Is $\cap_{c \in \mathbb{C}} \langle x_1-c \rangle=0$ in $\mathbb{C}[x_1,\ldots,x_n]$, $n \geq 2$?

Question

It is well-known that the Jacobson radical, the intersection of all maximal ideals, in $\mathbb{C}[x_1,\ldots,x_n]$ is zero, $n \geq 1$, see this.

In particular, in $\mathbb{C}[x_1]$ we have $\cap_{a \in \mathbb{C}}\langle x_1-a \rangle=0$.

Now let us concentrate on $R_n:=\mathbb{C}[x_1,\ldots,x_n]$ with $n \geq 2$. By Hilbert's Nullstellensatz, a maximal ideal is of the form $\langle x_1-a_1,\ldots,x_n-a_n \rangle$, for some $a_1,\ldots,a_n \in \mathbb{C}$.

Of course, $\langle x_1-a_1 \rangle$ is not a maximal ideal in $R_n$, because it is strictly contained in $\langle x_1-a_1, x_n \rangle$ etc.

Regardless of non-maximality, we can ask about the intersection of all such ideals: $I:=\cap_{c \in \mathbb{C}}\langle x_1-c \rangle$, where each ideal in the intersection is considered as an ideal of $R_n$ (not as an ideal of $R_1$).

Question: Is $I=0$?

Theoretically, this intersection can be nonzero, since each ideal in this new intersection in $R_n$, $n \geq 2$, is strictly bigger then each 'same' ideal in the old intersection in $R_1$.

I think that the answer is positive, namely $I=0$. Maybe if we assume by contradiction that there is a nonzero element in the intersection, and then substitute $x_2=\ldots=x_n=0$ and reduce to $R_1$.

Thank you very much!

Answer 1

Because $x_1-c$ and $x_1-c'$ are coprime when $c,c'$ are distinct, by Chinese Remainder Theorem, $\langle x_1-c\rangle \cap \langle x_1-c'\rangle=\langle x_1-c\rangle \langle x_1-c'\rangle=\langle (x_1-c)(x_1-c')\rangle$. The same can be said about any finite number of such ideals.

Hence if $p\in \cap_c \langle x_1-c\rangle$, we have $x_1(x_1-1)(x_1-2)\cdots (x_1-n) \mid p$. Therefore the highest degree of $x_1$ in $p$ can be arbitrarily large.

Another way is to use the (much harder) results that the intersection of all maximal ideals of $\mathbb C[x_1, \cdots, x_n]$ is trivial, as well as Nullstellensatz. Indeed, for each maximal ideal $\langle x_1-c_1, \cdots, x_n - c_n\rangle$, it contains the ideal $\langle x_1-c_1\rangle$ hence the intersection $\cap_c\langle x_1-c\rangle$.

Answer 2

Note that $I\subseteq(x_1-c)\subseteq(x_1-c,x_2-a_2,\cdots,x_n-a_n)$ for all $c,a_2,\cdots,a_n\in\mathbb{C}$. So $I$ will be contained in the intersection which is $0$.