This is just an after thought about my question. At the end it is really a linear algebra problem. Starting with the polynomials $g_1,g_2,\ldots,g_n,g_{n+1}=f$ one may ask if it is possible to find an array of coefficients $$ a_{\underline{i}}^{(k)} $$ where $\underline{i}=(i_1,i_2,\ldots,i_n)$ are $n$-tuples and $1\leq k\leq n+1$ such that $$ \sum_{k=1}^{n+1}\left(\sum_{\underline{i}} a_{\underline{i}}^{(k)}x_1^{i_1}\ldots x_{n}^{i_n}\right)g_k=cte $$ This is then a linear algebra problem. Now thanks to the Nullstellensatz, since $Z(g_1,g_2,\ldots,g_n,g_{n+1};\overline{\mathbf{Q}})=\emptyset$, we know that this linear algebra problem adimts a solution (even over $\mathbb{Q}$ since the system is linear). So really, what the Nullstellensatz says is that unless there is some commun solution to your set of polynomials in an algebraic closure, then the set of linear equations above always admit a solution.

This is just an after thought about my question. At the end it is really a linear algebra problem. Starting with the polynomials $g_1,g_2,\ldots,g_n,g_{n+1}=f$ one may ask if it is possible to find an array of coefficients $$ a_{\underline{i}}^{(k)} $$ where $\underline{i}=(i_1,i_2,\ldots,i_n)$ are $n$-tuples and $1\leq k\leq n+1$ such that $$ \sum_{k=1}^{n+1}\left(\sum_{\underline{i}} a_{\underline{i}}^{(k)}x_1^{i_1}\ldots x_{n}^{i_n}\right)g_k=cte $$ Now thanks to the Nullstellensatz, since $Z(g_1,g_2,\ldots,g_n,g_{n+1};\overline{\mathbf{Q}})=\emptyset$, we know that this linear algebra problem adimts a solution (even over $\mathbb{Q}$ since the system is linear). So really, what the Nullstellensatz says is that unless there is some commun solution to your set of polynomials in an algebraic closure, then the set of linear equations above always admit a solution.

This is just an after thought about my question. At the end it is really a linear algebra problem. Starting with the polynomials $g_1,g_2,\ldots,g_n,g_{n+1}$ one may ask if it is possible to find an array of coefficients $$ a_{\underline{i}}^{(k)} $$ where $\underline{i}=(i_1,i_2,\ldots,i_n)$ are $n$-tuples and $1\leq k\leq n+1$ such that $$ \sum_{k=1}^{n+1}\left(\sum_{\underline{i}} a_{\underline{i}}^{(k)}x_1^{i_1}\ldots x_{n}^{i_n}\right)g_k=cte $$ This is then a linear algebra problem. Now thanks to the Nullstellensatz, since $Z(g_1,g_2,\ldots,g_n,g_{n+1};\overline{\mathbf{Q}})=\emptyset$, we know that this linear algebra problem adimts a solution (even over $\mathbb{Q}$ since the system is linear). So really, what the Nullstellensatz says is that unless there is some commun solution to your set of polynomials in an algebraic closure, then the set of linear equations above always admit a solution.

This is just an after thought about my question. At the end it is really a linear algebra problem. Starting with the polynomials $g_1,g_2,\ldots,g_n,g_{n+1}=f$ one may ask if it is possible to find an array of coefficients $$ a_{\underline{i}}^{(k)} $$ where $\underline{i}=(i_1,i_2,\ldots,i_n)$ are $n$-tuples and $1\leq k\leq n+1$ such that $$ \sum_{k=1}^{n+1}\left(\sum_{\underline{i}} a_{\underline{i}}^{(k)}x_1^{i_1}\ldots x_{n}^{i_n}\right)g_k=cte $$ This is then a linear algebra problem. Now thanks to the Nullstellensatz, since $Z(g_1,g_2,\ldots,g_n,g_{n+1};\overline{\mathbf{Q}})=\emptyset$, we know that this linear algebra problem adimts a solution (even over $\mathbb{Q}$ since the system is linear). So really, what the Nullstellensatz says is that unless there is some commun solution to your set of polynomials in an algebraic closure, then the set of linear equations above always admit a solution.