Linear Algebra Done Right: Approach for 1a doesn't make much sense (to me)

Let $$F$$ either be $$\mathbb R$$ or $$\mathbb Z/2\mathbb Z$$. For each of the following subsets of $$F^3$$, determine whether it is a subspace. Be sure to explain your answer.

$$U_1=\{(x_1,x_2,x_3)\in F^3: x_1+2x_2+3x_3=0\}$$; $$U_2=\{(x_1,x_2,x_3)\in F^3: x_1+2x_2+3x_3=4\}$$;

For 1a, what does the solution mean by "properties of addition/multiplication for any field"? It's puzzling because not every field has 2,3 as elements in it. Multiplication/addition are not defined for ANY field; the field itself has its own rule for those 2 operations, right?

For 1b, I notice that 1a uses "properties ... for any field." Why can't you do the same general approach for 1b like so.

Choose an arbitrary $$(x_1,x_2,x_3) \in U_1$$. We know $$(0,0,0) \notin U_1$$, since $$0 + 2(0)+3(0) \neq 4$$. Since we have used properties of addition/multiplication for any field, this statement holds over $$\mathbb R$$ and $$\mathbb Z/2\mathbb Z$$.

Here, I used the 1a approach by taking an approach for any field. But this doesn't work for $$\mathbb Z/2\mathbb Z$$. In short, I'm confused when I should use operations for ANY field vs a specific field; basically, I'm confused on the why the 1a approach is justified.

Any field contains $$1$$, and hence contains elements $$n := 1 + ... + 1$$, where there are $$n$$ copies of $$1$$, for all $$n$$.
In 1b, you're using properties that are not true in all fields: "$$0 \neq 4$$" is not a field axiom, nor a consequence of the field axioms.
• No, that's true in any field: we know that $x0 = x(1 - 1) = x1 - x1 = x - x = 0$ for any $x$ (where each of those equals signs is an application of one of the field axioms, or two applications of the same axiom), and so $0 + 2(0) + 3(0) = 0 + 0 + 0 = 0$ (where the first equality uses the above result (three times) and the second is two applications of the additive identity axiom of a field. Oct 27, 2020 at 23:26