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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 $\lbrace x\cdot r: r\in R\rbrace$?

W. G.
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  • What do you mean by, "the power here is just addition"? Do you mean that $x^m$ is really $x+x+\cdots+x$, sum of $m$ terms? Which is equal to $xm$? – Gerry Myerson Apr 27 '18 at 12:34
  • $x^m$ is just $x+\cdots +x=x\cdot m$, so for $R=\mathbb{Z}$ both is the same. The first is written multiplicatively, the second additively. – Dietrich Burde Apr 27 '18 at 12:35
  • That is what I mean by $x^m$ is really $x+x+...+x$ (I had a hard time notating it lol). It makes sense when $R=\mathbb{Z}$, but I do not know why it holds true for other commutative rings. – W. G. Apr 27 '18 at 12:38

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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.

Eric Towers
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  • So, what is the definition of things generated by $x$ in a ring? Then, how would I write it for this example? Is the generator in this context somehow related to span of the vector space? I have not dealt with generators in the context of vector spaces before. – W. G. Apr 27 '18 at 12:44
  • You have already written the definition of (principally) generated (right ideal) in a ring, $\langle x \rangle = {x \cdot r : r \in R}$. Generally, you would have the set of sums of such things, but left distributivity ensures that sums sums are already in your set as written. I guarantee you have looked at bases of vector spaces, that is, generating sets of vector spaces, in linear algebra. – Eric Towers Apr 27 '18 at 12:52