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Specifically, I wish to show the non-existence of a field $(\mathbb{R}^3, +, \cdot)$ such that:
a) $(x_1, y_1, z_1) + (x_2, y_2, z_2) = (x_1+x_2, y_1+y_2, z_1+z_2)$,
and
b) $(x_1, 0, 0) \cdot (x_2, 0, 0) = (x_1 x_2, 0, 0)$.
Note that nothing is said about the linearity (as in scalar multiplication) of such a structure --- this is where I am stuck. All the similar questions I could find involve division algebras, in which linearity is assumed.
The furthest I ever got was that $(1, 0, 0)$ is the multiplicative identity (which makes $r \mapsto (r, 0, 0)$ a clearly injective field homomorphism), and that $(q, 0, 0) \cdot (x, y, z) = (qx, qy, qz)$ for $q \in \mathbb{Q}$ by induction, but I have no idea how (and if) this is supposed to imply the general case for real numbers.
Any help/clues would be greatly appreciated. I have been stuck on this problem for days and it is crippling my mental health :)

Robert Shore
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Lennox
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    Hint: Argue that the only irreducible polynomials over $\mathbb R$ have degree $≤2$. – lulu Sep 06 '24 at 14:50
  • Where would I construct a polynomial in this problem tho... (I do know this fact) – Lennox Sep 06 '24 at 14:53
  • It's would be a finite field extension. Every element in the "field" $\mathbb R^3$ would have to be algebraic over $\mathbb R$. And, from general facts about finite extensions in characterisic $0$, the extension would have to be generated by a single element. – lulu Sep 06 '24 at 14:55
  • omg okay I have yet to learn about field extensions, will look into those, thx! – Lennox Sep 06 '24 at 14:57
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    Oh, I assumed you wanted to retain the real vector space structure. What are you assuming $\mathbb R^3$ is? If you aren't assuming it is a real vector space, then my argument won't work. – lulu Sep 06 '24 at 14:57
  • There is no vector space structure going on in this problem, I don't think. $\mathbb{R}^3$ means 3-tuples of real numbers, thats all. I'd certainly like to prove $(a, 0, 0)\cdot (x, y, z) = (ax, ay, az)$ though. – Lennox Sep 06 '24 at 15:00
  • Then my argument does not apply. – lulu Sep 06 '24 at 15:03
  • @rschwieb, the duplicate you linked to is about division algebras, while this question is explicit about not being about division algebras. – Mees de Vries Sep 06 '24 at 15:41
  • @MeesdeVries Doesn't the assumption that $\mathbb R$ is a subfield together with associativity of the field indirectly imply linearity? (I'm no opposed to reversing what I did.) – rschwieb Sep 06 '24 at 15:46
  • @rschwieb That is not a duplicate (I made the same error). The OP has written $\mathbb R^3$ just to denote triples....there is no assumed vector space structure. – lulu Sep 06 '24 at 15:47
  • @lulu Hm, well to be on the safe side I'll back it out. – rschwieb Sep 06 '24 at 15:48
  • @rschwieb At least, it seems hard to prove that the very weak assumptions on the structure imply that the field laws respect some version of the usual vector space structure. – lulu Sep 06 '24 at 15:50
  • The field is certainly a vector space over $\mathbb R$ but yeah i guess I cannot guarantee it respects the tuples already established. – rschwieb Sep 06 '24 at 16:01

2 Answers2

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With the axiom of choice, such a field structure does exist. Seeing $(\mathbb R, +)$ as an abelian group, it is isomorphic to a $\mathfrak c$-dimensional vector space over $\mathbb Q$. In particular, $(\mathbb R^2, +) \cong (\mathbb R, +)$, using any bijection $\mathfrak c^2 \cong \mathfrak c$; call an isomorphism $\phi$. Then $(a, b, c) \mapsto a + i\phi(b, c)$ gives a bijection from $\mathbb R^3$ to $\mathbb C$, and you can check that using the normal field structure of $\mathbb C$ will give you the kind of structure you are looking for.

Mees de Vries
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  • huh? I am a freshman stumbling across this mathematical analysis homework problem... Could you perhaps refer to a textbook where I can learn about this stuff, especially the c-dimensional vector space stuff? – Lennox Sep 06 '24 at 15:06
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    If this is a mathematical analysis homework problem you either missed some of the details or the question is incorrectly stated. If you want to understand this answer you need to know basic but rigorous linear algebra ($(\mathbb R, +)$ is a vector space over $\mathbb Q$, so pick some basis $B$, and that gives you an isomorphism $(\mathbb Q^{|B|}, +) \cong (\mathbb R, +)$), and a small amount of set theory (that $|B| = |\mathbb R|$, and that $|\mathbb R| = |\mathbb R^2|$). – Mees de Vries Sep 06 '24 at 15:20
  • The interesting thing is that your bijection in the end is an isomorphism of Q-vectorspaces (in a canonical way) and enhances to the complex field (including R, making it to a R-vectorspace), while not even the scalar multiplication is beeing continued "naturally". I think i begin to understand why some people call the AC a cheat mode. – Tina Sep 06 '24 at 16:36
  • @TimonKolt, indeed my answer is perhaps better understood as "you have made the wrong assumptions, because I can create a contrived counterexample that's obviously not of interest but technically fits your requirements", rather than "this is the example you're looking for". – Mees de Vries Sep 19 '24 at 11:06
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The question you ask should be treating $\mathbf R^3$ as a real vector space in the usual way together with some multiplication giving $\mathbf R^3$ a field structure, so its $\mathbf R$-dimension is $3$.

Suppose $K$ is a field extension of $\mathbf R$ with odd dimension over $\mathbf R$. We will show every element of $K$ is in $\mathbf R$, so $K = \mathbf R$. Thus $\mathbf R^3$ with its usual vector space structure can’t be made into a field.

Let $a$ be in $K$ with minimal polynomial $f(x)$ in $\mathbf R[x]$. Then $f$ is irreducible. Also $\deg f$ is odd since $\deg f = [\mathbf R(a):\mathbf R]$ and that field degree divides $[K:\mathbf R]$, which is odd by hypothesis (you wanted $K/\mathbf R$ to have degree $3$).

By the nature of real numbers, every odd-degree polynomial has a real root. So $f(x)$ has a root $r$ in $\mathbf R$. Thus $f(x)$ has factor $x-r$ in $\mathbf R[x]$. Being irreducible in $\mathbf R[x]$ implies $f(x)$ is linear, so $r$ is its only root. Thus $a = r \in \mathbf R$, so $K = \mathbf R$.

KCd
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  • The thing is: preserving the natural vector space structures is not required, check the comments of the question. So you could simply take a bijection $\mathbb R \to \mathbb R^3$ and "shift the structures"/translate it via the bijection to $\mathbb R^3$ – Tina Sep 06 '24 at 17:37
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    @TimonKolt The technical term is transport of structure. – Arturo Magidin Sep 06 '24 at 17:48
  • @TimonKolt that is an unreasonable way to interpret the question. We don't know what the original "homework problem" was, but surely it was implicit in that problem to keep $\mathbf R^3$ as a $3$-dimensional real vector space, as that is exactly what you'd expect to reach the standard result in field theory that there is no field extension of $\mathbf R$ that is a $3$-dimensional real vector space. – KCd Sep 06 '24 at 18:06
  • I thought so too, but check out what @Lennox wrote: https://math.stackexchange.com/questions/4967972/how-does-one-prove-that-there-is-no-field-structure-on-bbb-r3-with-bbb-r/4968061?noredirect=1#comment10635300_4967972 – Tina Sep 06 '24 at 18:13
  • @TimonKolt yes, I saw that before posting my answer, but since the "homework" problem was to show $\mathbf R^3$ does not have a field structure, the only sensible way to give meaning to that problem is to make $\mathbf R^3$ as a field be a cubic extension field of $\mathbf R$, just as $\mathbf C$ is a field structure on $\mathbf R^2$. This is exactly the task that tormented Hamilton ("well, papa, can you multiply triples?" as his children asked him before he discovered the quaternions). – KCd Sep 06 '24 at 23:38
  • Yeah this is probably what my prof meant, I'm writing to check with him right now. Still, this is way too hard for freshman analysis, it's literally the first problem set lol – Lennox Sep 07 '24 at 00:55
  • Really appreciate all of you trying to make sense of this stuff, thx a lot! – Lennox Sep 07 '24 at 00:56