Suppose $F = \mathbb{Q}^2$ is a field equipped with the following operations:
Addition: Defined entrywise i.e.
$$(x_1, x_2) + (y_1, y_2) = (x_1 + y_1, x_2 + y_2)$$
Multiplication:
$$(x_1, x_2) \cdot (y_1, y_2) = (x_1y_1 + 2x_2y_2, x_1y_2 + x_2y_1)$$
Given $(x_1, x_2), (y_1, y_2) \in F$, define $(x_1, x_2) < (y_1, y_2) \iff x_1 + x_2 < y_1 + y_2$, as defined on $\mathbb{R}$. Show that this is an ordered relation on $F$.
My attempt:
Trichotomy: Choose $(x_1, x_2), (y_1, y_2) \in F$.
Case 1: $x_1 + x_2 < y_1 + y_2 \implies (x_1, x_2) < (y_1, y_2)$
Case 2: $x_1 + x_2 > y_1 + y_2 \implies y_1 + y_2 < x_1 + x_2 \implies (y_1, y_2) < (x_1, x_2) \implies (x_1, x_2) > (y_1, y_2)$
Case 3: $x_1 + x_2 = y_1 + y_2 \implies (x_1, x_2) = (y_1, y_2)$
Therefore, trichotomy is satisfied.
Transitivity: Consider $(x_1, x_2), (y_1, y_2), (z_1, z_2) \in F$. Assume that $(x_1, x_2) < (y_1, y_2)$ and $(y_1, y_2) < (z_1, z_2)$. We have that:
$(x_1, x_2) < (y_1, y_2) \implies x_1 + x_2 < y_1 + y_2$, and
$(y_1, y_2) < (z_1, z_2) \implies y_1 + y_2 < z_1 + z_2$
$\implies x_1 + x_2 + y_1 + y_2 < y_1 + y_2 + z_1 + z_2$
$\implies x_1 + x_2 < z_1 + z_2$
$\implies (x_1, x_2) < (z_1, z_2)$
Therefore, transitivity is satisfied.
This is an ordered relation because the axioms above are both satisfied.
Is the solution above correct? I feel as though I made a mistake or just made an incorrect conclusion somewhere. Any assistance is appreciated.