Can a vector space over finite field be written as union of finite number of proper subspaces?

Question

Recently, I solved a problem that says-
If $V$ is a vector space over an infinite field. Prove that, V cannot be written as set-thoretic union of a finite number of proper subspaces.
But is this result true in case of finite field?
. I can't get such an example where a vector space over finite field can be written as union of finite number of proper subspaces.
Can anybody give such an example? Thanks for assistance in advance.

Answer 1

The answer is yes; if $V$ is a finite vector space (so over a finite field and of finite dimension), then it has only finitely many elements. Let $\langle v\rangle$ denote the span of a vector $v\in V$. Then clearly $$V=\bigcup_{v\in V}\langle v\rangle,$$ because $v\in\langle v\rangle$ for every $v\in V$, and the union is finite because $V$ is finite. Hence $V$ is the union of finitely many proper subspaces (if $\dim V>1$, otherwise the subspaces aren't proper).

For a very concrete example, consider $\Bbb{F}_2^2$, a $2$-dimensional vector space over the finite field $\Bbb{F}_2$ of two elements. Then \begin{eqnarray*} \Bbb{F}_2^2&=&\{(0,0),(1,0),(0,1),(1,1)\}\\ \bigcup_{v\in\Bbb{F}_2^2}\langle v\rangle&=&\{(0,0)\}\cup\{(0,0),(1,0)\}\cup\{(0,0),(0,1)\}\cup\{(0,0),(1,1)\}. \end{eqnarray*}

Answer 2

It seems to me that the case of an infinite dimensional space is creating extra headache. Spelling out an example to show what can be done.

Consider the field $\Bbb{F}_2$. Let $V=\Bbb{F}_2[x]$ be the space of univariate polynomials. Let $U$ be the subspace of polynomials with zero constant and linear terms. Consider the following three subspaces of $V$: $$ V_1=\langle 1\rangle\oplus U=\{a_0+a_1x+a_2x^2+\cdots +a_nx^n\mid n\in\Bbb{N}, a_i\in\Bbb{F}_2, a_1=0\}, $$ $$ V_2=\langle x\rangle\oplus U=\{a_0+a_1x+a_2x^2+\cdots +a_nx^n\mid n\in\Bbb{N}, a_i\in\Bbb{F}_2, a_0=0\}, $$ and $$ V_3=\langle 1+x\rangle\oplus U=\{a_0+a_1x+a_2x^2+\cdots +a_nx^n\mid n\in\Bbb{N}, a_i\in\Bbb{F}_2, a_0=a_1\}. $$ It is clear that every polynomial of $V$ belongs to at least one of the subspaces $V_1,V_2,V_3$ according to what its constant and linear terms look like. Therefore $$V=V_1\cup V_2\cup V_3.$$

---

What just happened?

Consider the quotient space $V/U$. It is 2-dimensional. Its non-zero elements are the cosets $1+U,x+U$ and $1+x+U$. So we can write $V/U$ as a union of three 1-dimensional subspaces following the recipe of Servaes' answer each spanned by one of those cosets. The resulting subspaces are exactly the spaces $V_i/U, i=1,2,3$. No wonder that $V_1,V_2,V_3$ cover all of $V$!