Is this a field?

Let $S$ be the set of all the ordered pairs in the cartesian plane. That is: $$S=\{(x,y)|\ \ x, y \in \Bbb{R}\}$$ Then, If $a=(a_1, a_2)$ and $b=(b_1, b_2)$ are two arbitrary elements of $S$, the following operations are defined: $$a+b=(a_1+b_1, a_2+b_2)$$ $$a\times b=(a_1\cdot b_1, a_2 \cdot b_2)$$

Also, let's state that $a=b$ iif $a_1=b_1$ and $a_2=b_2$. Associativity and commutativity of the operations are straight forward to prove. The multiplicative neutral element is $u=(1,1)$ and the additive neutral element is $o=(0,0)$. It's trivial to show why thse two elements are the respective neutral elements.

Now, on additive inverses. For the arbitrary element $a\in S$, we can take $-a=(-a_1, -a_2)$, therefore:
$$a+(-a)=(a_1+(-a_1), a_2+(-a_2))=(0,0)=o$$ This proves there are additive inverses.

On multiplicative inverses, let's consider $b\neq o$, then if we take $b^{-1}=(\frac{1}{b_1}, \frac{1}{b_2})$, then:

$$b\times b^{-1}=\bigg(b_1\cdot\frac{1}{b_1}, b_2\cdot \frac{1}{b_2}\bigg)=(1,1)=u$$ which shows the existence of multiplicative inverses.

Finally, the distributive property: Let $c=(c_1, c_2)$ an element of $S$. Then:

$a\times (b+c)=(a_1,a_2)\times (b_1+c_1, b_2+c_2)=(a_1b_1+a_1c_1, a_2b_2+a_2c_2)$

On the other hand,

$a\times b+ a\times c =(a_1b_1, a_2b_2)+(a_1c_1, a_2c_2)=(a_1b_1+a_1c_1, a_2b_2+a_2c_2)$. Which means that $a\times(b+c)=a\times b+a\times c$.

Did I make a mistake or fail to notice something? On the other hand, I'd love some help with the writing or style.

1 Answer

Hint: What's the multiplicative inverse of $(1, 0)$?

• That's right! All elements that contain at least one zero will fail to have an inverse multiplicative. Could this be solved by removing all such elements and leaving the $(0,0)$ from $S$? – NotAMathematician Apr 10 '18 at 6:31
• @NotAMathematician If you do that, then what is $(1, 1) + (1, -1)$? – Arthur Apr 10 '18 at 6:33
• Imagine you remove (b,0), b≠0. Take (b,a) and (b,-a), a in \mathbb{R}. Then (1/2b,1/2b)*((b,a)+(b,-a))=(b,0). Would you be able to solve this problem? – Ángela Flores Apr 10 '18 at 6:37
• To mend the situation by removal of elements, the "only" way is to remove all $(x,y)$ with $x\ne y$. To mend the situation with keeping $S$, the "only" way us to define $a\times b=(a_1b_1-a_2b_2, a_1b_2+a_2b_1)$ instead. (Even if there are different ways possible, there is justification in me writing "only") – Hagen von Eitzen Apr 10 '18 at 6:38
• @HagenvonEitzen One can, however, also mend the situation by quotienting out an ideal, or localizing. – Arthur Apr 10 '18 at 6:49