Show that there exist $a_1,\cdots,a_n\in k$ such that $f(a_1,\cdots,a_n)\ne 0$.

Question

Let $k$ be an infinite field, and let $f$ be a nonzero polynomial in $k[X_1,\cdots,X_n]$. Show then, that there exist $a_1,\cdots,a_n\in k$ such that $f(a_1,\cdots,a_n)\ne 0$.

Is there something illuminating about this exercise?

We say that $f\ne 0$ in $k[X_1,\cdots,X_n]$, and hence there exist $b_1,\cdots,b_n\in k$ such that $f(X_1,\cdots,X_n)=b_1X_1+\cdots+b_nX_n\ne 0$

Then the result follows immediately. Is there something I am missing?

Your argument does not work. Take $X + X^2$ over the field with 2 elements. — J.-E. Pin, Jan 10 '16 at 23:05
What are you doing? You should find elements form $k$ on which $f$ doesn't vanish. How do you get the $b_i$? — , Jan 10 '16 at 23:05
@sigmabe That's my point, the polynomial is nonzero, so there is immediately some points $b_1,\cdots,b_n\in k$ such that the function is nonzero. — Skies burn, Jan 10 '16 at 23:06
Precisely ! You have to use this hypothesis, as shown by my example, and you don't. — J.-E. Pin, Jan 10 '16 at 23:07
Not every polynomial is a linear combination of the $X_i$. How do you immediately get the $b_i$ for $f(X,Y)=X^7Y^2-X^3-7$? — , Jan 10 '16 at 23:10
@sigmabe You're right. Thanks. I am too practiced in single variable rings! — Skies burn, Jan 10 '16 at 23:10
@J.-E.Pin Thanks, I see now, a nonzero polynomial may evaluate to zero at all points, but still as a polynomial in an indeterminate may be nonzero — Skies burn, Jan 10 '16 at 23:15

score 3 · Accepted Answer · answered Jan 11 '16 at 02:43

In one variable this follows from the fact that a polynomial of degree $n$ has at most $n$ roots. In multiple variables we can use induction. Say we have a polynomial $f(x_1,x_2,\ldots,x_n)$. If $f(x_1,x_2,\ldots,x_{n-1},0)$ is nonzero then this reduces to the case of $n-1$ variables. If instead $f(x_1,x_2,\ldots,x_{n-1},0)$ is zero, write $f$ as a polynomial in $x_n$: $$f(x_1,\ldots,x_n)=\sum_{i}{f_i(x_1,\ldots,x_{n-1})x_n^i}$$ Then there is some $i$ such that $f_i(x_1,\ldots,x_{n-1})$ is nonzero. Use the induction hypothesis to specialize $x_1,\ldots,x_{n-1}$ so that $f_i$ evaluates to something other than $0$. This brings us back to the one variable case, a nonzero polynomial in $x_n$. A value for $x_n$ making this nonzero exists as before, and the result follows.

Show that there exist $a_1,\cdots,a_n\in k$ such that $f(a_1,\cdots,a_n)\ne 0$.

1 Answers1