Conditions under which $\Bbb{R}^2$ is a field.

Question

On my assignment one of the questions asks me to prove that $\Bbb R^2$, a set containing ordered pairs of real numbers, with the operators:

$(a,b)+(c,d)=(a+c,b+d)$

$(a,b)\cdot(c,d)=(ac,bd)$

is not a field.

This is because of the multiplicative inverse correct? My understanding is that the inverse of an ordered pair $(a,b)$ would be $(b,a)$, and in the event that $a=0$ or $b=0$ no inverse exists to make this true. However, the axiom of multiplicative inverses states that every non-zero element has a multiplicative inverse in a field, contradicting my above statement.

So what is the actual definition of a multiplicative inverse in $\Bbb R^2$?

Any help is appreciated.

Before you think about multiplicative inverses, you should work out what the multiplicative identity of $\Bbb{R}^2$ is. — Chris Eagle, Jan 12 '13 at 16:14
Think of other fields you know... and the problem will solve itself. — Z. L., Jan 12 '13 at 16:19
I think you typed your multiplicative inverse wrongly. Please check. — Calvin Lin, Jan 12 '13 at 16:23
Possible duplicate: http://math.stackexchange.com/questions/229836/explain-why-mathbbz-times-z-and-mathbbr-times-r-is-not-a-field — Julian Kuelshammer, Jan 12 '13 at 16:54

Thomas · Answer 1 · 2013-01-12T16:24:47.427

3

If $\mathbb{R}^2$ was a field under the operations that you have listed, then (clearly) $(1,1)$ would be the (multiplicative) identity and $(0,0)$ would be the (additive) identity. But for $\mathbb{R}^2$ to be a field, you would need that for every $(a,b) \neq (0,0)$ there is a multiplicative inverse, that is given $(a,b)\neq (0,0)$ you should be able to find a unique $(c,d)$ such that $(a,b)(c,d) = (1,1)$. That is $(ac, bd) = (1,1)$.

So to answer the specific question that you ask, the multiplicative inverse of $(a,b)$ is (should be) $(a^{-1}, b^{-1})$.

Now consider for example an element of the form $(a,0) \neq (0,0)$ (as you do in your question).

edited Jan 12 '13 at 16:24

answered Jan 12 '13 at 16:19

Thomas

44,491

1

Isn't that exactly what he states in the question? – Z. L. Jan 12 '13 at 16:21
@ZangoLotino: Hmmm.. yes, you are right. Let me just add a bit more... – Thomas Jan 12 '13 at 16:23

Alexander Gruber · Answer 2 · 2013-01-12T16:26:00.023

To be a field $\mathbb{R}^2$ must first be an integral domain, and thus have no zero divisors; consider $(a,0)\cdot(0,b)$ for $a,b\not= 0$.

Alternatively we see that $(1,1)=1_{\mathbb{R}^2}$, so we must be able to solve $(a,0)\cdot x=1_{\mathbb{R}^2}$. However from the definition of $\cdot$, we see there is no $x$ which will take that $0$ to $1$ in the second slot, so no multiplicative inverse of $(a,0)$ exists.

Conditions under which $\Bbb{R}^2$ is a field.

2 Answers2