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Let $K = \mathbb{Q}(\alpha)$ where $\alpha^3 - 50\alpha- 10= 0$.

Prove that $\{1, \alpha, \alpha^{2}\}$ is an integral basis of $\mathcal O_{K}$.

I know that the minimal polynomial is

$$m_\alpha(x)=x^3 - 50x -10$$

but I'm not sure where to even begin. I have looked at plenty of resources but none of them seem to have concrete examples of how to solve a problem like this. I've tried to understand general examples but I'm not sure how to solve a specific problem like this one. Thanks in advance.

2 Answers 2

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Eisenstein easily verifies the polynomial is irreducible. Next, consider that the discriminant of the polynomial $x^3+ax+b$ is $-4a^3-27b^2=497300$ which is $100$ times a prime. So if we can show that $2$ and $5$ are ramified, we will see that this is an integral basis as otherwise the true discriminant would differ from ours by dividing by $4, 25,\,$ or $100$ using Dedekind's theorem on ramification.

But then this is simple as reducing $x^3-50x-10$ mod $2$ and $5$ both give $x^3$ (and both $2$ and $5$ only divide the constant term once each see theorem 3.1 for example) so we see that $(2)$ and $5$ are totally ramified as $(2,\alpha)^3, (5,\alpha)^3$ respectively, so $10|\Delta_K$, which is what we set out to show.

As KCd notes in the comments, you an also end this a bit earlier by appealing to theorem 2.3 in the linked notes from above. I usually prefer the full ramification information myself, because that's the form I usually use things in, but in your case if all you care about is showing it's a power basis you can save a little time by going straight there.

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    Since you don't yet *know* before solving the problem that the ring of integers is $\mathbf Z[\alpha]$ where $\alpha$ is a root of $x^3 - 50x - 10$, what is the basis for saying the way $x^3 - 50x - 10 \bmod p$ factors reflects how $p$ factors in $\mathcal O_K$? Sure, this is true if $p$ doesn't divide $[\mathcal O_K:\mathbf Z[\alpha]]$, but you don't yet know that index. That index divides the discriminant, which is divisible by 2 and 5, so I don't think you have justified by the reduction mod 2 or 5 being $x^3$ says $(2)$ and $(5)$ factor in $\mathcal O_K$ as the cube of a prime ideal. – 2017-02-24
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    @KCd it's a well-known fact that polynomials which reduce completely modulo a prime have this feature when $p^2\not | a_0$, but you're right I should note that this is a relevant fact. – 2017-02-24
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    Suppose $\beta$ is a root of $f(x) = x^3 - x^2 - 2x - 8$. Since $f(x) \equiv x^2(x+1) \bmod 2$, does that means $(2) = \mathfrak p^2\mathfrak q$ in the ring of integers of $\mathbf Q(\beta)$? Nope. The ring of integers is not $\mathbf Z[\beta]$ and in fact $(2)$ splits in the integers of $\mathbf Q(\beta)$ as a product of three distinct primes. – 2017-02-24
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    @KCd there $2^2 | 8$, so that's not the same case. – 2017-02-24
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    I was writing my second comment while you posted your own, so I did not see it at the time. By the way, when citing a 7-page file it'd be good to point directly to the relevant part: Theorem 2.3. It also lets you simplify the first paragraph in your answer: since $[\mathcal O_K:\mathbf Z[\alpha]]^2$ divides $497300$, once 2 and 5 don't divide the index all that's left for the index to be is 1. – 2017-02-24
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    @KCd ah, that makes sense. I thought it was odd your example was out of the relevant context. I also agree on the referencing. I read those notes for myself back when I was in grad school so I had forgotten it might take a bit longer for a student to jump to the right part. Cheers! – 2017-02-24
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First, compute the discriminant of $1,\alpha,\alpha^2$ (I assume you know how to do that?). That's $2^2 5^2 4973$. Next, the discriminant of ${\cal O}_K$ must divide that, and the quotient must be a square. However, the polynomial is 2-Eisenstein as well as 5-Eisenstein, which implies that $2$ and $5$ ramify, which implies that they divide the discriminant of ${\cal O}_K$. That leaves only 1 possibility for the discriminant of ${\cal O}_K$, namely, it has to be $2^2 5^2 4973$ as well. Hence $1,\alpha,\alpha^2$ is a $\mathbb{Z}$-basis of ${\cal O}_K$.

For more complicated examples you should simply use a computer algebra system (Sage, Magma, Maple, etc.) to compute the integral basis. Trying to do this by hand makes little sense. If you want to know how those programs work, there are various methods, some of which are not too difficult to understand (e.g. there is a book by Cohen on computational number theory, and there is also an online book by William Stein).