Prove that a specific ring of integers is not monogenic 
I'm trying to prove that the ring of integers of $K=\mathbb{Q}(\sqrt7, \sqrt13)$ is not of the form $ \mathbb {Z}[a]$ for some $a$. 

Unfortunately I can not figure out where to start. I tried to reason with the absurd, finding contradictions with the theorem Kummer-Dedekind but I did not find them. Is there someone that can to give me a detailed demonstration of this fact ?
Many thanks to everyone who give me this help !
 A: The following argument is due to an exercise in Daniel Marcus' book - Number Fields.
Let $ \alpha = \sqrt{7}, \beta = \sqrt{13} $. Suppose $ \mathcal{O}_K = \mathbb{Z}[a] $ for some $ a \in K $. Let $ f $ be the minimal polynomial of $ a $ (which has integer coefficients) and for any polynomial $ g \in \mathbb{Z}[X] $, denote by $ \bar{g} $ its canonical image in $ \mathbb{F}_3[X] $ by reducing coefficients modulo $3$. Clearly, $ 3 | g(a) $ in $ \mathbb{Z}[a] $ iff $ \bar{f} | \bar{g} $ in $ \mathbb{F}_3[X] $. Now, consider the four algebraic integers: $$ a_1 = (1+\alpha)(1+\beta) $$ $$ a_2 = (1+\alpha)(1-\beta) $$ $$ a_3 = (1-\alpha)(1+\beta) $$ $$ a_4 = (1-\alpha)(1-\beta) $$ Observe that all of them are conjugates of $ a_1 $ and that $ 3 | a_ia_j $ in $ \mathbb{Z}[a] $ for $ i,j $ distinct ($ \star $). However, $ 3 \nmid a_i^n $ in $ \mathbb{Z}[a] $ for all $ i $. For if this was true, then $ \mbox{Tr}^K(a_i^n/3) \in \mathbb{Z} $, i.e $$ 3 | \mbox{Tr}^K(a_i^n) = a_1^n + a_2^n + a_3^n + a_4^n $$ in $ \mathbb{Z} $ and hence, also in $ \mathbb{Z}[a] $. But observation ($ \star $) implies that $$ 3 | (a_1 + a_2 + a_3 + a_4)^n = 4^n $$ in $ \mathbb{Z}[a] $, hence $ 3 $ divides $ 1 $ in $ \mathbb{Z}[a] $, a contradiction. So if we write $ a_i = f_i(a) $ for each $ i $, the $ f_i$s having integer coefficients, then $ \bar{f} | \bar{f_i} \bar{f_j} $ for distinct $ i,j $ and $ \bar{f} \nmid \bar{f_i}^n $ in $\mathbb{F}_3[X] $. Thus, for each $ i $, $ \bar{f} $ has an irreducible factor $  h_i $ in $ \mathbb{F}_3[X] $ such that $ h_i \nmid \bar{f_i} $ and $ h_i | \bar{f_j} $, $ j \neq i $. Clearly the $ h_i $s are pairwise distinct, hence $ \bar{f} $ has atleast four irreducible factors over $ \mathbb{F}_3[X] $, but there are only three irreducibles of degree one, hence $ \deg{\bar{f}} \ge 5 $. But $ \deg{\bar{f}} = \deg f \le [K:\mathbb{Q}] = 4 $, a contradiction.
