A commutative ring is prime if and only if it is a domain.

Question

Lemma: A commutative ring is prime if and only if it is a domain.

Remark : Let $R$ be a prime ring with $char(R) \neq 0$. We first remark that if $0 \neq a \in R$ and $n \in \mathbb{N} $ are such that $na = 0$, then $aR(nb) = 0$ for every $b \in R$, and hence $nR = 0$. Secondly, if $nR = 0$ and $n = rs$ for some $r, s \in \mathbb{N}$, then $rR.sR = 0$. Since $rR$ and $sR$ are ideals of $R$, we must have $rR = 0$ or $sR = 0$. This implies that $char(R)$ is a prime number $p$. Therefore we can consider $R$ as an algebra over $\mathbb{Z_{p}}$ in the natural way. That is, for $k \in \mathbb{Z_{p}} = \{0, 1, . . . ,p -1\}$ and $x \in R$ we define $kx$ as $x+ ........ +x$ (k times).

1: Can you give me more information about these statements?

" we can consider $R$ as an algebra over $\mathbb{Z_{p}}$ in the natural way."

and

"we define $kx$ as $x+ ........ +x$ (k times)."

2: Why is $char(R)$ a prime number $p$?

Answer 1

For 1., we can consider $R$ an algebra over $\mathbb{Z}_p$ as defined, where $kx$ = $x+\dots+x$ k times with $k \in \mathbb{Z}_p$ since the characteristic of $R$ is $p$: this way, $(k+np)x = kx$ and we can consider it an algebra over $\mathbb{Z}_p$ rather than an algebra over $\mathbb{Z}$, since we have this equality. If we didn't, we couldn't restrict to $\mathbb{Z}_p$ in the natural way. The definition is the same as for an algebra over $\mathbb{Z}$, we're just 'restricting' it in a suitable sense to $\mathbb{Z}_p$.

For 2., the characteristic is prime because we show that if we have $nR = 0$ for $n$ composite, then we can factor $n$ and show that both of its factors kill $R$, so $n$ cannot be the characteristic, and thus the characteristic cannot be composite and must be prime.

Answer 2

A ring is prime if and only if $\{0\}$ is a prime ideal, by definition. Since $\{0\}$ is a prime ideal if and only if the ring is a domain, the lemma is proved.

The second part can be dealt with in a much simpler way. Every ring $R$ has a minimum subring, precisely the image of the unique ring homomorphism $\mathbb{Z}\to R$; the kernel of this homomorphism is of the form $k\mathbb{Z}$ for a unique $k\ge0$ and $k$ is the characteristic of $R$.

If commutative prime ring (domain) $A$ has characteristic $k$, then $\mathbb{Z}/k\mathbb{Z}$ is a subring of a domain, hence a domain itself. Therefore $k=0$ or $k$ is prime.