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over the sub group $A = \{(x,1)|x \in \Bbb R\}$ of $\mathbb R^2$ we will define two operations:

Addition marked with $\oplus$: for every $ x,y \in \Bbb R$: $(x,1) + (y,1) = (x + y,1)$
Multiplication marked with $*$: for every $ x,y \in \Bbb R$ is defined as $(x,1) * (y,1) = (xy, 1)$.

I'm being requested to answer if $(A, \oplus, *)$ is a Field ?

I have few points which confuse me which I'll appreciate if you can share your view.

The operation is conducted on the horizontal line which pass along Y=1 for any X $\in\mathbb R$. when you multiply the two vectors the answer is correct (x,1)(y,1) = (xy,1) as x,y,1 $\in\mathbb R$ and xy = xy and 1*1 = 1. But when you sum the vectors (x,1) + (y,1) = (x+y, 1+1) = (x+y, 2) since x,y,1 $\in\mathbb R$. However, this is not a normal sum but a $\oplus$ and on the sub group A. but should sum of two vectors on A should be as in $\mathbb R^2$ ? another aspect, if the sum should be indeed different, then 1+1 means that 1 is the identity for both the * operation and the $\oplus$ operation, and this can not be, to have the same identity element for different operations in a field.

I'll appreciate your point of view on the matter, or if I got it all wrong.

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  • $\begingroup$ Incorrect, you got the trivial field where one element only exists. $\endgroup$ – Zelos Malum Oct 25 '16 at 3:25
  • $\begingroup$ Why do you say that $1$ acts as identity for addition? Note that $1$ is not an element of $A$. $\endgroup$ – nombre Oct 25 '16 at 6:52
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The confusion you are having is that $\oplus$ is not the usual addition that you are used to in $\mathbb{R}^2$. Instead of $(x,1)+(y,1)=(x+y,2)$ as would normally be defined, you are asked to consider $(x,1)\oplus (y,1)=(x+y,1)$. My advise would be to try to check that $A$ is a field with the operations $\oplus, *$. However, in contrast to what you wrote in your first line, the additive group $(A,\oplus)$ will not be a subgroup of $(\mathbb{R}^2,+)$ since the addition $+$ on $\mathbb{R}^2$ does not restrict to the addition $\oplus$ on $A$.

If you want to be less confused as a first step, try proving that $\{(x,0)|x\in \mathbb{R}\}$ is a field, in fact the addition and multiplication restrict from the ones on $(\mathbb{R}^2,+,\cdot)$.

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