non isomorphism between $A^1$ and zero set of $y^2-x^3=0$'s coordinate ring by cohomology argument?

Question

Consider $A_k^1=\mathrm{Spec}(k[s])$ and $V=\mathrm{Spec}(k[x,y]/(y^2-x^3))$. Suppose $\mathrm{char}(k)\neq 2,3$. There are various ways (dimension of tangent space, integral closure and constructing ring homomorphism...) to show $A^1$ not isomorphic to $V$ as schemes.

Q: Is there a cohomological argument to show $A^1$ not isomorphic to $V$?

I know $A^1$ is normalization of $V$ for sure. My guess is that ring $K(-)$ functor will identify the difference between the two schemes. $K(k[s])$ should be isomorphic to $Z$ as every f.g. projective is free. What is $K(k[x,y]/(y^2-x^3))$ then? What are other alternatives to see two schemes different?

Answer 1

Here is a cohomological argument: we have for $\mathbb A^1_k=\operatorname {Spec}(k[s])$ $$H^1(\mathbb A^1_k,\mathcal O^*)=\operatorname {Pic} \mathbb A^1_k=0$$ whereas for $V=\operatorname {Spec}(k[x,y]/\langle y^2-x^3\rangle)$ $$H^1(V,\mathcal O^*)=\operatorname {Pic} V=k.$$ The triviality of $\operatorname {Pic}$ in the first displayed line follows from $\mathbb A^1_k$ being the spectrum of $k[s]$, a principal ring.
The result in the second displayed line is explained in Chapter I, Example 3.10.1 of Weibel's The $K$-book.
Weibel generously allows you to download his book for free here.

Remarks
1) The field $k$ in the above is completely arbitrary: it needn't be algebraically closed and it can have arbitrary characteristic, including $2$ or $3$.
2) Since you ask, we have (Weibel, Chapter II,Example 2.9.1) $$K_0(k[x,y]/\langle y^2-x^3\rangle)=\mathbb Z\oplus k$$

Answer 2

Another cohomological argument: compute their Yoneda coalgebras (i.e. Tor with trivial coefficients). The polynomial ring $\mathbb k[t]$ is Koszul and has vanishing Tor in dimensions two and higher. On the other hand, the Yoneda coalgebra of the cusp has nonvanishing Tor in degree two.