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I know that a subring of a PID isn't necessarily a PID, but what about ideals?

Suppose $R$ is a PID, and $I$ is its ideal. Now take $J$ to be an ideal of $I$ (note that $J$ isn't necessarily an ideal of $R$), is $J$ necessarily generated by one element?

Sorry about the edit.

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No One
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    A domain is usually defined to have a unity. – MooS Mar 20 '17 at 19:28
  • Isn't a PID an unitary ring? – Xam Mar 20 '17 at 19:30
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    But $I$ is not even a ring as it does not necessarily contain a $1$. Even if you belong to the faction that allows rings without unity, then consider the following. If $I$ were a principal ideal of $I$ then we would need $I=Ia$ for some $a\in I$. This implies that $a=ia$ for some $i\in I$. This leads to the equation $a(1-i)=0$ in $R$. As $R$ is a domain we need either $a=0$ or $1=i\in I$. Neither alternative looks good :-) – Jyrki Lahtonen Mar 20 '17 at 19:30
  • Well, let us only consider the case when rings don't include $1$ whenever it doesn't have to...... – No One Mar 20 '17 at 19:35
  • Hmm. Above I equated "principal ideal" with "cyclic submodule". May be that's not how you define a principal ideal in a rng? – Jyrki Lahtonen Mar 20 '17 at 19:39
  • @JyrkiLahtonen sorry, I forgot the $1$ is included in the definition of $PID$. See my edit. – No One Mar 20 '17 at 19:43

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Typically $J$ will not be generated by one element as an ideal of $I$. For instance, let $R=\mathbb{Q}[x]$, $I=(x)$, and $J=I$. Then no element of $J$ generates it as an ideal of $I$. Indeed, if $f(x)\in J$ has the form $$f(x)=ax^n+\text{terms with higher powers of $x$},$$ then every element of the ideal in $I$ generated by $f(x)$ will have a coefficient of $x^n$ that is an integer multiple of $a$. Since not every rational number is an integer multiple of $a$, the ideal generated by $f(x)$ cannot be all of $J$.

Eric Wofsey
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  • Nice! My argument was losing its footing if we define principal ideal generated by $x$ = the smallest ideal containing the element $x$, but yours works! – Jyrki Lahtonen Mar 21 '17 at 14:53