Let $F=\mathbb{R} , E=\mathbb{F}_2$ . Does $E$ subfield of $F$

Question

Let $F=\mathbb{R} , E=\mathbb{F}_2$ .

so to check it we need:

$0_F=0 \in E - true$

$1_F=1 \in E - true$

$a,b \in E,\:\: a +_Fb \in F $ - not ture because $1+1=2\notin E$

so $E$ not a subfield of $F$

is my solution correct ?

Answer 1

It's sort of correct. However, as it is usually defined, $\mathbb E$ is not even a subset of $\mathbb F$. Although its elements are written as $\{0,1\}$, these should not (or at least not necessarily) be understood as the real numbers 0, 1, but rather as two abstract elements which are the identity element of addition and multiplication respectively -- the 0 and 1 that are present in any field. So, saying that $0_{\mathbb F} \in \mathbb E$ is not really true or even sensible.

Thus, the question should be understood, probably, as "Is there an (injective) field homomorphism $\phi: \mathbb E \to \mathbb F$?" Then, you can say: well, if such $\phi$ existed, we would have that $\phi(1_{\mathbb E}) = 1_{\mathbb F}$, yet $$ 0_{\mathbb F} = \phi(0_{\mathbb E}) = \phi(1_{\mathbb E} + 1_{\mathbb E}) = \phi(1_{\mathbb E}) + \phi(1_{\mathbb E}) = 1_{\mathbb F} + 1_{\mathbb F} = 2_{\mathbb F}, $$ which is obviously false; so such a $\phi$ cannot exist. If you would phrase this argument very informally, you would get something like in your question, but you should understand that to make it more precise you have to do what I just wrote.

Answer 2

A subfield needs to have the same $+$ operation. Here you will have to involve "wrap-around" logic, which essentially means the function must be different. So, no, it is not.