# Zero divisors, nilpotents and units in the ring of functions $\mathbb{R} \to \mathbb{R}$

Let $R$ be the set of all real valued functions defined for all real numbers under function addition and multiplication. i have to show that

1. all the zero divisors of $R$
2. all nilpotent elements of $R$
3. every non zero element is either a zero divisor or a unit.
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Tav, it is not considered polite here to command other users to do something. Your question does not show that you have thought about the problem. Please explain what you've tried so far, and where you are stuck. –  Zev Chonoles Aug 29 '11 at 21:25
And of course a more descriptive title would be a good idea also ;) –  Adrián Barquero Aug 29 '11 at 21:27
Hint for $1$: Suppose that $f$ is not identically $0$, but is $0$ at $x=a$. Can you think of a function $g$ such that $g$ is not identically $0$, but $fg$ is identically $0$? Remember that $g$ need not be given by a simple single "formula." –  André Nicolas Aug 29 '11 at 22:52
More hints: Suppose that $f(a) \ne 0$ for some $a$. Can $(f(a))^n = 0$? Also, building on André's hint, suppose that $f(x) \ne 0$ for all $x$. Can you find $g$ such that $f(x)g(x)=1$? –  Ilmari Karonen Aug 30 '11 at 2:33
Tav: If you add a blank after the @-sign @Zev (and Asaf) aren't notified. –  t.b. Aug 31 '11 at 10:39

The ring $R$ of all functions from $\mathbb{R}$ to $\mathbb{R}$ may be identified with the infinite Cartesian product $\prod_{x \in \mathbb{R}} \mathbb{R}$. This suggests consideration of properties of an arbitrary Cartesian product $R = \prod_{i \in I} R_i$ of commutative rings: that is, $I$ is some index set and for each $i \in I$, $R_i$ is a commutative ring. In this level of generality, it is straightforward to show:

1) An element $x \in R$ is a zero divisor iff at least one of its coordinates $x_i$ is a zero divisor in $R_i$.
2) An element $x \in R$ is nilpotent iff there exists $N \in \mathbb{Z}^+$ such that $x_i^N = 0$ for all $i \in I$. In particular, every coordinate $x_i$ of a nilpotent element is nilpotent.
3) An element $x \in R$ is a unit iff $x_i$ is a unit in $R_i$ for all $i \in I$.

In the case where each $R_i$ is a field, these observations imply that there are no nonzero nilpotent elements, and also: an element $x$ is a unit iff $x_i \neq 0$ for all $i \in I$; otherwise $x$ is a zero divisor.

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I'm a little worried about the case where $I$ has but one element. When the ring is a field, your last statement makes zero a zero divisor. Actually, I'm worried about 1), as well, in the arbitrary case. Isn't $x$ a zero divisor even if no coordinate is a zero divisor, so long as at least one coordinate is zero, and at least one coordinate isn't zero? –  Gerry Myerson Aug 31 '11 at 13:25
@Gerry: zero is a zero divisor, at least according to me. (But I think you'll find this makes things work out nicely, e.g. the answer above. The only drawback is that instead of saying "without zero divisors" you need to say "without nonzero divisors of zero".) –  Pete L. Clark Aug 31 '11 at 16:11
Pete, the other drawback to the Humpty Dumpty approach to definition ("When I use a word ... it means just what I choose it to mean") is of course that you confuse the world of Alices who haven't been initiated into your private language. Are there any other words that don't mean the same to you as to the rest of us? –  Gerry Myerson Sep 1 '11 at 5:36
@Gerry: well, first of all one could ask who is playing the role of Humpty Dumpty. My definition of a zero divisor x in a (say commutative) ring is that there exists y not equal to zero such that xy = 0. This makes zero a zero divisor in any nonzero ring. Moreover when you study commutative algebra things like "the set of all zero divisors" comes up naturally, and naturally it includes zero. Why wouldn't you want zero to be a zero divisor? Anyway, I am sure that I am not the only person who uses this terminology: when I get the chance, I'll try to look up some references. –  Pete L. Clark Sep 1 '11 at 14:29
@Gerry: okay, see the discussion here. ncatlab.org/nlab/show/zero-divisor –  Pete L. Clark Sep 1 '11 at 14:31