Localization $(R_{\mathfrak p})_{\mathfrak q}$ for an Artinian ring. 
Let $R$ be an Artinian ring, and let $\mathfrak p$ and $\mathfrak q$ be distinct prime ideals of $R$. I have to prove that $(R_{\mathfrak p})_{\mathfrak q}=0$.

What I have done is the following: Since $R$ is Artinian we have $\dim R=0$, so every prime ideal of $R$ is maximal, so $\mathfrak p\not\subset\mathfrak q$ and vice versa. If we localize at $\mathfrak p$ we invert all elements outside of $\mathfrak p$, so we invert at least one element of $\mathfrak q$. Then $\mathfrak qR_{\mathfrak p}$ contains a unit and thus equals $R_{\mathfrak p}$, so if we localize at $\mathfrak qR_{\mathfrak p}$ we invert everything outside $\mathfrak qR_{\mathfrak p}=R_{\mathfrak p}$, so we invert seemingly invert..nothing. This doesn't make sense.
Can somebody offer some help? It is much appreciated.
 A: 
Let $f:R\to R_{\mathfrak p}$ be the canonical homomorphism and $S=f(R\setminus\mathfrak q)\subset R_{\mathfrak p}$. Then $S$ is a multiplicative set, and $S^{-1}R_{\mathfrak p}=0$. (By definition, $(R_{\mathfrak p})_{\mathfrak q}=S^{-1}R_{\mathfrak p}$.)  

In order to prove this we want to show that $\frac01\in S$, that is, there is $a\in R\setminus\mathfrak q$ and $b\in R\setminus\mathfrak p$ such that $ab=0$.
Denote by $\mathfrak m_1,\dots,\mathfrak m_t$ the maximal ideals of $R$ others that $\mathfrak p$ and $\mathfrak q$. Now we can pick $x\in\mathfrak p\setminus\mathfrak q$ and $y\in(\mathfrak q\cap\mathfrak m_1\cap\cdots\cap\mathfrak m_t)\setminus\mathfrak p$. Then $xy\in\mathfrak p\cap\mathfrak q\cap\mathfrak m_1\cap\cdots\cap\mathfrak m_t$, so it is nilpotent, that is, there is $k\ge1$ such that $x^ky^k=0$. Now set $a=x^k$ and $b=y^k$. 
A: I realize that this answer comes some 2 years too late, but I thought I'd post it nontheless for completeness sake now that I've stumbled upon this question. Here is an alternative way to see what is going on.
One can decompose Artinian rings as finite products of local Artinian rings, whose localizations are trivial, and taking localization commutes with taking the product. One possible explicit realization of this is as follows. 
First suppose that $R=(R,\mathfrak{m})$ is local Artinian. Then $\mathfrak{m}$ is the only prime of $R$ and $R\setminus\mathfrak{m} = R^{\times}$, hence $R_\mathfrak{m}\cong R$. Next, for a general Artinian ring $R$, let $R=\prod_{i=1}^n R_i$ be its decomposition into Artin local rings $(R_i,\mathfrak{m}_i)$. The primes of $R$ have the form $\mathfrak{p}_i:=R_1\times\dots\times R_{i-1}\times\mathfrak{m}_i\times R_{i+1}\times\dots\times R_n$, $1\leq i\leq n$. Hence $R_{\mathfrak{p}_i}\cong R_i$ and if you take a different prime $\mathfrak{p_j}$, $j\neq i$, then $(R_{\mathfrak{p}_i})_{\mathfrak{p}_j} \cong R_i^{-1}R_i=0$ because $R_i\ni 0$.
