I have just started out proofs so I am not fully grasping this concept yet.
Question:
Is the set $\mathbb{R}^2$, with addition and multiplication defined below a field? Explain.
-
$(a, b) + (c, d) = (a + c, b + d)$
-
$(a, b) (c, d) = (ac, bd)$
Attempted solution:
We know: $f\colon F \times F \rightarrow F$
$$f(x, y) = x + y$$
$g\colon F \times F \rightarrow F$
$$g(x, y) = xy$$
So solving 1:
$$(a, b) + (c, d) = (a + c, b + d)$$
$$
\begin{align*}
f(a, b) + (c, d) &= (a + b) + (c + d) &&\text{definition of field} \\
&= (a + c) + (b + d) \\
&= (a + c, b + d)
\end{align*}
$$
The second part I did almost the same thing using $g(x, y) = xy$.
Sorry for the formatting, I still don't know how to use MathJaX but any help is appreciated. If you can also explain for a beginner that would be great, thanks.
Best Answer
Your attempted solution apparently amounts to substituting the definition of addition and multiplication, but these are given to you so there's no need to "find them". In fact, there are lots of different ways to define addition and multiplication. Most of them would not result in a field (think of $(a,b)+(c,d) = (0,0)$). On the other hand, there might be more than one which will result in a field. So the question really depends on the definition of addition and multiplication.
In order to check whether this object is a field, you have to verify all the field axioms. For example, in a field $x + y = y + x$. In your case, we need to verify that $$(x_1,x_2) + (y_1,y_2) = (y_1,y_2) + (x_1,x_2).$$ All we need to do is substitute the definition of addition: $$(x_1,x_2) + (y_1,y_2) = (x_1+y_1,x_2+y_2)$$ whereas $$(y_1,y_2) + (x_1,x_2) = (y_1+x_1,y_2+x_2),$$ and both expressions are equal by the commutativity of real addition.
Some of the field axioms require you to identify a zero element and a unit element, i.e. the $0$ and $1$ of the field. What should these be? Do they satisfy all the required axioms? For example, is every non-zero element invertible?