Quotient of ring by radical ideal.

Question

Let $R$ be a commutative ring with identity, and let $I$ be an ideal of $R$. Let $J$ be the radical of $I.$

What can we say about the quotient ring $R/J?$ Does it have any special properties? In particular, if $J$ is the nilradical of $R,$ what can we say about $R/J?$

I couldn't find any special properties of this. Please help me with this.

Answer 1

Consider a commutative unital ring $R$ with a proper ideal $I.$ Consider the ideal $$J = \operatorname{rad}(I) = \{r \in R \,|\, r^n \in I \text{ for some integer } n \geq 1 \}.$$ Observe that in $R / J,$ the nilpotent elements are precisely the cosets whose representatives belong to $J.$ Explicitly, we have that $r + J$ is nilpotent if and only if $r^n + J = (r + J)^n = 0 + J$ for some integer $n \geq 1$ if and only if $r^n$ is in $J$ for some integer $n \geq 1$ if and only if $r \in J.$ Consequently, the nilradical of $R / J$ is precisely the set of elements in $J,$ i.e., the only nilpotent element of $R / J$ is the coset $0 + J.$ Consequently, the ring $R / J$ is reduced, i.e., there are no nonzero nilpotent elements in $R / J.$

On the other hand, the nilradical of a commutative unital ring $S$ is the intersection of all prime ideals in $S.$ Considering that every proper ideal (and hence every prime ideal) is contained in a minimal prime ideal, and every prime ideal contains a minimal prime ideal, it follows that the nilradical of $S$ is the intersection of all minimal prime ideals of $S,$ i.e., $\operatorname{nil}(S) = \bigcap_{P \in \operatorname{MinSpec}(S)} P,$ where we define$$\operatorname{MinSpec}(S) = \{P \,|\, P \text{ is a minimal prime ideal of } S \}.$$ By the previous paragraph, we have that $J = \bigcap_{P \in \operatorname{MinSpec}(R) \,|\, P \supseteq J} P = \bigcap_{P \in \operatorname{MinSpec}(R) \,|\, P \supseteq I} P.$

Answer 2

Of course, even for noncommutative rings, the radical of $J$ is a semiprime ideal. (The semiprime ideals are exactly the intersections of prime ideals. In particular the Jacobson radical is a semiprime ideal, and in a commutative ring the radical of any ideal is semiprime.)

That makes $R/J$ a semiprime ring, that is, it has no nonzero nilpotent ideals.

For commutative rings, this is equivalent to having no nonzero nilpotent elements. This condition describes a reduced ring.

Answer 3

One more related fact for commutative rings with identity proved here:

An ideal $J$ is a radical ideal if and only if the quotient ring $R/J$ is a reduced ring.