$K = \mathbb{Z} / 2 \mathbb{Z}$. How many subspaces does the $K$-vector space $K^2$ have? 
Let $K = \mathbb{Z} / 2 \mathbb{Z}$, How many subspace does the $K$-vector space $K^2$ have?

I (hope to) already know the following

*

*When diving a whole number by 2, we can only obtain 0 or 1, so $K = \{0, 1\}$ and $K^2 = \{0, 1\} \times \{0, 1\}$. Therefore $K^2$ has only four elements, $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$.


*Every vector space has the two trivial subspaces: itself and the zero vector space.


*A subspace has to be not empty and closed under addition and multiplication with scalars from $K$.
My Question
Is $V = \{0\} \times \{0,1\} \not= \emptyset$ a $K$-vector subspace? Using the addition from $K$, we obtain $(0,0) + (0,1) = (0,1) \in V$ And $\alpha \cdot (0,0) = (0,0)$ for all scalars $\alpha \in K$ and we obtain $0 \cdot (0,1) = (0,0) \in V$ as well as $1 \cdot (0,1) = (0,1) \in V$ etc.
Short Summary of answers given
Besides the two trivial subspaces mentioned above there are three other subspaces, the reasoning being analogous to the one for $V = \{0\} \times \{0,1\}$ above. They are

*

*$V_1 = \{0\} \times \{0,1\}$

*$V_2 = \{0, 1\} \times \{0\}$

*$V_3 = \{ (0,0), (1,1) \} $

Yes, $\{0\}\times\{0,1\}$ is indeed a $K$-vector subspace. You almost completed your proof, but it has some flaws you should still fix. Remember, to prove that $U\subset V$ is a vector subspace (over some field $F$), you need to prove two things:


*

*For every pair $u_1, u_2\in U$, the element $u_1+u_2$ is also in $U$. In other words, $$\forall u_1, u_2\in U: u_1+u_2\in U$$

*For every $u\in U$, and every $\alpha\in F$, the element $\alpha u$ is also in $U$. In other words, $$\forall u\in U, \alpha\in F: \alpha u\in U$$



So far, you have proven that $(0,0) + (0,1)$ is indeed an element of $V$, but you still have to prove that all pairs of elements from $V$ add up to elements of $V$. This includes pairs of identical elements!
Also, I don't like the sentence

$\alpha \cdot (0,1)$ will be either $(0,0)$ or $(0,1)$ because of the way addition in $K$ works.

which I think you should rewrite. First of all, it's not because of the way addition works, it's about how multiplication works (sure, the two are related, but it's better form). Second, it's better if you just write down all the possibilities, since there are only two options for $\alpha$ anyway.
