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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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    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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    "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