Question: > 
> I wish to show that the injection $k[y^2, y^3] \rightarrow k[y]$ is not flat. 
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I know of geometric ways to see this, but I wish to see explicitly $k[y^2, y^3]$-modules (or localizations thereof) $0 \rightarrow M' \rightarrow M$ which does not remain injective upon tensoring with $k[y]$. Is such a search futile?

Answer: 
The "geometric proof" automatically produces an algebraic proof. The singularity on the cusp corresponds to the maximal ideal $(y^2,y^3) \subseteq k[y^2,y^3]$.

The inclusion $(y^2,y^3) \subseteq k[y^2,y^3]$ doesn't stay injective when tensored with $k[y^2,y^3] \to k[y]$. In fact, $y^2 \otimes y$ and $y^3 \otimes 1$ have the same image under $(y^2,y^3) \otimes\_{k[y^2,y^3]} k[y] \to k[y]$, but are not equal.
I'd like to add a short algebraic proof (although this is not exactly what you asked for) which shows [again](https://math.stackexchange.com/questions/385364/integral-closure-tildea-is-flat-over-a-then-a-is-integrally-closed) the usefulness of Exercise 7.2 in Matsumura, *Commutative Ring Theory*: 

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> If $A\subset B$ are integral domains with the same field of fractions and $B$ is faithfully flat over $A$, then $A=B$. 
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In your case $A=k[y^2,y^3]$ and $B=k[y]$. If the extension is flat, then it is faithfully flat (since it is integral) and therefore $A=B$, a contradiction.