I'm trying to do exercise 27 in chapter of Marcus but it seems to me there is a typo or maybe it's me not understanding.
The exercise is the following
Let $\alpha^5=5(\alpha+1)$ R=$\mathbb{A}\cap \mathbb{Q}[\alpha]$.
Let $p\neq 3$ be a prime of $\mathbb{Z}$.
Show that the prime decomposition of pR can be determined by factoring $x^5-5x-5 \; mod\; p$
Do it for p=2
The hint is to use a previous exercise that tells us that the discriminant of $\alpha$, root of the irreducible polynomial $x^5+ax+b$ is $disc(\alpha)=4^4a^5+5^5b^4$ so in our case the discriminant is $5^5*3^3*41$ (isn't it?).
Another theorem (27 chapter 3 of Marcus Number Fields) tells us that we can decompose pR factoring the minimal polynomial of $\alpha$ if $p\not||S/R[\alpha]|$ where S is the integer ring of L and R is the integer ring of K, with L:K.
If I'm not wrong in our case we have $|S/R[\alpha]|=|\mathbb{A}\cap\mathbb{Q}[\alpha]/\mathbb{Z}[\alpha]|$.
A last corollary tells us that if $p^2\not| disc(\alpha)$ then the hypotesis of the theorem are satisfied.
This allows me to say that all the primes but maybe 3 and 5 satisfy the theorem hypotesis, however I don't know how to say that actually 5 is good but 3 is not.
My last option is to compute an integral basis but it seems a long process so I'm asking if there is another way to do it.
Best Answer
There is also a useful theorem: let $\alpha$ be an algebraic integer, let $L=\mathbb{Q}(\alpha)$. Assume the minimial polynomial of $\alpha$ is $p$-Eisenstein. Then $p\nmid [O_L:\mathbb{Z}[\alpha]]$, where $O_L$ is the ring of integers of $L$.
IN your case, the minimal polynomial is $X^5-5X-5$, which is $5$-Eisenstein. So the decomposition of $5$ is reflected by the decomposition of s $X^5-5X-5\mod 5.$
Note that the exercise does not ask to show that it does not work ofr $3$. It indeed does not work: modulo $3$, $X^5-5X-5$ has a decomposition into irreducble factors of the form $f_1^2f_2$, while $3$ actually does not ramify in $L$. A way to see htat would be to compute the discriminant, which is $5^2\cdot 41$, but I on't see an easy argument to do so at this moment.