$K$ is a field and $K[X_1,\dots, X_n]$ and $K[X_1,\dots, X_m]$ are polynomial algebras in $n$ and $m$ indeterminates respectively.
Suppose there is a $K$-algebra homomorphism $\phi:K[X_1,\dots, X_n] \to K[X_1,\dots, X_m]$. Prove that:
If $\phi$ is injective, $n\leq m$.
If $\phi$ is surjective, $n\geq m$.
My idea so far has been to try and exploit the vector space structure of polynomials but I haven't gotten anything concrete yet.
An idea for injectiveness: If $\phi$ is injective, then there exists a sub-algebra $A$ of $K[X_1, X_2,\dots, X_m]$ that is isomorphic to $K[X_1, X_2,\dots, X_n]$. $A$ can be in at most $m$ indeterminates because it is a subalgebra of an algebra generated by $m$ elements. This forces $n\leq m$. Does this work?
Best Answer
Let $\phi$ be a surjection from $K[X_1,\dots,X_n]$ to $K[X_1,\dots,X_m]$. Let $n<m$. Then there is a natural surjection $\psi$ with non-zero kernerl from $K[X_1,\dots,X_m]$ to $K[X_1,\dots,X_n]$. Composing the two, we get a surjective endomorphism of $K[X_1,\dots,X_m]$ that is not an injection. We know that $K[X_1,\dots,X_m]$ is Noetherian. Hence, surjective endomorphisms of $K[X_1,\dots,X_m]$ are automorphisms. Contradiction. Hence, $n\geq m$.
Let $\phi$ be an injection. We know that the polynomials of degree less than are equal to $r$ in $m$ variables form a vector space of dimension ${m+r}\choose{m}$. Let $d$ be the maximum degree of $\phi(X_i)$ for all $i$. Then the polynomials with degree less than or equal to $r$ map to a vector space of dimension ${n+rd}\choose{n}$. Because $\phi$ is injective, ${{m+r}\choose{m}}\leq {{n+rd}\choose{n}}$. Looking at this as an inequality of an $m$ dimensional and $n$ dimensional polynomials in $r$ and $rd$ respectively, we can conclude that $n\leq m$.