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A ring $A$ is principal if every ideal in it is of the form $Ax$. $\mathbb{Z}$ is said to be a principal ideal, but it seems to me that a set $I = \{z \in \mathbb{Z} : |z| > n\} \cup \{0\}$ for some $n > 1$ is an ideal and cannot be written as $\mathbb{Z}x$. Where am I wrong?

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What is your definition of ideal? –  lhf May 11 '12 at 15:52
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up vote 5 down vote accepted

The set is not an ideal; if $n=2$ then $4,-3\in I$, but $4-3=1\notin I$.

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Thanks, I forgot ideals had to be groups.. –  Karolis Juodelė May 11 '12 at 16:05
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Hint $ $ Nonzero ideals $\rm I\subset \mathbb Z$ are generated by their least positive element $\rm\:n,\:$ therefore $\rm\:I = n\mathbb Z.\:$ Hence if $\rm\:n>1\:$ then $\rm\:1+n\not\in I,\:$ since otherwise $\rm\:1 = (1+n)-n\in I\:$ contra minimality of $\rm\:n.\:$ Generally, the elements in $\rm I$ differ by a multiple of $\rm n,\:$ and $\rm n>1$ is the least difference, in other words $\rm n$ is the "unit" in the scaled "number line" module $\rm n\mathbb Z = \{\ldots,-2n,\:-n,\:0,\:n,\:2n,\:\!\ldots\}.$

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