$i=1,\ldots,r$, are all irreducible over $\K$.

In particular, by the famous Hilbert’s irreducibility theorem, any finite extension of $\Q$ is a Hilbertian field. Furthermore, by [@Sch Theorem 49] every finitely generated infinite field and every finitely generated transcendental extension of an arbitrary field are Hilbertian.

We also need the following simple fact.

\[lem:multconst\] Let $\K$ be an arbitrary field. The polynomials $F_1,\ldots,F_{s}\in\K[X_1,\ldots,X_{\ell}]$ 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, where $F_i^*(\vec{0}) =1$ if $F_i(\vec{0})=0$ and $F_i^*(\vec{0}) =F_i(\vec{0})$ otherwise.

Assume that $F_1,\ldots,F_{s}\in\K[X_1,\ldots,X_{\ell}]$ are multiplicatively independent in $\K(X_1,\ldots,X_{\ell})^*/\K^*$, but there exist integers $i_1,\ldots,i_{\ell}$ not all zero, such that $$\(\frac{F_1}{F_1^*(\vec{0})}\)^{i_1}\cdots\(\frac{F_{s}}{F_s^*(\vec{0})}\)^{i_s}=1.$$ Then, $F_1^{i_1}\cdots F_s^{i_s}=a$, where $a=\(F_1^*(\vec{0})\)^{i_1}\cdots\(F_s^*(\vec{0})\)^{i_s}$, and thus we obtain a contradiction.

For the other implication, assume $F_1^{i_1}\cdots F_s^{i_s}=a$ for some $a\in\K^*$ and integers $i_1,\ldots,i_s$ not all zero. From here we get again that $
a=\(F_1^*(\vec{0})\)^{i_1}\cdots\(F_s^*(\vec{0})\)^{i_s}$ (we note that if $F_k(\vec{0})=$ for some $k$, then we need to have $i_k=0$ as otherwise we get $a=0$, which is a contradiction). Thus we obtain again a contradiction with the fact that the polynomials $F_1/F_1^*(\vec{0}),\ldots, F_{s}/F_s^*(\vec{0})$ are multiplicatively independent.

We note that the conclusion of Lemma \[lem:multconst\] holds with any ${{\boldsymbol{\alpha}}}\in\K^{\ell}$, that is, $F_1,\ldots,F_{s}\in\K[X_1,\ldots,X_{\ell}]$ are multiplicatively independent in $\K(X_1,\ldots,X_{\ell})^*/\K^*$ if and 