# Behavior of kernels of homomorphisms of polynomial rings under a change of the field of coefficients

Suppose $k,k'$ are fields such that $k\subset k'$ and $x_1,x_2,...,x_n,t$ are indeterminates over $k'$. Suppose we have a homomorphism of rings given by $f:k[x_1,x_2,...,x_n]\to k[t]$, where $$x_1\to f_1(t)$$ $$x_2\to f_2(t)$$ $$\vdots$$ $$x_n\to f_n(t)$$ Let the kernel of this map be $I$.

What can be said about the kernel $I'$ of the map $f':k'[x_1,x_2,...,x_n]\to k'[t]$, where restriction of $f'$ to $k[x_1,...,x_n]$ is $f$ (note this completely determines $f'$ since it determines its action on the indeterminates).

The ideal $I$ represents the "ideal of relations" for the polynomials $f_1,...,f_n$. So it seems there may be a compact description for $I'$ in terms of the generators of $I$.

In particular what can we say when $k=\mathbb{Q}$ and $k'=\mathbb{C}$. This case is important because Macaulay2 can compute kernels of such maps when the field of coefficients is $\mathbb{Q}$. So we can use those results to compute the kernels of such maps when the field is $\mathbb{C}$. (For the same reasons what can be said when $k$ is a finite field?)

-

If $I$ is the original kernel, you have a short exact sequence $$0\to I\to k[x_1,\dots,x_n]\to k[t]\to0$$ Apply the $(\mathord-)\otimes_k{k'}$ functor, which is exact.

-
Thank you. This is perfect. It turned out to be easier than I thought. – Timothy Wagner Dec 9 '10 at 4:16