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Let F be a field and $G=F\times F$ Define addition by $(a,b)+(c,d)=(a+c,b+d)$ and multiplication by $(a,b)\cdot(c,d)=(ac,bd)$

Does these operations define a field on G?

I'm fairly comfortable with the addition part, however its the multiplication part that trips me up. Surely $(ac)^{-1}$ exists since F is a field, and $\frac{1}{a}, \frac{1}{c}$ are each in F since we assumed that F is a field.

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Watch out for zero divisors! – Adam Saltz Nov 30 '12 at 3:40
$(a,c)^{-1}$ does not exist when either $a=0$ or $b=0$. – Lior B-S Nov 30 '12 at 7:24
up vote 0 down vote accepted

Check what happens with $\,(1,0)\cdot (0,1)$ ...

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Ah, right. The inverse would have been division by 0. Thank you, what gave you the hint to check that particular example? – MathScratch Nov 30 '12 at 3:48
Well, experience (the almost unique advantage of being older than someone else): the cartesian product of any ammount of fields is never a domain, let alone a field, because there are zero divisors. – DonAntonio Nov 30 '12 at 3:49
could you tell me the difference between a domain and a field? – MathScratch Nov 30 '12 at 3:51
In a field any non-zero elements has a multiplicative inverse, which is far from being true in a domain (e.g., the integers $\,\Bbb Z\,$ or any polynomial ring over a field) – DonAntonio Nov 30 '12 at 3:55

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