only if $\frac{F_1}{F_1^*({{\boldsymbol{\alpha}}})},\ldots,\frac{F_{s}}{F_s^*({{\boldsymbol{\alpha}}})}$ are multiplicatively independent for any ${{\boldsymbol{\alpha}}}\in\K^{\ell}$, where as before $F_i^*({{\boldsymbol{\alpha}}}) =1$ if $F_i({{\boldsymbol{\alpha}}})=0$ and $F_i^*({{\boldsymbol{\alpha}}}) =F_i({{\boldsymbol{\alpha}}})$ otherwise.

The following result is easily derived from Lemma \[lem:multconst\].

\[lem:multindep\] Let $\K$ be a Hilbertian field and $F_1,\ldots,F_s\in\K[X_1,\ldots,X_{\ell}]$ multiplicatively independent polynomials in $\K(X_1,\ldots,X_{\ell})^*/\K^*$. Then, there exists a specialisation $(\alpha_2,\ldots,\alpha_{\ell})\in\K^{{\ell}-1}$ such that the polynomials $F_i(X_1,\alpha_2,\ldots,\alpha_{\ell})$, $i=1,\ldots,s$, are multiplicatively independent in $\K(X_1)^*/\K^*$.

By Lemma \[lem:multconst\] the polynomials $F_1,\ldots,F_s$ are multiplicatively independent in $\K(X_1,\ldots,X_{\ell})^*/\K^*$ if and only if $\frac{F_1}{F_1^*(\vec{0})},\ldots,\frac{F_{s}}{F_s^*(\vec{0})}$ are multiplicatively independent.

We denote $G_i=\frac{F_i}{F_i^*(\vec{0})}$, $i=1,\ldots,s$. Let $P_1,\ldots,P_r\in\K[X_1,\ldots,X_{\ell}]$ be the distinct irreducible factors of $G_1,\ldots,G_s$, that is, we have the factorisation $$G_i=P_1^{e_{i,1}}\cdots P_r^{e_{i,r}},\quad i=1,\ldots,s.$$

We note that the polynomials $G_1,\ldots,G_s$ being multiplicatively independent is equivalent with the matrix $(e_{i,j})_{\substack{1\le i\le s\\1\le j\le r}}$ having full rank.

Since $\K$ is Hilbertian, there exists a specialisation $(\alpha_2,\ldots,\alpha_{\ell})\in\K^{{\ell}-1}$ such that $P_j(X_1,\alpha_2,\ldots,\alpha_{\ell})$, $j=1,\ldots,r$, are all irreducible over $\K$ and for $i=1,\ldots,s$, we have the factorisation $$G_i(X_1,\alpha_2,\ldots,\alpha_{\ell})=P_1(X_1,\alpha_2,\ldots,\alpha_{\ell})^{e_{i,1}}\cdots P_r(X_1,\alpha_2,\ldots,\alpha_{\ell})^{e_{i,r}}.$$ If the polynomials $G_i(X_1,\alpha_2,\ldots,\alpha_{\ell})$, $i=1,\ldots,s$, are multiplicative dependent over $\K$, then there exist integers $\ell_1,\ldots,\ell_s$, not all zero such that $$G_1(X_1,\alpha_2,\ldots,\alpha_{\ell})^{\ell_1}\cdots G_s(X_1,\alpha_2,\ldots,\alpha_{\ell})^{\ell_s}=1.$$ This is equivalent to the fact that the matrix $(e_{i,j})_{\substack{1\le i\le r\\1\le j\le s}}$ does not have full rank, which contradicts the fact that the initial polynomials $G_1,\ldots,G_s\in\K[X_1,\ldots,X_{\ell}]$ 