# Build a reduced ring starting from an ordinary one

UPDATE: After much thinking, with the help of the replies/comments below and of my algebra teacher, I've written a post blog on the matter: Relationship between reduced rings, radical ideals and nilpotent elements

This may be easier than I think, but still I can't seem to wrap my head around it.

I've learnt that if we take a ring $R$ and quotient it for a (two-sided) ideal $I \subset R$ which is radical, the resulting quotient ring is reduced, meaning it doesn't have any nilpotent elements.

First time I heard that, I thought the idea convinced me even intuitively. I thought to myself: "this makes sense, because if an element is in $I$ for some power, then its base is in $I$ as well. This way, if you had an element $a \in R$ nilpotent, then $a$ is not nilpotent anymore in the quotient, because its base becomes zero there."

However, stopping to think about it, it seems to me we are shooting a fly with a bazooka. In fact, we are not only getting rid of nilpotents in the quotient, because all elements which showed up in the ideal, with whatever power of them, are being deleted!

This has risen two questions for me:

1. Why, in the first place, a radical ideal does its job for all nilpotents? It's clear that all elements that show up in the ideal become zero in the quotient, but why would this guarantee that this gets us rid of all nilpotents is not clear to me. Why doesn't it happen that some nilpotents do indeed vanish in the quotient, but some of them don't? Is there something forcing a radical ideal to contain all nilpotent elements, or some power of them?

2. The fact that if we quotient a ring with a radical ideal is useful when we want to inspect the properties of a quotient without actually looking at it. This way, we can just look at the ideal we used to form the quotient ring. However, starting from an ordinary ring $R$, this really does not seem a good way to create a new ring which is as similar as possible to $R$ but stripped of its nilpotent elements, also keeping in mind the point before. So my question is: is there a better way to form a new ring which is close to what we started with, but doesn't have nilpotents?

I'd also like to point out that I have come up with an answer to the second point, although it's not satisfactory for me. We know that in a commutative ring, nilpotents form an ideal. We can then form a quotient ring with that ideal, and it is reasonable to believe that the quotient will comply with our request: i.e. it will not have any nilpotents and yet will not differ too much from $R$, a part from its nilpotents. Still, this is not satisfactory to me because it requires explicit knowledge of the nilpotents. It is true that we can often find a formula to describe a ring nilpotents, but this may not always be the case, so I was looking for something more general (and elegant).

• In a commutative ring nilpotents form a radical ideal. So this is indeed still consistent with what you are inquiring. For your first question, note that every ideal contains zero. So any (two-sided) radical ideal contains all nilpotent elements. Jun 29, 2016 at 22:21

The fact that $I$ is a radical ideal is enough to ensure $R/I$ has no nilpotents. For suppose $a^n+I=0+I=I$, so that $a^n\in I$. Then since $I$ is radical we must have $a\in I$, so we must have had $a+I=0+I$, hence $a$ was zero in $R/I$ in the first place. The maximal quotient without nilpotents is the quotient by the radical of the zero ideal, in the sense that every quotient that does not have nilpotent elements is a quotient of $R/\mathrm{Rad}\{0\}$. This ideal happens to consist solely of nilpotent elements in a commutative ring, so it seems you've done the best you can do.
• Thanks for the reply. I still have a doubt though, sparked from your reply. In a commutative ring, the ideal $I$ formed by nilpotents and a general radical ideal $J$ are not the same ring, right? My thought is that, as long as $J$ contains Rad{0}, you can build a radical ideal as big as you want, in particular bigger than $I$. Is this correct? Aug 5, 2016 at 13:02
• @Ste_95 every radical ideal $J$ contains the radical of the zero ideal. $J/\text{Rad}\{0\}$ is a radical ideal in $R/\text{Rad}\{0\}$, and $R/J$ is isomorphic to $(R/\text{Rad}\{0\})/(J/\text{Rad}\{0\})$ by the third isomorphism theorem. Aug 6, 2016 at 12:17