0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ another field: rational expressions (fractions of polynomials in $X$) $\endgroup$ – J. W. Tanner May 15 at 16:52
5
$\begingroup$

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 $a+b\sqrt{2}$.)

$\endgroup$
1
$\begingroup$

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 $0=0*(b,b')=((a,a')+(-a,a'))*(b,b')=(b,b')+(b,b')=(2b,b')$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.