Let $k$ be a field and $I$ be an infinite set such that $|k| > |I|$ . Let $R := k [X\_i : i \in I ] $ and $m$ be a maximal ideal of $R$ ; then is it true that $m \cap k[X\_i] \ne 0 , \forall i \in I$ ?


I can show that $R/m$ is an algebraic extension of $k$ ; I don't know whether it is helpful here or not . Perhaps we can try localizing $R$ at $S\_i := k[X\_i ] \setminus \{0\} $ ; but I can't see further ... 



On maximal ideals of $k [X_i : i \in I ] $ where $k$ is a field , $I$ is an infinite set with $|k| > |I|$
You are right that the algebraicity of $R/\mathfrak m$ is important. Indeed, suppose $\mathfrak m \cap k[X\_i] = 0$. That means that the map $k[X\_i] \to R/\mathfrak m$ is injective. But then (the image of) $X\_i$ is a transcendental element, which is impossible. (More generally, this shows that the intersection with any finitely generated subring is nonzero, unless the subring is $k$.)


The result is false if $R/\mathfrak m$ is transcendental; for example if $R/\mathfrak m \cong k(T)$, where $T$ is one of the $X\_i$, then $\mathfrak m \cap k[X\_i] = 0$. You rightly observed that for this you need at least $|k|$ generators (e.g. you need to invert all $X-\alpha$ for $\alpha \in k$). So as long as $|I| < |k|$, the extension $k \to R/\mathfrak m$ is algebraic.


