Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?)

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.