How do I know an element generates a coordinate ring K[W] as a vector space over K? I have an example which proves that a cuspidal cubic $W\subset \mathbb{A}^2$ defined by $y^2-x^3=0$ is not isomorphic to to $\mathbb{A}^1$. I'll start by defining a few things:
Let $V=\mathbb{A}^1$, $W\subseteq \mathbb{A}^2$ and let $\phi : W \rightarrow W$ be the morphism $t \rightarrow (t^2, t^3)$. We show $V$ and $W$ are not isomorphic by showing $K[W]$ and $K[V]$ are not isomorphic where these are the coordinate rings of $W$, $V$ respectively. It is given that $I[W] =  \langle x^3-y^2 \rangle$ so we have that $K[W]=K[x,y]/\langle x^3-y^2 \rangle$. I understand the overall reasoning of the question except one part:
What I dont understand is the following:
'$\overline{(x^a*y^b)}$ generates $K[W]$ as a vector space over $K$'
What does this mean? I understand $\overline{(x^a*y^b)}$ is a coset and element of $K[W]$ but how does it generate it? And what does it mean by as a vector space over $K$? Thanks!
 A: One can see that these two rings are not isomorphic because $I[W]=K[x]\oplus K[x]y\not\simeq K[z]$ since every time $y^2$ occurs, it can be replaced by $x^3$.    Now, to answer your questions.
Recall that a vector space over a field $K$ is a commutative group $V$ with a addition operation $+:V\times V\rightarrow V$ and a scalar multiplication operation $\ast:K\times V\rightarrow V$.  These operations are associative and distributive.
In the case of $K[x,y]$, the monomials $x^ay^b$ generate the ring as a vector space over $K$ because every polynomial can be written as a sum of scalars times monomials.  In the quotient, these elements still generate because you have the natural surjection
$$
K[x,y]\rightarrow K[x,y]/\langle x^3-y^2\rangle.
$$
The resulting space is also a vector space because you can multiply by scalars (elements of $K$) and get another element of the quotient (such a multiplication actually commutes with the quotient map).
Now, you must check the various vector space properties.  Most of the properties come directly from the fact that $K[x,y]$ is a vector space.
