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I understand that uni-variate polynomial rings with coefficients in a field only have principal ideals. For example, $\mathbb{C}[x]$. But how can I tell if an ideal of integer polynomial ring is principal, please? For example, a textbook claims that "the kernel of the map $\mathbb{Z[x]} \rightarrow \mathbb{Z[i]}$ sending $x \mapsto i$ is the principal ideal of $\mathbb{Z}[x]$ generated by $f=x^2+1$" without any justification. How to show this is true, please?

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In this case, it is fairly easy. Since the polynomial $f$ is monic, you can write every $p \in \mathbb{Z}[x]$ as

$$p = q\cdot f + r$$

with $r \in \mathbb{Z}[x]$ of degree at most one. Since - easily verified - $f$ maps to $0$, the kernel of $\pi \colon g \mapsto g(i)$ surely contains the principal ideal $(f)$ generated by $f$, and to show that it contains nothing more, by the above it suffices to show that no polynomial of degree $\leqslant 1$ except the zero polynomial is mapped to $0$. But $a + bx \mapsto a + bi$, so $a+bx \in \ker \pi \iff a = b = 0$.

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  • $\begingroup$ Thank you for your answers very much. $\endgroup$ – LaTeXFan Oct 26 '13 at 2:33
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Since you know that $\mathbb{R}[x]$ is principal, you can use that.

Consider the extension of the map $x\mapsto i$ to $\mathbb{R}[x]$. The kernel of this map is clearly the principal ideal $I=(x^2+1)\mathbb{R}[x]$. Now consider $I\cap\mathbb Z[x]$, which is the kernel of the original map. If $g\in \mathbb R[x]$ and $g(x)(x^2+1)\in\mathbb{Z}[x]$, then by expanding the product it is easy to see that in fact $g\in \mathbb Z[x]$.

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  • $\begingroup$ Thank you for your response. $\endgroup$ – LaTeXFan Oct 26 '13 at 2:33
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A prime ideal in ${\bf Z}[x]$ can only be $(0)$, principal, or maximal. Your ideal is not $(0)$ because it has an evident nonzero element. It's not maximal because ${\bf Z}[i]$ does not contain a field. That leaves one option.

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