Is this structure a field?

Question

I'm wondering whether the following addition and multiplication over the set $(\mathbb{R}\setminus\{0\}\times \mathbb{Z}) \cup \{0\}$ define a field:

$$ (a,a')+(b,b')= \begin{cases} (a,a') \text{ if } a'>b'\\ (b,b') \text{ if } b'>a'\\ (a+b,a') \text{ if } b'=a' \text{ and } a\neq -b\\ 0 \text{ if } b'=a' \text{ and } a= -b\\ \end{cases} $$ $$ (a,a')(b,b')=(ab,a'+b') $$ $$ -(a,a')=(-a,a') $$ $$ (a,a')^{-1}=(a^{-1},-a') $$

[$0$ is the additive unit, which fixes addition and multiplication with $0$. $(1,0)$ the multiplicative unit.]

If yes, does this field have a name? If no, which of the axioms fail?

I'm a bit confused because I thought there are only "relatively few" different fields, such as the rational, real, complex numbers, or finite fields.

Answer 1

No, addition is not associative. For instance, $$((1,0)+(-1,0))+(1,-1)=0+(1,-1)=(1,-1)$$ but $$(1,0)+((-1,0)+(1,-1))=(1,0)+(-1,0)=0.$$

Note that you can tell something must be wrong with just the additive axioms, since your operation $+$ does not allow cancellation and so cannot be a group operation. Since there clearly is an identity and inverses, associativity must fail.

By the way, there are lots and lots of different fields; there are just a few that are familiar in elementary mathematics. For instance, the following is a field: the underlying set is $\mathbb{Q}\times\mathbb{Q}$, addition is $$(a,b)+(c,d)=(a+c,b+d),$$ and multiplication is $$(a,b)\cdot(c,d)=(ac+2bd,ad+bc).$$ (To make this more familiar, this field is isomorphic to the subfield of real numbers consisting of numbers of the form $a+b\sqrt{2}$ with $a,b\in\mathbb{Q}$, sending $(a,b)$ to $a+b\sqrt{2}$.)

Answer 2

The most glaring issue I see is that your addition is not one-to-one. For instance, $(10,10)+(b,b')$ is $(10,10)$ for any $b'<10$. But addition has to be one-to-one: if we have $x+y=x+z$, then $x+y+(-x)=x+(-x)+z \rightarrow y=z$.

0 is the additive unit

But then $0=0*(b,b')$. If we take $(a,a')$ with $a'<b'$, by the distributive property we we have $0=0*(b,b')=((a,a')+(-a,a'))*(b,b')=(b,b')+(b,b')=(2b,b')$.