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Let $R = \Bbb C[x]/J$. Then the ideals of $R$ are in 1-1-correspondence with the ideals of $\Bbb C[x]$ that contain $J$. Since $\Bbb C[x]$ is a PID then every ideal is principal. Let $I=\langle f(x)\rangle$ be a principal ideal that contains $J$. Then this corresponds to the ideal $I/J$ in $R$.

My question is:

Is the ideal $I/J$ in $R$ the same as the principal ideal in $R$ written as $\langle f(x) + J\rangle$? Why or why not?

To be a bit more specific: Let $I=\langle x \rangle$ be an ideal in $\Bbb C[x]$ that contains $J$. Is $\langle x + J \rangle$ the corresponding ideal in $R = \Bbb C[x]/J$? Or should that be written $\langle x \rangle/J?$

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  • $\begingroup$ Sorry, should be complex numbers. I'll change that! $\endgroup$ – Rosters Dec 7 '13 at 20:34
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    $\begingroup$ I changed $<f(x)>$ to $\langle f(x)\rangle$. The whole reason why TeX and LaTeX and MathJax were invented was so that you could do things like that instead of writing things like $<f(x)>$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 7 '13 at 20:35
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    $\begingroup$ Yeah, sorry, just learning this way of writing. Didn't find that kind of parentheses... Thanks for correcting. $\endgroup$ – Rosters Dec 7 '13 at 20:36
  • $\begingroup$ It should be written $(x)+J$. Elements of the quotient ideal $R/J$ are written $x+J$, and the ideal should follow the same way. $\endgroup$ – Ian Coley Dec 7 '13 at 20:49
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    $\begingroup$ @IanColey: I + J is an ideal in $\mathbb C[x]$, you might be thinking $(I + J)/J$ which is an ideal in $\mathbb C[x]/J$ and when $J \subseteq I$ we have $I + J = I$ so $I/J$ is standard notation. $\endgroup$ – Jim Dec 7 '13 at 20:51
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Yes, the ideal $I/J$ is exactly the ideal $\langle f(x) + J\rangle$ in $\mathbb C[x]/J$. The reason is that the ideal $I/J$ is exactly the set of elements $I/J = \{g(x) + J \ | \ g(x) \in I\}$. Every $g(x) \in I$ can be written as $g(x) = a(x)f(x)$. This means every element of $I/J$ can be written as $g(x) + J = a(x)f(x) + J = (a(x) + J)(f(x) + J)$ hence $I/J$ is the principal ideal generated by $f(x) + J$.

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