0
$\begingroup$

Weak Hilbert's Nullstellensatz says the following: Let $P_1,\ldots,P_m \in k[x_1,\ldots,x_d]$ be polynomials. Then exactly one of the following statements holds: (1) There exists $a \in k^d$ such that $P_1(a)=\ldots=P_m(a)=0$. (2) There exist polynomials $Q_1,\ldots,Q_m \in k[x_1,\ldots,x_d]$ such that $P_1 Q_1 + \ldots + P_m Q_m = 1$.

Are the $Q_i$'s in (2) unique? Namely, if also there exist
polynomials $R_1,\ldots,R_m \in k[x_1,\ldots,x_d]$ such that $P_1 R_1 + \ldots + P_m R_m = 1$, is it true that $R_j=Q_j$, $1 \leq j \leq m$?

Clearly, $P_1 (Q_1-R_1) + \ldots + P_m (Q_m-R_m) = 0$, but I am not sure if this may help in answering my question.

$\endgroup$

1 Answer 1

3
$\begingroup$

No, the $Q_i$'s are, in general, not unique. Consider what would happen if you took the equality $P_1 Q_1 + \ldots + P_m Q_m = 1$ and squared both sides. The right-hand side is still $1$, but the left-hand side is completely different. However, it can still be written as a linear combination of the $P_i$'s, albeit with changed $Q_i$.

For a specific example, take $k = \Bbb R, d = 1, m = 2$ and $P_1 = x, P_2 = x-1$. Then we have $$ P_1\cdot 1 + P_2\cdot(-1) = 1 $$ so $Q_1 = 1, Q_2 = -1$ works. Squaring, we get $$ P_1^2 -2P_1P_2 + P_2^2 = 1\\ P_1\cdot P_1 + P_2\cdot(P_2-2P_1) = 1\\ P_1\cdot x + P_2 \cdot(-x-1) = 1 $$ so $Q_1 = x, Q_2 = -x-1$ also works.

$\endgroup$
2
  • $\begingroup$ Thanks! Can something interesting be said about the $Q_i$'s if we require minimal degrees (or at least that the sum of their degrees will be minimal). $\endgroup$
    – user237522
    Commented Apr 3, 2018 at 7:42
  • $\begingroup$ @user237522 It's possible. I don't know. $\endgroup$
    – Arthur
    Commented Apr 3, 2018 at 7:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .