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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.2011-08-29
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    And of course a more descriptive title would be a good idea also ;)2011-08-29
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    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."2011-08-29
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    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 $f(x)g(x)=1$?2011-08-30
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    Tav: If you add a blank after the `@`-sign @Zev (and Asaf) aren't notified.2011-08-31
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    @Zev I have edited the question now. Sorry to come off as impolite...no intention whatsoever.2011-08-31
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    @Asaf Karagila: Thank you !! I was not aware of this.2011-08-31
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    @Theo Buhler: Thanks :)2011-08-31
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    I'm unclear on your question. You say that "you have to show" and then just say "all the zero divisors of $R$". Normally, "I have to show that all the zero divisors of $R$" would be followed by a clause saying what it is you have to show *about* the zero divisors of $R$. Did you mean, you have to exhibit all zero divisors? You have to *characterize*/*describe* all zero divisors?2011-08-31

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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?2011-08-31
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    @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".)2011-08-31
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    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?2011-09-01
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    @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.2011-09-01
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    @Gerry: okay, see the discussion here. http://ncatlab.org/nlab/show/zero-divisor2011-09-01
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    @Gerry: to answer your last question: unfortunately the answer is probably yes, in ways that I may not be aware of. But I try not to do it intentionally, and I don't do it lightly...2011-09-01
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    I see no discussion at nlab, just an assertion, and the remarkable "Some authorities will differ on this point." I'd say most authorities will differ on this point. Lang, Algebra, page 61: Elements $x,y$ of $A$ are said to be *zero divisors* if $x\ne0$, $y\ne0$, and $xy=0$. Herstein, Topics In Algebra, page 88: If $R$ is a commutative ring then $a\ne0\in R$ is said to be a *zero-divisor* if there exists a $b\in R$, $b\ne0$, such that $ab=0$. Also Birkhoff and Mac Lane, A Survey of Modern Algebra, p. 6; Hungerford, Algebra, p. 116; Fraleigh, 7th ed., p. 178; etc., etc. I found one exception:2011-09-02
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    (cont.) Jacobson, Basic Algebra I, accepts zero as a zero-divisor.2011-09-02
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    @Gerry: that's a big exception. I just checked the three most standard commutative algebra texts -- Atiyah-Macdonald, Eisenbud and Matsumura -- and two out of the three allow zero as a zero divisor. But honestly, I don't think this is such a big deal: I taught a semester course on commutative algebra last spring and I don't even remember explicitly giving the definition of zero divisor, let alone any confusion arising from it. (But I did just search my own commutative algebra notes, and I do explicitly say that $0$ is a zero divisor.) What happens when Humpty Dumpty self-publishes?2011-09-02
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    Also Kaplansky's *Commutative Rings* on p. 3 says that the set of zero divisors is a union of prime ideals, which implies that $0$ is a zero divisor. I could probably continue, but I'm not sure there's profit in it. In summary: no, I did not come up with this definition myself; I followed several standard texts (and did not follow many other standard texts...). Perhaps my definition is more common in "commutative algebra" than "algebra"...and this question is tagged **commutative algebra**. (Anyway, I think by now you can see that calling this my "private language" is a little unfair.)2011-09-02
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    "Unfair" is my middle name...seriously, though, I grant your many points, and agree that my wording was unnecessarily combative. I guess I'll find no glory here (in the Humpty Dumpty sense of the word).2011-09-02