Commutative Rings: Principal Ideal Generator Question

34 Views Asked by At

Here is a definition: We say that an ideal, $I$, of a commutative ring $(R,+, \cdot) $ is principal iff $I=<x>=\lbrace x\cdot r: r\in R\rbrace$.

My question is how $<x>=\lbrace x^m :m\in \mathbb{Z}\rbrace$ (where the power just here is addition) ends up to be equal to $\lbrace x\cdot r: r\in R\rbrace$?

1

There are 1 best solutions below

2
On BEST ANSWER

You are conflating "the collection of things generated by $x$ in a group (written multiplicatively)" with "the collection of things generated by $x$ in a ring". You have successfully written the definitions of both of these things. But "generated by [set]" should always (although usually silently) be followed by "in a(n) [algebraic object]" so you know what set of operations and scalars you get to use.

For instance, $\langle \vec{x}, \vec{y} \rangle = \mathbb{R}^2$ must mean "generated as an $\mathbb{R}$-vector space".

Sometimes, but only very rarely, you will see a disambiguator on the presentation brackets, such as $\langle \dots \rangle_{\text{Ab}}$, which would mean "generated by [...] in an abelian group". One usually does not need to specify "in a whatever" because the algebraic objects under discussion are all of the same type.