Suppose that $F$ is a field of characteristic $\neq 2$ and non zero elements of $F$ form a cyclic group under multiplication. Prove that $F$ is finite

Question

$1.$ 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.

Answer 1

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:

• $(a+b\sqrt2)(a'+b'\sqrt2)$ with $a,a',b,b'$ rational is of this form, and
• the inverse $$\frac1{a+b\sqrt2}$$ is of this form, whenever either $a$ or $b$ or possibly both are non-zero. Hint: rationalize the denominator.

Answer 2

For the first question you are just considering positive powers of a what about the non positive ones?? $( a^0=1)$

But the second one is very easy.

We know $Q(\sqrt2)$ is subfield of R containing $\sqrt2$.

Let $S$ be any subfield of $R$ containing $\sqrt2$ , as in general 1 belongs to $S$ thus $Z$ is a subset of $S$ and as $\sqrt2$ is in $S$ hence $Z(\sqrt2)$ is contained in $S$.

Let $p$ and $q$ be rationals such that they don't belong to rationals.

$P=\frac{m_1}{m_2}, q=\frac{n_1}{n_2}$

Now $m_1n_2+(\sqrt2)n_1m_2$ belongs to $S$.

And as $S$ is a subfield and $n_2$ and $m+2$ are non zero elements thus $\frac{1}{n_2}$ and $\frac{1}{m_2}$ belongs to $S$.

Hence $(m_1n_2 + (\sqrt2) \frac{n_1m_2)}{n_2m_2}$ belongs to $S$.

Hence $p+\sqrt2 q$ belongs to $S$ for all $p, q$ belongs to rationals.

Hence if $S$ is a subfield containing $\sqrt2$

Then $Q[\sqrt2]$ is a subset of $S$.

Thus $Q[\sqrt2]$ is the smallest subfield of $R$ containing $\sqrt2$