Ring of integers of $\mathbb{Q}(a)$

Question

I am stuck with this exercise in my home work, could use a hand:

Let $a$ be a root of the irreducible polynomial $f(x)=x^3+x+1$.

Show that the ring of integers of $F:=\mathbb{Q}(a)$ is $\mathbb{Z} + \mathbb{Z}a + \mathbb{Z}a^2$.

Thanks in advance!

Answer 1

Here is just an ad hoc solution, there is probably a better one using algebraic number theory. Let $$x^3+x+1=(x-a)(x-b)(x-c)$$

and let $$x+ya+za^2$$ be an algebraic integer.

Multiplying by $a$ and then by $a^2$ and simpifying we get that $$-z+(x-z)a+ya^2$$ and $$-y-(y+z)a+(x-z)a^2$$

are also integers.

Now consider

$$x+ya+za^2$$ $$x+yb+zb^2$$ $$x+yc+zc^2$$

adding then together we have $3x-2z$ is an integer.

Where we use that $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=-2$

Applying the same trick to the other integers we get that $-3z-2y$ and $-3y-2x+2z$ are also integers.

This means that

$$\begin{pmatrix}3&0&-2\\ 0&-2&-3\\ -2&-3&2\\ \end{pmatrix}\begin{pmatrix}x\\y\\z\\ \end{pmatrix}$$ is an integer vector.

Using (integer!) row reduction we have that $$\begin{pmatrix}1&0&-11\\ 0&1&-14\\ 0&0&31\\ \end{pmatrix}\begin{pmatrix}x\\y\\z\\ \end{pmatrix}$$ is an integer vector.

Thus we see that $31z$ is an integer and $x=n+11z$, and $y=m+14z$.

Thus the whole problem reduces to the question of whether $$\alpha=\frac{a^2+14a+11}{31}$$ is an integer or not.

It is not an integer, as one sees by calculating $\alpha^3$.

Answer 2

The quicke answer, (the other answer is due to me also, so if you upvote one be sure to downvote the other !)

$\{1,a,a^2\}$ is an integral basis. As seen from calculating the discriminant of the equation $x^3+x+1$ using Cardano's formula $$\Delta=-27q^2-4p^2=-27-4=-31$$

(Oh lookie here where have I seen $31$ before ?)

Now if $D$ is the discriminant of the field, then $$\Delta=\pm k^2D$$ we see this is only possible if

$$\Delta=\pm D$$ and this implies that $\{1,a,a^2\}$ is an integral basis.