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Definition of Principal ideal:

Let $R$ be a commutative ring with unity and let $a \in R$ . The set $\langle a\rangle = \{ra\mid r \in R\}$ is an ideal of $R$ called the principal ideal generated by a.

Doubt

What is the principal ideal in case of a subset of $R[x]$ (set of all polynomials with real coefficients) with constant term zero.

$A=\langle x\rangle $ will work but I actually don't understand what $\langle x\rangle$ mean here. I think

$$A=\{f(x) \in R[x]\mid f(0)=0\}.$$

How all these things fit together?

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    $\begingroup$ You wrote yourself what $\langle x\rangle $ means just above it. $\endgroup$ Commented Sep 6, 2018 at 9:17
  • $\begingroup$ Ok. What if $\angle x^2+1\angle$ means then ? $\endgroup$ Commented Sep 6, 2018 at 9:19
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    $\begingroup$ Still, you have the definition right there, so I am not sure what you are really asking about. $\endgroup$ Commented Sep 6, 2018 at 9:21
  • $\begingroup$ @TobiasKildetoft $<x^2+1>={ f(x^2+1)|f(x) \in R[x]}$. $\endgroup$ Commented Sep 6, 2018 at 9:26
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    $\begingroup$ You would make things more clear for yourself if you stop writing $f(x)$ for the polynomial $f$, since that makes it hard to tell the difference between evaluating $f$ at $x^2+1$ and multiplying $f$ by $x^2+1$. $\endgroup$ Commented Sep 6, 2018 at 9:31

1 Answer 1

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$\langle x\rangle$ is the set of all polynomials that can be written as a multiple of the polynomial $f(x)=x$. Let's take a look at your example:

Every polynomial $f\in R[x]$ with $f(0)=0$ can be written as $f(x)=g(x)x$ for some $g\in R[x]$. Vice versa, every polynomial of the form $f(x)=g(x)x$ satisfies $f(0)=0$. Therefore $\langle x\rangle = \{f\in R[x]\;|\;f(0)=0)\}$.

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  • $\begingroup$ Ok. What if $\angle x^2+1\angle$ means then ? $\endgroup$ Commented Sep 6, 2018 at 9:30
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    $\begingroup$ It is the set of all polynomials that can be written as a multiple of the polynomial $x^2+1$. For example, $x^4+x^2=(x^2+1)x^2$ is such a polynomial, whereas for example $x$ or $x^2$ cannot be written in that form. $\endgroup$ Commented Sep 6, 2018 at 9:33
  • $\begingroup$ I get it now . Thanks. $\endgroup$ Commented Sep 6, 2018 at 10:43

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