Consider the polynomial ring of countable variables with coefficients in the real numbers, i.e, $S=\Bbb{R}[x\_1,x\_2,x\_3,...,x\_n,...]$. My Question is about the **depth** of this ring. 


**Question**: *Could we find any maximal ideal in $S$ such that it contains an **anti-chain** with uncountable number of prime ideals*? 


**PS**: We recall that a set $\{P\_{\alpha}\} \_{\alpha \in \mathcal{A}}$ is an anti-chain of prime ideals if for each $u ,v \in T$, neither $P\_{u} \subseteq P\_{v}$ nor $P\_{v} \subseteq P\_{u}$.


**PPS**: By this notation, I am seeking an anti-chain $\{P\_{\alpha}\} \_{\alpha \in \mathcal{A}}$ of prime ideals in $S$ for which $T$ is uncountable.



Depth of polynomial ring $S=\Bbb{R}[x_1,x_2,x_3,...,x_n,...]$
In the same spirit as the answer to your [earlier question](https://mathoverflow.net/questions/146157/chain-of-prime-ideals-in-polynomial-ring-s-bbbrx-1-x-2-x-n), pick a bijection $\phi$ from the natural numbers to the rationals. For each real number $\alpha$, let $I\_\alpha$ be the ideal generated by all $x\_n$ such that $\phi(n)<\alpha$ and let $J\_\alpha$ be the ideal generated by all $x\_n$ such that $\phi(n)>\alpha+1$. Let $P\_\alpha=I\_\alpha+J\_\alpha$. The $P\_\alpha$ form an uncountable anti-chain contained in the maximal ideal of $S$ generated by all the $x\_n$.


