Veronese embedding as ring isomorphism At the level of rings, the Veronese map corresponds to isomorphisms like
$$k[w,x,y,z]/(wz - xy, wy-x^2, xz - y^2) \cong k[a^3, a^2 b, a b^2, b^3].$$
This isomorphism is the statement that the relations satisfied by the monomials of degree 3 generate exactly the same ideal as $I = (wz - xy, wy-x^2, xz - y^2)$. One direction is obvious (corresponding to a well-defined map $\to$): the monomials of degree 3 certainly satisfy these three relations. 
The other direction seems hard. It encodes the idea that we have found all the relations, which is certainly an extremely hard general problem. How would one prove such a thing explicitly? 
I believe all of this can be formulated in much higher geometric language as an isomorphism of $\operatorname{Proj}$ constructions, but that language should surely encode a concrete argument.
 A: I suspect there are fancier more systematic ways to prove it, but here's a simple hands-on proof.   By the easy direction we have a surjective homomorphism $f:R\to S$ where $R=k[w,x,y,z]$ and $S=k[a^3, a^2 b, a b^2, b^3]$, and the kernel of $f$ contains $I$.  To show that the kernel of $f$ is contained in $I$, it suffices to show the following.

Claim: If monomials $m=w^i x^j y^k z^\ell$ and $m'=w^{i'}x^{j'}y^{k'}z^{\ell'}$ are such that $f(m)=f(m')$, then $m\equiv m'\pmod I$.

Why does this suffice?  Well, since $f$ maps monomials to monomials, if $p\in \ker f$, then the monomials in $p$ must map to identical monomials in $S$ whose coefficients cancel out (since monomials in $S$ are linearly independent, that's the only way $f(p)$ can be $0$).  But then the Claim would show that the monomials of $p$ also cancel out mod $I$, so $p\in I$.
So, let's prove the Claim. We have $f(m)=a^{3i+2j+k}b^{j+2k+3\ell}$ and similarly for $m'$, so we must have $$3i+2j+k=3i'+2j'+k'$$ and $$j+2k+3\ell=j'+2k'+3\ell'.$$
We will use induction on $i+i'+\ell+\ell'$.  Let us first suppose that $i+i'+\ell+\ell'=0$.  Then our equations are just $2j+k=2j'+k'$ and $j+2k=j'+2k'$.  This easily implies $j=j'$ and $k=k'$, and so $m=m'$ and we are done.
Now suppose $i+i'+\ell+\ell'>0$; we may assume without loss of generality that $i>0$.  If $i'>0$, we can cancel $w$ from both $m$ and $m'$ to reduce to a case where $i+i'+\ell+\ell'$ is smaller.  If $\ell>0$, we can use the relation $wz\equiv xy\pmod{I}$ on $m$ to decrease $i$ and $\ell$ by $1$ and increase $j$ and $k$ by $1$, reducing to a case where $i+i'+\ell+\ell'$ is smaller.  Similarly if $k>0$ we can use the relation $wy\equiv x^2\pmod{I}$ on $m$ to reduce to a case where $i+i'+\ell+\ell'$ is smaller.
So, we may assume that $i'=\ell=k=0$.  Our two equations then reduce to $$3i+2j=2j'+k'$$ and $$j=j'+2k'+3\ell'.$$  But now observe that the second equation implies $2j\geq 2j'+k'$ which makes the first equation impossible since $i>0$.  Thus this case is impossible, completing the proof.
