

I have a first question concerning this sentence in my book.

The ring is unital if, in addition, $(R\setminus\{0\}, \times)$ is a monoid.

I do not understand why we retrieve $0$, that is the additive identity, since for a monoid we do not have any problem at my knowledge by keeping $0$ ?

My second question is concerning the annihilator property of the additive identity $0$ in a ring, if the ring is not unital, that is it is not a monoid for the multiplicative operation, I do not see how we can prove that this property is still true: I suspect it is no more the case (since in my proof I use the multiplicative identity). Am I right ?

Thank you a lot for your help.

Answer 1

The intent of this definition is to disallow $0$ from being the multiplicative unit. However, this is actually not correct. The zero ring is unital but $0 = 1$ in that ring (and it is the unique ring in which $0 = 1$).

Edit: Actually this definition is even incorrect for a second reason: it requires that the product of two nonzero elements is nonzero, so it forbids zero divisors. This is defining a domain, not a unital ring.

I don't know what "this property" is referring to in your second question.

Answer 2

Perhaps the intent was to later draw a parallel about fields being such that $(R,+)$ and $(R\setminus\{0\},\times)$ are both abelian groups. Considering the case when $(R\setminus\{0\},\times)$ is a monoid is a natural relaxation of that, but as Qiaochu pointed out, $R$ has to be a domain for the binary operation to work right.

Of course, $(R,\times)$ is always a monoid (with absorbing element $0$ and identity that of $R$, if it exists.)

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My second question is concerning the annihilator property of the additive identity 0 in a ring, if the ring is not unital, that is it is not a monoid for the multiplicative operation, I do not see how we can prove that this property is still true:

The annihilator property does not depend on the ring having a multiplicative identity at all, just on the additive group structure and distributivity. See, if you have any element $r\in R$, then $r0=r(r-r)=r^2-r^2=0$ (or any other convenient variation thereof.) Not knowing what your proof is, I do not know why you feel the identity is indispensable.