Quotient ring is Prime iff the ideal is prime ideal

Question

I do not assume a ring is commutative.

My notes say a unital ring is prime if its zero ideal is a prime ideal.

It then says that for a ring $R$ and ideal $A$ of $R$ that the quotient ring $R/A$ is prime if and only if $A$ is a prime ideal.

But I cannot understand where this comes from as it doesn’t explain this well.

Would someone explain why this is true?

Answer 1

Let $R$ be a ring.

Then let $f:R \to R/A$ definef by $f(r)=r+A$ be natural ring homomorphism, with $\text{ker}(f)=A$

Hint: (Correspondence theorem) Note that ,for any prime ideal $J \subseteq R/A$, we have $f^{-1}(J)$ prime ideal in $R$, which contains $A$.

Note that, $ f^{-1}((0)+A)=A $, where (0) is zero ideal.Note that,

$R/A$ is prime ring

Iff

$(0)+A$ is prime ideal in $R/A$

Iff

$A$ is prime ideal in $R$.

Answer 2

Let $R$ be a unital prime ring (not necessarily commutative) and $P \unlhd R$ an ideal of $R$.

Assume $R/P$ is prime. Then $\{0\}= \{P\}$ is prime in $R/P$. We show that $P$ is a prime ideal of $R$ in the following way:

Let $a,b \in R$ with $aRb \subseteq P$. We show that $a\in P$ or $b \in P$. This is rather straightforward: going to the quotient, we get

$$\overline{a} (R/P) \overline{b} = \overline{0}$$

and as $\{\overline{0}\}$ is prime, we get $\overline{a} = \overline{0}$ or $\overline{b} = \overline{0}$, which is equivalent to $a \in P$ or $b \in P$.

The converse is equally easy.