Validity of Vector Space Comprised of Subspaces I have a homework question that is as follows:

Determine whether or not the following set V = (F, G) is a valid vector space.


Working through the axiomatic definition of a vector space, I believe that: 
1) Since P and Q are defined as vector spaces, they must both contain zero vectors.  So, there exists a g = (p, q) that results in the zero vector.
2) Addition of two elements in G is associative and commutative, based on the definition given above.
3) Multiplication is also associative and commutative.
However, I am struggling to wrap my head around the idea that the sum of any two elements in G will still be in G.  What if, given that the field F is the  set of ordered pairs in the cartesian plane, P and Q are the x-axis and y-axis, respectively?  Wouldn't some addition of any points p and q then lie outside of the set, having components in the x and y directions?
 A: In a sense, you're right: The addition of things in $P$ and $Q$ will cause us to leave both $P$ and $Q$, typically -- but that's not the same as leaving $G$! By considering ordered pairs $(p, 0)$ and $(0, q)$ we've already left $P$ and $Q$ to enter $G$. From this point on, we will stay in $G$.
In your example, we'll "embed" one copy, $P$, of the real line  into our new vector space $G$ by sending a vector (real number) $x \in P$ to the vector $\overline{x} = (x, 0) \in G$.
We'll do the same with $Q$, sending $y \in Q$ to the vector $\overline{y} = (0, y) \in G$. So for example the sum $\overline{x} + \overline{y} = (x, y)$ is still something in $G$, the Cartesian plane.
Note also that in your example, the field $F$ is still the field of real numbers. The set of ordered pairs in the Cartesian plane form a vector space (over $\Bbb R$) but do not themselves form a field (namely, in a field we need to be able to multiply; we're not going to multiply any ordered pairs).
You should also make sure you can write down the $0$ vector of this new space; it's exactly what you'd expect it to be. I only comment on this because you say "there exists a $g = (p, q)$ that results in the zero vector," when it's not hard to say precisely what that zero vector must be.
