I am studying the wikipedia proof of the Frobenius theorem. At a certain point it is shown that the divisionalgebra $D$ can be composed by a direct sum: $D=\mathbb{R}\oplus V$, as $\mathbb{R}$-vector spaces where $V$ is a linear (vector) subspace of $D$. Then in the final part of the proof they say this:
Let $W$ be a subspace of $V$ that generates $D$ as an algebra and which is minimal with respect to this property.
My question now revolves around this: How can a subspace of $V$ still manage to generate D? It would seem to me that $W$ could never generate the entrire algebra $D$, since $D=\mathbb{R}\oplus V$ implies you need the whole of $V$ to accomplish that?
Could someone please explain to me how this is possible?
-Edit- After reading the first answer and comment it has become clear to me that I need to expand my question. I would like to understand what it means when someone says a vector space generates an algebra.
It's true that no subspace of $V$, not even all of $V$ itself, can generate $D$ as a vector space. That's because the only operations available are addition and the vector space "scaling," and $V$ is closed under both of these.
However, if we throw in the algebra multiplication, things change completely. By taking sums and products of things in $V$, it is still possible to generate all of $D$.
For (a slightly tangential) example, $\Bbb R[x]=x\Bbb R\oplus W$ as $\Bbb R$ vector spaces for some subspace $W$, and yet the smallest $\Bbb R$ algebra containing $x\Bbb R$ is $\Bbb R[x]$, so it generates the entire polynomial ring as an algebra.
"$X$ generates $Y$ as a something" is usually defined to mean that $Y$ is the smallest something containing $X$ as a subset. Most algebraic structures are stable under intersection, so this is usually equivalent to saying "the intersection of all somethings containing $X$ is the something $Y$." $X$ can be any subset whatsoever, and if $Y$ is an algebra, then $X$ could be a subspace (or not).
For example, the smallest vector space containing $x$ in $\Bbb R[x]$ is just $x\Bbb R$. Thus $x$ generates $x\Bbb R$ as a vector space. But you can't get $x^2$ using only vector space operations: there is no way to escape x$\Bbb R$ with addition and scaling alone.
But the smallest $\Bbb R$ algebra in $\Bbb R[x]$ containing $x$ is all of $\Bbb R[x]$. Why should this be? Well unlike the last paragraph, an algebra containing $x$ must also contain $x^2$, $x^3$ and all the other powers, so any polynomial can be generated with $x$ and $\Bbb R$.