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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) \} $
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    $\begingroup$ The set $V=\langle (1,0)\rangle$, vector subspace of $K^2$ generated by $(1,0)$, is a subspace. $\endgroup$
    – Wuestenfux
    Sep 17, 2018 at 9:51
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    $\begingroup$ $V=\{(0,0),(1,0)\}$. $\endgroup$
    – Wuestenfux
    Sep 17, 2018 at 16:20
  • $\begingroup$ $\mathbb{Z} / 2 \mathbb{Z} =\{2\mathbb{Z},1+2\mathbb{Z}\}\neq\{0,1\}$. I think it could have been mentioned here before the usual notation abuse. $\endgroup$ Oct 9, 2022 at 6:52
  • $\begingroup$ $\mathbb{Z} / 2 \mathbb{Z} \neq\{0,1\}$; for example, if we are interested in $(\mathbb{Z} / 3 \mathbb{Z})^2$, it will be preferable in my opinion to write $\mathbb{Z} / 3 \mathbb{Z}=\{-1,0,1\}$. $\endgroup$ Oct 9, 2022 at 7:48

2 Answers 2

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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.

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  • $\begingroup$ So the non-trivial subspaces are $\{0\} \times \{0,1\}$, $\{0,1\} \times \{0\}$ and which other one? $\endgroup$ Sep 17, 2018 at 16:03
  • $\begingroup$ @ViktorGlombik What's the third non-trivial element of $K^2$? $\endgroup$
    – 5xum
    Sep 17, 2018 at 17:22
  • $\begingroup$ $(1,1)$, but $(1,1) + (1,1) = (0,0)$, so $\{1\} \times \{0,1 \}$ can't be a subspace, since it's not closed under addition in $K$, right? $\endgroup$ Sep 17, 2018 at 18:12
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    $\begingroup$ The set $\left\{(0,0),(1,1)\right\}$ is a subspace. The set $\left\{1\right\}\times \left\{0,1\right\}$ can't be a subspace as any subspace contains the neutral element of the underlying group. Hence any subspace must contain $(0,0)$. $\endgroup$ Sep 17, 2018 at 19:39
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    $\begingroup$ @ViktorGlombik It might help to write down $\{0\}\times \{0,1\}$ as $\{(0,0), (0,1)\}$ and $\{0\}\times\{1,0\}$ as $\{(0,0), (1,0)\}$. This gives you a better idea what the third subspace should be. $\endgroup$
    – 5xum
    Sep 18, 2018 at 7:55
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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.

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