# Why aren't all prime subfields equal to {0,1}?

the definition of a prime subfield that I have been provided with is the following:

It can be shown that for any subset $S$ of $F$ there exists a minimal subfield $K$ of $F$ which contains $S$. One very important case is when the set $S$ is empty, i.e., every field $F$ contains a unique minimal subfield. This subfield is called the prime subfield.

I further know that any subfield $K$ of a field $F$ contains the elements $0$ and $1$ and is closed under addition, multiplication, and taking negatives and inverses of non-zero elements.

And since the prime subfield is the intersection of all possible subfields across $F$, why is it not the case that the prime subfield is always equal to $0$ and $1$?

Thank you for the help!

• Isn't the prime subfield equal to $\{0, 1\}$ if and only if the field has characteristic 2? – Charlie Aug 28 '16 at 21:40
• Don’t forget: “closed under addition” means that you also get an element of the field by adding $1$ to itself. – Lubin Aug 28 '16 at 21:45

The subset $\{0,1\}$ may not even be a field. Is that a subfield of $\mathbb{R}$ for instance?
Prime subfields of course contain $0$ (the additive identity and multiplicative absorption element) and $1$ (the multiplicative identity), but they also contain $1+1$ and $1+1+1$, etc. (because fields of course must be closed under the operation of addition) and their additive inverses. In other words, the prime subfield (and in general the prime subring of any unital ring) will be the image of the unique unital ring homomorphism $\mathbb{Z}\to R$, since the image is precisely the set generated by $0$ and $1$ under addition and additive inverses.
• Oh, thank you very much. I feel silly for not realizing the operations $+$ and $\cdot$ are inherited lol oops – sos Aug 28 '16 at 21:44