 Solution

Yes. Let $\mathfrak{m}$ be maximal and setThen $L$ is a field generated as a $\mathbb{C}$-algebra by countably many elements. Since the polynomial ring in countably many variables has a countable monomial basis, $L$ has at most countable dimension as a $\mathbb{C}$-vector space.

Suppose $t\in L$ were transcendental over $\mathbb{C}$. The familywould be linearly independent over $\mathbb{C}$. Indeed, after multiplying a finite relation by ${\prod }_j(t-x_j)$, evaluation at $t=x_i$ forces the $i$th coefficient to vanish. This would be an uncountable linearly independent subset of the countable-dimensional vector space $L$, a contradiction.

Thus $L/\mathbb{C}$ is algebraic. Since $\mathbb{C}$ is an algebraically closed field, $L=\mathbb{C}$. If $x_i$ is the image of $T_i$, the quotient map is evaluation at $(x_1,x_2,\dots )$ and