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This is part of an attempt to understand what multiplicative structure a Hamel basis of the reals over the rationals can have.

Does there exist a Hamel basis $\mathcal B$ for $\mathbb R$ over $\mathbb Q$ such that $a,b \in \mathcal B \implies \dfrac ab \in \mathcal B$ ?

Additionally, as proposed by Noah Schwerber, if the answer to the above is negative, what if the restriction that $a \neq b$ is imposed, that is:

Does there exist a Hamel basis $\mathcal B$ for $\mathbb R$ over $\mathbb Q$ such that $a,b \in \mathcal B $ distinct $ \implies \dfrac ab \in \mathcal B$ ?

The following earlier question showing that such a Hamel basis cannot be closed under multiplication by a (non-trivial) constant could be helpful Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ cannot be closed under scalar multiplication by $a \ne 0,1$

A recent related but distinct question Hamel basis for the vector space of real numbers over rational numbers and closedness of the basis under inversion focuses on whether a Hamel basis can be closed under taking inverses.

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    $\begingroup$ More generally, if $F$ is a proper field extension of a field $K$ with an extension basis $B$, then $B$ is not a semigroup under the multiplication of $F$ (per i707107's answer). Hence, $B$ is not a group (noting that the condition that $a,b\in B$ implies $a/b\in B$ forces $B$ to be a group). $\endgroup$ Commented Aug 12, 2016 at 21:48

2 Answers 2

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The solution follows closely from the standard argument in group theory.

Suppose that a Hamel basis $\mathcal{B}$ satisfies the property $$ a, b\in \mathcal{B} \Longrightarrow \frac ab\in \mathcal{B}. $$ From this, we have $$ a\in\mathcal{B} \Longrightarrow \frac aa =1\in \mathcal{B}. $$ Then $$ b\in\mathcal{B} \Longrightarrow \frac 1b \in \mathcal{B}. $$ Now, $$ a,b \in \mathcal{B} \Longrightarrow a, \frac 1b \in \mathcal{B} \Longrightarrow \frac a{1/b} = ab \in \mathcal{B}. $$ Since $\mathcal{B}$ contains some element $a\neq 0, 1$, a contradiction comes from the same argument as in Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ cannot be closed under scalar multiplication by $a \ne 0,1$.

For the new condition by Noah Schweber

If $1\in\mathcal{B}$ then it can be proven the same way as above. So, we suppose that $1\notin \mathcal{B}$. Then we have $1$ as a $\mathbb{Q}$-linear combination of elements in $\mathcal{B}$, say $$ 1=\sum_{i=1}^{k} \epsilon_i x_i $$ where $x_i\in\mathcal{B}$, $\epsilon_i\in\mathbb{Q}$.

Choose $y_1\in \mathcal{B}$. We modify the argument in Jonathan Golan's answer to Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ cannot be closed under scalar multiplication by $a \ne 0,1$.

Let $\mathcal{B}_s= \mathcal{B} - \mathbb{Q}(y_1, x_1, \ldots, x_k)$ be a selected basis which are not rational in $y_1$, $x_i$'s. Let $\alpha: \mathbb{R} \rightarrow \mathbb{Q}$ defined by $$ r=\sum_{x\in\mathcal{B}} r_x x \mapsto \alpha(r)=\sum_{x\in\mathcal{B}_s} r_x. $$ Then for any $x\in \mathcal{B}_s$, we have $\frac x{y_1}\in\mathcal{B}_s$. Let $a=\frac1{y_1}$. For any $r\in \mathbb{R}$, we have $\alpha(r) = \alpha(ar)$.

For any $x\in \mathcal{B}_s$, put $ r=x(a-1)^{-1}. $

Then $$ 1=\alpha(x) = \alpha( r(a-1) ) = \alpha( ar - r) = \alpha(ar)-\alpha(r) = 0. $$ This is a contradiction.

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    $\begingroup$ What if we restrict the condition only to distinct pairs of elements - that is, $a\not=b, a, b\in \mathcal{B}\rightarrow{a\over b}\in\mathcal{B}$? The answer should be the same, but I don't immediately see how to fix the argument. $\endgroup$ Commented Aug 12, 2016 at 21:43
  • $\begingroup$ If $a,b\in B$ are such that $a,b\neq 1$, $a\neq b^2$, $b\neq a^2$, and $ab\neq 1$, then $\frac{a}{b}$, $\frac{b}{a}$, $\frac{a^2}{b}=\frac{a}{b/a}$, and $\frac{a}{b^2}=\frac{a/b}{b}$ are in $B$. That is, $$ab=\frac{a^2/b}{a/b^2}$$ must be in $B$. This may help. $\endgroup$ Commented Aug 12, 2016 at 22:18
  • $\begingroup$ Hmmm, I don't know why I put $a,b\neq 1$ there, when I actually meant $a\neq b$. $\endgroup$ Commented Aug 12, 2016 at 22:29
  • $\begingroup$ @Batominovski I am not sure how this helps. Although $ab\in B$, this is not for all $a,b\in B$. You have several restrictions on $a, b$. $\endgroup$ Commented Aug 12, 2016 at 23:07
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    $\begingroup$ @quid, As I saw $ab^{-1}\in\mathcal{B}$, I instantly recalled the standard argument in group theory. You are right about multiplication by $b_0^{-1}$ and using the quoted result. Also, the second part is written many hours later than the first part. In fact, the argument in the second part alone covers the first part too. $\endgroup$ Commented Aug 13, 2016 at 14:24
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This is not an answer. It is an attempt at Noah Schweber's new condition. See his comment under i707107's solution.

Let $F$ be a field extension of a field $K$ with $n:=[F:K]\geq 4$. Suppose that $B$ is a Hamel basis of $F$ over $K$ with the divisibility property, namely, for all $a,b\in B$ with $a\neq b$, $\dfrac{a}{b}\in B$. Then, it holds that $ab\in B$ for all $a,b\in B$ with $ab\neq 1$.

Fix $a,b\in B$ with $ab\neq 1$. Then, choose $c\in B\setminus\{a\}$ and $d\in B\setminus\left\{c,bc,\dfrac{c}{a}\right\}$. Hence, $\dfrac{c}{a}\in B$ and $\dfrac{ad}{c}=\dfrac{d}{c/a}\in B$. Furthermore, $\dfrac{d}{c}\in B$ and $\dfrac{d}{bc}=\dfrac{d/c}{b}\in B$. Since $ab\neq 1$, we have $\dfrac{ad}{c}\neq \dfrac{d}{bc}$. Consequently, $$ab=\frac{ad/c}{d/bc}\in B\,.$$


Involutive Hamel Bases

A subset $S$ of $F\setminus\{0\}$ is said to be involutive if $S$ is invariant under inversion (that is, $\dfrac{1}{s}\in S$ for all $s\in S$). Now, if we can show that every Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ is not involutive, then it follows that a Hamel basis with the divisibility property does not exist. I don't think that an involutive Hamel basis for the extension $\mathbb{R}>\mathbb{Q}$ exists, but I have no idea about a proof. More generally, I would like to know whether there exist a field $K$ and an infinite field extension $F$ of $K$ with a Hamel basis $B$ over $K$ such that $B$ is involutive. As in i707107's solution, involutivity of $B$ is not needed if $[F:K]>|\bar{K}|$, where $\bar{K}$ is the algebraic closure of $K$, but it is an interesting question nonetheless.

If $n=[F:K]$ is finite and odd, then $$B=\left\{\bar{x}^{-\left\lfloor\frac{n}{2}\right\rfloor},\bar{x}^{-\big(\left\lfloor\frac{n}{2}\right\rfloor-1\big)},\ldots,\bar{x}^{+\big(\left\lfloor\frac{n}{2}\right\rfloor-1\big)},\bar{x}^{+\left\lfloor\frac{n}{2}\right\rfloor}\right\}$$ is an involutive basis of $F$ over $K$, where $F=K[x]\big/\big(f(x)\big)$ for some irreducible polynomial $f(x)\in K[x]$. If $n$ is even, then there exists an irreducible polynomial $f(x)$ of degree $n$ in $K[x]$ such that the coefficient of the term $x^{n/2}$ is nonzero and that $F=K[x]\big/\big(f(x)\big)$, making $$B=\left\{\bar{x}^{-1},\bar{x}^{-2},\ldots,\bar{x}^{-(n/2)}\right\}\cup\left\{\bar{x}^{+1},\bar{x}^{+2},\ldots,\bar{x}^{+(n/2)}\right\}$$ an involutive Hamel basis of $F$ over $K$. (For example, with $F=\mathbb{C}$ and $K=\mathbb{R}$, we can take $f(x)=(x+1)^2+1$.)


Existence of $B$ with the Divisibility Property

For $n=2$, take $K:=\mathbb{F}_2$ and $F:=\mathbb{F}_4=K[x]\big/\left(x^2+x+1\right)$. Then, $B:=\left\{\bar{x},\bar{x}+1\right\}$ satisfies the condition, where $\bar{x}$ is the image of $x$ under the canonical projection $K[x]\to K[x]\big/\left(x^2+x+1\right)$. In fact, if such a Hamel basis $B$ exists for the case $n=2$, then $x^2+x+1$ is an irreducible polynomial in $K[x]$ and $F=K[x]\big/\left(x^2+x+1\right)$, in which case $B=\left\{\bar{x},-\bar{x}-1\right\}$.

For $n=3$, it turns out that $B$ cannot exist. Suppose contrary that $B=\{a,b,c\}$ exists. Then, we can easily see that $B$ does not contain $1$ (or the extension would be of index at most $2$). Hence, $\dfrac{a}{b}\neq a$ and $\dfrac{a}{c}\neq a$. If $\dfrac{a}{b}=c$, then $c\notin K$, whence $\dfrac{b}{a}=\dfrac{1}{c}\neq c$ so that $\dfrac{b}{a}=a$ is the only possibility. Furthermore, we also have $\dfrac{a}{c}=b$, which leads to $b\notin K$ and $\dfrac{c}{a}=a$. Therefore, $b=a^2=c$, a contradiction. Hence, $\dfrac{a}{b}=b$ and $\dfrac{a}{c}=c$. However, this gives $b^2=a=c^2$, or $b=\pm c$, which is absurd.

As i707107 shows, $B$ does not exist if $[F:K]>|\bar{K}|$. Replace $\mathbb{Q}$ in i707107's solution by $K$ and $\mathbb{R}$ by $F$ to get a proof of this claim. We are left to deal with the case where $[F:K]>3$ is finite and the case when $[F:K]$ is infinite but $[F:K]\leq|\bar{K}|$.

Case 1: The index $n=[F:K]$ is an odd integer greater than $3$. Suppose $B$ exists. It is evident that $1\notin B$ and $B$ must be involutive. Since $n=|B|$ is odd, $B$ has an involutive element $u\in B$ with $u=\dfrac{1}{u}$. Because $1\notin B$, we have $u=-1$ (whence the characteristic of $K$ cannot be $2$). Thus, for any $a\in B\setminus\{u\}$, we have $\dfrac{u}{a}=-\dfrac{1}{a}$ must lie in $B$. This contradicts the fact that $B$ is involutive (which means $\dfrac{1}{a}$ is in $B$).

Case 2: The index $n=[F:K]$ is an even integer greater than $2$. If $B$ exists, then $B$ can be partitioned into $\left\{t^{+j},t^{-j}\right\}$ for some $t\in F$ and $j=1,2,\ldots,\frac{n}{2}$. Ergo, $t^p=1$ for some integer $p>0$. If $p$ is not prime, then we can see that $B$ is not $K$-linearly independent, which is absurd. Hence, we see that $[F:K]=n=p-1$ for some odd prime natural number $p$ and $$B=\left\{t,t^2,\ldots,t^{p-1}\right\}=\bigcup\limits_{j=1}^{\frac{p-1}{2}}\,\left\{t^{+j},t^{-j}\right\}$$ for some primitive $p$-th root of unity $t\in F\setminus K$. This is possible only if $\text{char}(K)\neq p$. In summary, for the case $[F:K]$ is even, $B$ exists if and only if $F$ is a cyclotomic field extension over $K$ generated by a primitive $p$-root of unity with $p$ being an odd prime.

Case 3: The index $n=[F:K]$ is infinite and $n\leq|\bar{K}|$. If $F$ is algebraic over $K$, then we can follow the two former cases that such a Hamel basis $B$ would have to contain only primitive $p$-roots of unity for odd prime natural numbers $p$, but this is absurd as products of most pairs in $B$ are also in $B$. Hence, $B$ does not exist if $F$ is algebraic over $K$. The subcase where $F$ is not purely algebraic over $K$ seems to be very difficult.

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  • $\begingroup$ Nice observation! I thought about showing this a bit, but didn't see how to do so. $\endgroup$ Commented Aug 12, 2016 at 23:27
  • $\begingroup$ @NoahSchweber Do you know anything about involutive Hamel bases? (See my definition in the answer.) Do they exist for the extension $\mathbb{R}>\mathbb{Q}$? $\endgroup$ Commented Aug 12, 2016 at 23:34
  • $\begingroup$ No idea, sadly. I've never heard about them before. $\endgroup$ Commented Aug 12, 2016 at 23:36
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    $\begingroup$ @Batominovski My solution might not generalize to finite extensions, because it might not be possible to have such $\mathcal{B}_s$. What if every basis elements turn out to be rational on $y_1, x_1, \ldots, x_k$? In my solution, it is possible to have such $\mathcal{B}_s$ since $\mathcal{B}$ is uncountable and $\mathbb{Q}(y_1, x_1,\ldots, x_k)$ is countable. $\endgroup$ Commented Aug 13, 2016 at 1:40
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    $\begingroup$ I know that $\mathbb{C}$ is spanned over $\mathbb{Q}$ by the unit circle $C$ subtracted $\{-1,+1\}$. However, does there exist a Hamel basis $B$ of $\mathbb{C}$ over $\mathbb{Q}$ which is closed under complex conjugation and contained in $C\setminus\{-1,+1\}$? $\endgroup$ Commented Aug 13, 2016 at 12:51

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