# How is the prime field “generated”?

so I was studying about field extensions and I came up with this innocent question. Let $F$ be a field and $1$ its identity, then the prime field $G$ of $F$ is the field generated by $1$, i.e., $G=\langle 1\rangle$. Now, I know that $G$ is the smallest field in $F$ containing $1$ so $G=\mathbb{Q}$ or $\mathbb{Z}_p$ with $p$ prime depending on the characteristic of $F$, so I'm ok with that in that perspective. My problem is that, when I read "generated", I get stuck about how does this happen, like I understand that when you have a ring it's just taking sums and products in some order, but in the case of when you have a field I'm not quite sure how to "generate" $G$ properly when seen in this fashion. Thanks a lot

• You add all inverses, products and sums of the elements of set generating the field. – bjn Oct 3 '16 at 16:16

If I understand your question correctly, you are looking for a sequence of "steps" that ensure you have obtained all elements in the prime subfield.

If the characteristic is positive, then you only need to take all finite sums of $1$'s.

If the characteristic is zero, you could do this: first include all finite sums of $1$'s (get $\mathbb{Z}$), then include all quotients of one element by another nonzero element (get $\mathbb{Q}$).

In general, if $X$ is a subset of a field $F$, then you can get the subfield generated by $X$ in these steps:

• if $X$ does not contain $1$, include it;
• include all finite sums of finite products of elements of $X\cup \{1\}$ (get the subring generated by $X$);
• then include all quotients of one element by another nonzero element (here the elements are taken amongst the ones obtained in the previous step).
• Shouldn't we also include negatieve sums? Otherwise I don't see how we could get negative elements. Or at least include $-1$ as well. – Sha Vuklia May 7 '19 at 10:00
• @ShaVuklia Indeed. Finite sums should be of $1$'s and $-1$'s. – Pierre-Guy Plamondon May 7 '19 at 22:13