A field as a vector space over itself is not the direct sum of two proper subespaces Let $K$ be a field. Then $K$ over $K$ is not the direct sum of two proper subspaces.
Assume $K = V_1 \oplus V_2$ with $V_1,V_2$ proper subespaces, then, for every $k \in K$, $k = v_1 + v_2$ then $k-v_2 \in V_2$ but also $k-v_2 +v_2 \in V_2$, i.e $k \in V_2$ for all $k$, so $V_2$ is not actually proper.
This seems wrong to me (but I'm not sure where), I didn't even use that it was a field over itself.
Is this correct? If not, can you give me a hint?
 A: Write $1$ as a sum $x+y$, with each summand in each direct summand. If you multiply by $x$ the equality $1=x+y$, you get $x=x^2+xy$, so that $x-x^2=xy$. Now the left hand side of this equality is in one of the direct summands and the right hand one is in the other. Since the sum is direct, we have that one of $x$ or $y$ is zero. It follows that $1$ is in one of the direct summands and therefore that summand is in fact all of $k$.
A: If $K=V + U$, internal direct sum as $K$-vector spaces, then $V$ and $U$ are also ideals of $K$ hence, since $K$ is a field, either $V=K$ and  $U=0$, or $V=0$ and $U=K$.
A: Kind of a weird answer that uses the concept of 'dimension'. Maybe I can't use 'dimension' because this question's answer is a prerequisite to defining the concept of 'dimension'. Here goes:


*

*$\dim_K K = 1$

*$\dim_K K = \dim_K V_1 + \dim_K V_2$

*For any $K$-subspace $X$ of $K$, $X$ is proper if and only if $\dim_K X \ge 1$

*$V_i$ is proper if and only if $\dim_K V_i \ge 1$

*Both $V_i$ are proper if and only if $1 = \dim_K K = \dim_K V_1 + \dim_K V_2 \ge 2$
