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.


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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