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I know this is a very basic question but I am always unsure what exactly is meant by "generating".

For example, consider the polynomial ring $k[x]$, and the ideal generated by $f(x)$, denoted by $\langle f(x) \rangle = k[x] \cdot f(x)$.

Furthermore, in the context of a group it means taking all integral powers of the generator.

However, I am reading about prime subfields which are defined to be the subfield of a field $F$ $\textbf{generated}$ by the multiplicative identity $1_F$ of $F$. In particular, the prime subfield of $\mathbb{R}$ is $\mathbb{Q}$.

What exactly is meant by the subfield $\textbf{generated}$ by $1_F$ in this context?

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"The [thing] generated by [stuff]" just means "The smallest [thing] which contains all the [stuff]."

So the ideal in a commutative ring $R$ generated by an element $a$ is the smallest ideal of that ring containing $a$. It's a theorem that such an ideal exists, and that moreover it's just the set $\{ra: r\in R\}$.

Similarly, the subfield generated by $1_F$ is the smallest subfield of $F$ containing $1_F$.

  • Exercise 1 (easy): The subfield generated by $1_F$ contains $1_F, 1_F+1_F, 1_F+1_F+1_F$, etc., as well as the multiplicative inverses $(1_F+...+1_F)^{-1}$, additive inverses and sums of such, etc.

  • Exercise 2 (harder): That's exactly the subfield generated by $1_F$. More concretely: the subfield generated by $1_F$ is the smallest subset of $F$ which contains $1_F$ and is closed under $+, \times$, and additive and multiplicative inverses (that is: everything you can get by starting with $1_F$ and applying $+, -, \times,$ and $^{-1}$).

HINT for exercise 2: by exercise 1, it's enough to just show that that set is indeed a subfield!

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  • $\begingroup$ The multiplicative identity in $\mathbb{R}$ is $1$. How can the set $\{ 1, 1+1, \dots \}$ ever be equal to $\mathbb{Q}$. I think we would still need to close it under inverses. $\endgroup$ – GhTU Jun 14 '16 at 21:15
  • $\begingroup$ @GhTU Wow that was stupid of me. I'm tired, and switched from "subfield" to "subsemigroup" in my head for absolutely no good reason. I've fixed it now. $\endgroup$ – Noah Schweber Jun 14 '16 at 21:17
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I imagine it means the subgroup generated by 1_F under the addition of the field and the inverses of the elements of that subgroup. So 1_F in R would generate the integers, and then throwing in their inverses would give the rationals.

It would make sense that this is the smallest subfield containing the multiplicative identity of the initial field, because the subfield has to be closed under addition (meaning it contains all integral multiples of 1_F) and has to contain multiplicative inverses.

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