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I'm new to Ring Theory and I'd like to check if I'm on the correct track with this.

Wolfram defines a PID as "an integral domain in which every proper ideal can be generated by a single element".

As I understand it, an ideal is something like a normal subgroup in group theory. So, if let's say I have a ring defined by all one-variable real coefficients of polynomials denoted $k[x]$ where $k \in \mathbb R$, an ideal $I$ would be addictive subset of $k[x]$ when $k.i \in I$, where $k \in K$ and $i \in I$. How can I go about thinking of such an ideal? Just off my mind, there seems to be no suitable candidates (other than the trivial ones, $0$ and $k[x]$) that can form an ideal.

Also, is my thinking correct on this? $k[x]$ is a PID because for $x \in k[x]$, I can generate all the possible polynomials in $k[x]$ by simply multiplying $x$ by itself to reach the degree of $k[x]$. As I can add $x$ of whichever power I like together, I can get any polynomial in $k[x]$. For instance, if you gave me something like $x^3+2x$, it can be generated by the polynomial $x$ by mutiplying $x$ by itself $3$ times and adding that to $2x$. Is my intuition correct for this? Otherwise, how can you prove that $k[x]$ is a PID?

Thanks for all the help.

koifish
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  • The standard argument (surely found in any book on ring theory, and probably everywhere online) is to use the polynomial division algorithm. It enables you to prove that the polynomial in an ideal which has leading coefficient $1$ and least degree actually generates everything else in the ideal. – rschwieb Dec 09 '18 at 14:57
  • In any commutative ring $R$ with identity, for any element $x\in R$ you have that the set $xR={xr\mid r\in R}$ is an ideal of $R$. YOu can check the axioms. These are, in fact, exactly the principal ideals of $R$. But sometimes there can be more ideals than just these. In the case of $F[x]$, there are no more. Thinking of ideals as "the ring theoretic analogue of normal subgroups" is a good way to start. – rschwieb Dec 09 '18 at 14:58

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