$(a,b) \mathbin\# (c,d)=(a+c,b+d)$ and $(a,b) \mathbin\&(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)$. Multiplicative inverse?

Question

Let $r\in \mathbb{R}$ and let $0\neq s \in \mathbb{R}$. Define operations $\#$ and $\&$ on $\mathbb{R}$ x $\mathbb{R}$ by $(a,b) \mathbin\#(c,d)=(a+c,b+d)$ and $(a,b) \mathbin\&(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)$ as addition and multiplication, respectively. Do these operations form a field?

I have shown that every axiom of the field is satisfied except for the "existence of multiplicative inverses" axiom. This is where I need help. We either need to show that this axiom fails and so we don't have a field or that the axiom $aa^{-1}=1$ is satisfied for every $a\neq 0 \in \mathbb{R}$

The multiplicative identity in this problem is $(1,0)$.

Any solutions or hints are greatly appreciated.

Answer 1

The addition and multiplication formulas somewhat remind of the formulas for complex number. We suspect that a map of the form $$ \phi\colon (x,y)\mapsto \alpha x+\beta y$$ for suitable $\alpha,\beta\in\mathbb C$ turns out to be a field isomorphism. As we want $(1,0)\mapsto 1$, we conclude $\alpha=1$. Then $$ \phi((a,b)\mathop{\&}(c,d))=\phi(a,b)\phi(c,d)=(a+\beta b)(c+\beta d) = ac+\beta(bc+ad)+\beta^2bd$$ and on the other hand $$\phi(ac-bd(r^2+s^2),ad+bc+2rbd)=ac-bd(r^2+s^2)+\beta(bc+ad+2rbd)$$ so that we obtain an identity by chosing $\beta$ with $\beta^2=2r\beta-(r^2+s^2)$, i.e., one of the roots of the polynommial $X^2-2rX+r^2+s^2$. For $s\ne 0$, this garantees $\beta\in\mathbb C\setminus\mathbb R$. Therefore the image of $\phi$ in $\mathbb C$ is more than just $\mathbb R$; as $\phi$ is linear $\mathbb C$ and $F$ both have diemnsion $2$, we conclude that $\phi$ is a vector space isomorphism. As the choice of $\beta$ guaranteed compatibility with multiplication, $\phi$ is indeed a ring isomorphism and as $\mathbb C$ is a field, so is $F$.

Answer 2

This looks similar enough to the pair-based definition of the complex numbers that it looks promising to explore how far that analogy will take us.

First, it is easy to see that multiplying any pair by $(a,0)$ simply scales both of its elements by $a$. If we identify $(a,0)$ with the usual real numbers (that is sort of handwavy but bear with me for a moment) and we give $(0,1)$ a name $x$, we can represent the pair $(a,b)$ as $a+bx$, and you already know that the ring axioms hold, so we can handle such expressions by the usual rules.

Since you have checked that the distributive law holds, multiplication has to work like $$ (a+bx)(c+dx) = ac + (ad+bc)x + bdx^2 $$ and by comparison with your multiplication rule we see that $$ x^2 = -(r^2+s^2) + 2rx $$ This gives us a quadratic equation that $x$ has to satisfy, and if we just pretend the quadratic formula will work in this setting, we can solve to get $$ x = r \pm |s|\sqrt{-1} $$

Now all of the above is not very rigorous, but it does strongly suggest that what you have your hands on are just the complex numbers under a non-standard representation. So what we should do to get back on a rigorous footing is to verify from first principles that $$ f(a,b) = a + b(r+si) $$ is a ring isomorphism $(\mathbb R^2,\#,\&)\to\mathbb C$. It is not difficult to see that it is a ring homomorphism, and proving that it is bijective is a simple matter of linear algebra.

Since $(\mathbb R^2,\#,\&)$ is isomorphic to a ring that we already know is a field, it must itself be a field.

(You don't need to work out the inverse mapping explicitly, but if you want to you now have enough information to do it relatively painlessly).

---

Note that once you know that $f$ is a bijection that preserves addition and multiplication in the sense that $f(p)+f(q)=f(p\#q)$ and $f(p)f(q)=f(p\&q)$, you can get all of the field axioms for free by "pulling them backwards" along $f$ -- not only multiplicative inverse. So the explicit verifications of, say, the associative and distributive laws you have already done will not be strictly necessary anymore.