We will assume that is a commutative ring with We will allow the case i.e. where The goal is to describe the Krull dimension of in terms of elements of and not prime ideals of The reference is this paper: A Short Proof for the Krull Dimension of a Polynomial Ring. Recall that the **Krull dimension** of is the largest integer for which there exist prime ideals of such that Then we write If there is no such integer, then we define and if we define

**Notation**. Let and set Clearly is multiplicatively closed. Let be the localization of at If then we define

**Problem 1**. Let and suppose that is a maximal ideal of Then Moreover, if is a prime ideal and then

**Solution**. if then and we are done. Otherwise, and thus for some Then For the second part, suppose to the contrary that for some integer and Then, since we have and thus because

**Problem 2**. if and only if for all

**Solution**. As we already mentioned, means Suppose that for some Then there exists a prime ideal of such that because is multiplicatively closed. By Problem 1, is not a maximal ideal and thus Conversely, suppose that for all and is a prime ideal of Let be a maximal ideal of which contains Choose By Problem 1, contradicting

**Problem 3**. Let be an integer. Then if and only if for all

**Solution**. The case was done in Problem 2. So we’ll assume that Suppose that and let A prime ideal of is in the form where is a prime ideal of and Also, if and are two prime ideals of with then Now, suppose to the contrary that and consider the chain of prime ideals of By Problem 1, is not a maximal ideal and so there exists a prime ideal such that and that will give us the chain of prime ideals of contradicting Conversely, suppose that for all Suppose also, to the contrary, that there exists a chain of prime ideals of Let By the second part of Problem 1, and thus is a chain of prime ideals of contradicting

**Problem 4**. Let be an integer. Then if and only if for every there exist integers and such that

**Solution**. By Problem 3, if and only if Therefore if and only if