Let $R$ be a Noetherian commutative ring and let $\dim(R)$ denote the Krull dimension of $R$. For the polynomial ring $R[x]$, we have $\dim(R[x]) = 1 + \dim(R)$. In fact, the same is true if we replace the polynomial ring by the power series ring: again $\dim(R[[x]]) = 1 + \dim(R)$.

The situation is a little different for the Laurent series ring $R((x))$. Recall that the Laurent series ring is obtained by inverting $x$ in the power series ring $R[[x]]$. That is,

$$R((x))\cong R[[x]]/(tx-1).$$ What is the Krull dimension of $R((x))$ in relation to $R$? If $R$ is a field, then so is $R((x))$. Therefore, $R((x))$ has Krull dimension zero just like $R$.

In fact, the obvious generalization holds:

**Theorem.**If $R$ is a Noetherian commutative ring then $\dim(R) = \dim[R((x))]$.

*Proof.*For each $t = \sum_{i=n}^\infty t_ix^n\in R((x))$ for $t_n\not=0$, define the order of $t$ to be $n$ and $\lambda(t) = t_i$. If $I$ is an ideal of $R$ then define $\lambda(I)$ to be the ideal of $R((x))$ generated by

$$\{ \lambda(t) : t\in I\}.$$ It is clear that $\lambda(I)\subseteq \lambda(J)$ whenever $I\subseteq J$. In fact, we claim that if $I\subseteq J$ and $\lambda(I) = \lambda(J)$ then $I = J$ so that $\lambda$ preserves strict inclusions.

Indeed, suppose $I\not=0$, for otherwise the result is trivial. Since $R((t))$ is Noetherian, we can choose $a_1,\dots,a_m\in I$ such that $\lambda(a_1),\dots,\lambda(a_m)$ generate $\lambda(I)$, and choose each $a_i$ such that the order of $a_i$ is zero, which is possible because $x$ is invertible. Let $b\in J$. Without loss of generality, assume that the order of $b$ is zero. We will prove that $b\in I$.

We claim that we can construct $s_{ij}\in R$ for $i=1,\dots,m$ and $j=0,1,2,\dots$ such that

$$b = \sum_{i=1}^m\sum_{j=0}^n a_is_{ij}x^j$$ has order strictly greater than $n$ for all $n=0,1,2,\dots$. We proceed inductively on $j$. Since $\lambda(b)\in\lambda(J) = \lambda(I)$, there exists $s_{i0}$ for $i=1,\dots,m$ such that

$$\lambda(b) = \lambda(a_1)s_{10} + \cdots + \lambda(a_m)s_{m0}.$$ Since each of $a_i$ and $b$ has order zero, the element

$$b-(a_1s_{10} + \cdots + a_ms_{m0})$$ has order strictly greater than zero. Now assume we have constructed $s_{ij}$ for $i=1,\dots,m$ and $j=0,\dots,n$ such that

$$c = b-\sum_{i=1}^m\sum_{j=0}^n a_is_{ij}x^j$$ has order strictly greater than $n$. We note that $c\in J$. Let $c_{n+1}$ denote the coefficient of $x^{n+1}$ in $c$. Either $c_{n+1} = 0$ or $c_{n+1} = \lambda(c)$, and therefore $c_{n+1}\in \lambda(J) = \lambda(I)$. Again, we find $s_{i,n+1}$ such that

$$c_{n+1} = \lambda(a_1)s_{1,n+1} + \cdots + \lambda(a_m)s_{m,n+1}$$ and by construction the element

$$c-(a_1s_{1,n+1}x^{n+1} + \cdots + a_ms_{m,n+1}x^{n+1})$$ has order strictly greater than $n+1$, which is the induction step of the construction.

Set $d_i = \sum_{j=0}^\infty s_{ij}x^j$ for $i=1,\dots,m$. By construction,

$$b = a_1d_1 + \cdots + a_md_m\in I.$$ Thus, we have shown that $\lambda$ maps the lattice of ideals in $R((x))$ to the lattice of ideals of $R$, respecting strict inclusion. Therefore,

$$\dim R((x))\leq\dim R.$$ The prove the reverse inequality, for each ideal of $R$, write $I((x))$ for the ideal of $R((x))$ of Laurent series whose coefficients are in $I$. Then $I\mapsto I((x))$ maps the lattice of ideals of $R$ into the lattice of ideals of $R((x))$, respecting strict inclusion. Thus

$$\dim R \leq \dim R((x)).$$

This is essentially the proof given by Goodearl and Small in their paper [1]. It seems somewhat complicated but actually the notation is making it more complicated than it really is. It comes down to the fact that we can choose generators of ideals of any degree, something which is impossible in the power series or polynomial rings. I feel like there is probably an easier proof, but I have not been able to find it.

Also, it is important to realize that there is nothing special about $R$ being commutative. By eliminating ‘commutative’ and replacing ‘Noetherian’ with ‘right Noetherian’ and Krull dimension with ‘right Krull dimension’, the proof works exactly the same way. I just omitted these extra words for brevity, but in fact Goodearl and Small state the result without the commutativity condition. Goodearl and Small also prove an analogous result for global dimension, and I might talk about that in another post.

## References

1. Goodearl, K. R.; Small, L. W. Krull versus global dimension in Noetherian P.I. rings. Proc. Amer. Math. Soc. 92 (1984), no. 2, 175–178.