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
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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 $(1,1)$.
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Every vector space has the two trivial subspaces: itself and the zero vector space.
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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) \} $
Best Answer
This answer is not feedback on your work but a nice conceptual way to think about this. For feedback see the answer by 5xum.
Let's work a bit more general: Let $K$ be a field with $q$ elements.
As the dimension of $K^2$ as $K$-vector space is two, any non-trivial subspace will have dimension at most $1$. Now any non-zero vector $v\in K\setminus \left\{0\right\}$ will generate a one-dimensional subspace (and there are $q^2-1$ such vectors) but beware you might be counting several subspaces multiple times. Indeed, any non-zero multiple of $v$ will generate the same subspace (and there are $q-1$ such non-zero multiples). It follows that the number of one-dimensional subspaces of $K^2$ is given by
$$\frac{q^2-1}{q-1}=q+1.$$
Hence in your case there are $3$ one-dimensional subspaces and two trivial subspaces. Hence $5$ subspaces in total.