How can I prove that the Krull dimension of the polynomial ring $R=K[X_1,X_2,…]$ in countably many variables ($K$ a field) is infinity ? I have already proved that $R$ is an integral domain but not Noetherian, because $(X_1)\subsetneq(X_1,X_2)\subsetneq\cdots$ is a chain of ideals that does not stabilise at some point. I know that the Krull dimension is the supremum of the heights of all prime ideals of the ring and that the Krull dimension of the polynomial ring $K[X_1,…,X_n]$ in finitely many variables is equal to $n$. How can I efficiently apply this knowledge in order to compute the Krull dimension for countably many variables ? Thanks for your help !
Krull dimension of polynomial ring in countably many variables
commutative-algebrakrull-dimensionpolynomial-rings
Related Solutions
I'll prove the following result:
$$K[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>\simeq K[s^3,s^2t,st^2,t^3],$$ where $K$ is a field.
Let $\varphi: K[x_1,x_2,x_3,x_4]\to K[s, t]$ be the ring homomorphism that maps $x_1\mapsto s^3$, $x_2\mapsto s^2t$, $x_3\mapsto st^2$ and $x_4\mapsto t^3$. Obviously $\operatorname{Im}\varphi=K[s^3,s^2t,st^2,t^3]$; this is a subring of $K[s,t]$ and the extension $K[s^3,s^2t,st^2,t^3]\subset K[s,t]$ is integral, hence $\dim K[s^3,s^2t,st^2,t^3]= \dim K[s,t]=2.$
It remains to prove that $\ker\varphi=\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>$. By definition $\varphi(f(x_1,x_2,x_3,x_4))=f(s^3,s^2t,st^2,t^3)$. In particular, this shows that the polynomials $g_1=x_1x_3-x_2^2$, $g_2=x_2 x_4-x_3^2$ and $g_3=x_1x_4-x_2 x_3$ belong to $\ker\varphi$.
Conversely, let $f\in\ker\varphi$, i.e. $f(s^3,s^2t,st^2,t^3)=0$. We want to show that $$f\in\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>.$$ The initial monomials of $g_1$, $g_2$, resp. $g_3$ with respect to the lexicographical order are $x_1x_3$, $x_2x_4$, resp. $x_1x_4$. The remainder on division of $f$ to $G=\{g_1,g_2,g_3\}$, denoted by $r$, is a $K$-linear combination of monomials none of which is divisible by $x_1x_3$, $x_2x_4$, resp. $x_1x_4$. This shows that the monomials of $r$ can have one the following forms: $x_1^ix_2^j$ with $i\ge 1$ and $j\ge 0$, $x_2^kx_3^l$ with $k\ge 1$ and $l\ge 0$, respectively $x_3^ux_4^v$ with $u,v\ge 0$. In order to prove that $f\in\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>$ it is enough to show that $r=0$. But we know that $f(s^3,s^2t,st^2,t^3)=0$ and therefore $r(s^3,s^2t,st^2,t^3)=0$. The monomials (in $s$ and $t$) of $r(s^3,s^2t,st^2,t^3)$ are of the following types: $s^{3i+2j}t^j$, $s^{2k+l}t^{k+2l}$, respectively $s^ut^{2u+3v}$. Now check that there is no possible cancelation between these monomials (because they can't be equal), so $r=0$.
Now it follows that $\dim R=2$.
This is true - in a Noetherian ring, the only possible heights for which there are finitely many primes of that height are either $0$, or maximal ideals. This follows from the following fact:
Proposition: If $R$ is Noetherian and $\mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \mathfrak p_2$ is a chain of distinct prime ideals in $R$, then there are infinitely many primes $\mathfrak q$ such that $\mathfrak p_0 \subsetneq \mathfrak q \subsetneq \mathfrak p_2$.
Proof: Localizing at $\mathfrak p_2$ and quotienting by $\mathfrak p_0$, it suffices to show that any Noetherian local domain of dimension $2$ has infinitely many primes. If there were only finitely many height $1$ primes, then since $\mathfrak p_2$ is not contained in any of them, by prime avoidance it is not contained in their union, so there exists $x \in \mathfrak p_2$, but not in any height $1$ prime. But this contradicts the Principal Ideal Theorem, since $\text{ht}(x) \le 1$.
Best Answer
Hint:
For any $n\in\mathbf N$, there exists a prime ideal of height $n$. Namely, $$(X_1, X_2,\dots, X_n)\varsupsetneq(X_1, X_2,\dots, X_{n-1})\varsupsetneq\dots\varsupsetneq(X_1, X_2)\varsupsetneq (X_1)\varsupsetneq\{0\}$$ is a chain of prime ideals with length $n$.