$1.$ Suppose that $F$ is a field of characteristic $\neq 2$ and that the non zero elements of $F$ form a cyclic group under multiplication. Prove that $F$ is finite
Attempt : Let $F=\{0,a,a^2,a^3,……….\}$
Given that non zero elements of $F$ form a cyclic group under multiplication. Now, since, by definition, a field must contain the unity, $\implies a^i = 1 $ for some $i$. As a result, This $i$ must be finite. Only then, can we say that a unity $\in F$
$\implies H = \{a, a^2, …., a^i\}$ are all distinct elements and all powers of $a$ with index $<i$ belong to this set $H$. For all, indices $>i$, the elements get repeated and belong to $H$ itself.
Since, $i$ is finite $\implies$ $H$ must be finite and as a result $F$ obtained by adding an extra element $0$ must be finite as well.
Hence, $F$ must be finite.
Is my proof correct? What was the use of stating that $char~ F \neq 2$ ?
$2.$ Describe the smallest subfield of the field of real numbers containing $\sqrt 2$
Attempt: A field is a commutative ring with unity whose each element is a unit (i.e. possesses an inverse )
The sub field test states that $H$ is a sub field of $F$ if and only if $H$ is a field in itself.
(a) Closure under multiplication – $(\sqrt 2) \in H \implies (\sqrt 2)^2 , (\sqrt 2)^3, …. \in H $
(b) Existence of inverse for each element – (a) $\implies (\sqrt 2)^{-1},(\sqrt 2)^{-2},(\sqrt 2)^{-3}, ….. \in H$
(c) Closure under addition –
$…+ \alpha_{-2} (\sqrt 2)^{-2} + \alpha_{-1} (\sqrt 2)^{-1} + \alpha_{0} (\sqrt 2) + \alpha_{1} (\sqrt 2)^{1} + \alpha_{2} (\sqrt 2)^{2}+..\in H$ where $\alpha_i \in Z$
Hence, the smallest sub field of real numbers containing $\sqrt 2$ is of the form $\alpha (\sqrt 2)^\beta ; \alpha, \beta \in Z$
Edit: Due to closure under addition, $ (c_1 + c_2 (\sqrt 2)^m)$ should have an inverse as well. Hence, I think , the smallest sub field should be of the form $\{a + b (\sqrt 2)^m \}$ where $a,b \in Q$
Is my attempt Correct?
Thank you for your help.
Best Answer
I'm afraid there are shortcomings in your attempts.
(i) The choice $i=0$ gives $a^i=1$, so you cannot conclude that $a$ is of finite order in thise way. The question has been covered many times recently. I recommend this answer by blue.
(ii) I don't think the number $1+\sqrt2$ is in your set $H$, so $H$ is not closed under addition and hence not a subfield. Have you, for example, done an example in class, where the smallest subfield of the real numbers containing $i$ was described? Failing that, have you covered a result that shows what the prime fields are? These are the minimal fields: $\Bbb{Z}_p$, $p$ a prime number and $\Bbb{Q}$. The edited attempt is much better. The answer is, indeed, that the set $$ F=\{a+b\sqrt2\mid a,b\in\Bbb{Q}\} $$ is the smallest subfield containing $\sqrt2$. It is easy to show that this is an additive subgroup of $\Bbb{R}$ and that $1\in F$. What remains to be shown is that: