I'm trying to show the following:

Let $F_1,\dots,F_m$ be forms of degree one in $K[x_1,\dots,x_{n+1}]$ with $K$ an algebraic closed field and $m\leq n$. Then all the forms of the same degree can not be in the ideal $I=(F_1,\dots,F_m)$, for all degrees.

I showed the result for degree one, by means of this simple argument: If a form $G$ of degree $n$ is in $I$ then $G=\sum_{i=1}^m G_iF_i$ with $G_i\in K[x_1,\dots,x_{n+1}]$ then

$$ G=\sum_{i=1}^m G_iF_i=\sum_{i=1}^m\sum_{j}G_{j,i}F_i $$

With $G_{k,l}$ a form of degree $k$. By uniqueness of the descomposition in forms, and the fact that $G_{k,l}F_l$ are forms; we obtain:

$$ G=\sum_{i=1}^m G_{n-1,i}F_i $$

It means that if a form of some degree is generated by these polynomials, iff is a combination of the polynomials with forms of one degree less. Thus in the case of degree one, we get that all the forms of degree one are in $I$ iff all the forms of degree one are in the $K-$vector space generated by the $F_i$. As $\dim_K\langle F_1,\dots,F_m\rangle\leq m\leq n$ and we have that the dimension of the forms of degree one is $n+1$ over $K$; is not possible that all the forms of degree one are in $I$.

I tried with this argument to show the result for larger degrees, but without success. Lastly, I know that the result is true because it's equivalent to a simple problem of elementary algebraic geometry, namely: In the projective space $\mathbb P^n$ the intersection of $m$ hyperplanes is not empty if $m\leq n$; which is solved easily with Cramer's rule.

I would be very grateful if someone gives an answer that follows the above path. Thank you.


1 Answer 1


It is not totally clear to me what your question is, but let me show the following statement.

Let $S = K[x_0,\dots, x_n]$ and $S_+ = (x_0,\dots, x_n)$. Let $F_1,\dots, F_m$ be forms of degree $d$ and $m \le n$. If $(F_1,\dots, F_n) \neq S$, then $S_+ \nsubseteq \sqrt{(F_1,\dots, F_m)}$. In particular, $S_+^l \nsubseteq (F_1,\dots, F_n)$ for all $l$.

Let $I = (F_1,\dots, F_m)$. Assume to the contrary $S_+ \subseteq \sqrt{I}$. Then $S_+^l \subseteq I$ for some $l$. This implies that codimension of $I$ is $n+1$. Since $I$ can be generated by $m$ elements, by Krull's Generalized Principal Ideal Theorem, $n+1 =$ codimension $I \le m \le n < n+1$. This is a contradiction.

I believe that your question is the case when $d = 1$ in the statement.


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