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Let $\frac{R}{P}$ be a domain ( not necessarily commutative ). Show that $P$ is prime.

In previous questions asked on this site, integral domain ( commutative ) is assumed, but I don't think this is necessary.

My attempt, using a hint given involving the correspondence theorem for rings.

Let $ \pi : R \rightarrow \frac{R}{P}$ be the canonical surjection. The kernel of this is $P$ and the image is $\frac{R}{P}$. Since this is also a homomorphism, there is a one-to-one correspondence between the ideals of $R$ that contain $P$, and the ideals of $\frac{R}{P}$.

So to show that $P$ is prime, let $A$ and $B$ be ideals of $R$ and suppose $AB \subseteq P$. Suppose that $A$ is not contained in $P$, must show $B$ is contained in $P$.

So far I've been using this hint, but not sure if I'm onto something.

Could I try this? Since $A$ and $B$ are ideals of $R$, $\frac{A}{P}$ and $\frac{B}{P}$ are ideals of $\frac{R}{P}$. Then $\frac{A}{P} \cdot \frac{B}{P} \subseteq \frac{P}{P} = 0 \implies \frac{A}{P} \cdot \frac{B}{P} = 0$. Since $\frac{R}{P}$ is a domain, and $\frac{A}{P}$ is non-zero, it follows $\frac{B}{P} = 0 \implies B \subseteq P$.

turnip_man
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  • $$\begin{align} a,b\not\in P&\Rightarrow ab\not\in P\ \iff\ a,b\not\equiv 0\ &\Rightarrow ab\not\equiv 0!!!\pmod{!~!!P}\end{align}\qquad\qquad$$ – Bill Dubuque Jul 14 '22 at 14:01

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You can do it easier. Suppose there are some $a\in A\setminus P$ and $b\in B\setminus P$. Then the equality $(a+P)(b+P)=P$ in $R/P$ contradicts the assumption that $R/P$ is a domain.

Mark
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