Let $K = \mathbb{Q}(\alpha)$ where $\alpha$ is a root of the polynomial $x^3 + 2x + 1$, and let $g(x) = x^3 + x + 1$. Does $g(x)$ have a root in $K$?

Question

Problem: Let $K = \mathbb{Q}(\alpha)$ where $\alpha$ is a root of the polynomial $f(x) = x^3 + 2x + 1$, and let $g(x) = x^3 + x + 1$. Does $g(x)$ have a root in $K$?

My Attempt: I have proved that $f(x)$ and $g(x)$ are irreducible and found that their discriminates are not squares, so the Galois group is $S_3$ for both of them. Then it has to be that $[\mathbb{Q}(\alpha):\mathbb{Q}] = 2$ and $[\mathbb{Q}(\beta):\mathbb{Q}]$. So, $\alpha$ is a square root of something, but I am stuck here. I don't know what to do. The discriminant is not in the extension $\mathbb{Q}(\alpha)$, so we cannot work with it. And the formula for the roots of cubic polynomial gives alpha as a sum of two very complicated cubic roots.

Answer 1

Here’s an approach that is every bit as advanced as @BrunoJoyal’s, but goes at the question from an entirely different direction. And it is not complete, as you will see below.

Let $\alpha$ be a root of your $f$, and $\beta$ be a root of your $g$, let $K=\mathbb Q(\alpha)$ and $L=\mathbb Q(\beta)$. The question is simply whether $K=L$, since both are cubic fields (from irreducibility of $f$ and of $g$) and if $\beta\in K$ then $L=\mathbb Q(\beta)\subset K$. Multiplicativity of degrees would then show equality.

Now, both $\alpha$ and $\beta$ are units in their respective integer rings, and if the fields are the same, it must be that there are (small) integers $m$ and $n$ such that $\alpha^m=\beta^n$. The reason is that both $K$ and $L$ are cubic fields with only one real embedding and one complex. This follows from the fact that both polynomials cross the $x$-axis only once. The consequence for the groups of units is that each is of the form $\mathbb Z\oplus C_2$, where the second is just the cyclic group $\{\pm1\}$. Now, if I was very ambitious, I’d calculate the real roots of $f$ and $g$, and find the ratio of their logs. If this is not obviously a rational number, I’d be completely convinced that the two groups of units were incommensurable, and feel confident that the fields thus were different.

EDIT:

Here’s another approach that is just as advanced, but complete and not dubious. Look at the situation locally at the prime $3$. Since $f(x-1)=x^3-3x^2+6x-3$, Eisenstein, $K$ has just one prime above $3$, totally ramified. On the other hand, modulo $3$, we have $x^3+x+1=(x-1)(x^2+x-1)$, so that in $L$, $3$ splits into two primes, one of degree $1$ and one of degree $2$ with residue field extension degree $2$. That is, $L$ is unramified at $3$. So the fields are different, and therefore $\beta\notin K$.

Answer 2

The discriminant of $f$ is $-59$ and the discriminant of $g$ is $-31$. These two numbers are square-free, so they are actually the discriminants of the ring of integers of the corresponding number fields $K$ and $L=\mathbf Q[x]/g(x)$. Since $(31,59)=1$ it follows that $g$ has no root in $K$ (because $K$ ramifies only over the prime $59$, and $L$ ramifies only over the prime $31$).

Suggestion for a more elementary approach: pick an indeterminate element $y=u\alpha^2 + v\alpha + w \in K$, where $u,v,w \in \mathbf Q$. By calculating a linear dependence relation between $y^2, y$ and $1$, obtain a (not necessarily monic) polynomial $p(x) \in \mathbf Z[u,v,w][x]$ such that $p(y) = 0$. Then, calculate the resultant of $p$ and $g$ by writing it as the determinant of the appropriate Sylvester matrix. This will give you a polynomial $R\in \mathbf Z[u,v,w]$. If you can prove that $R(a,b,c) \neq 0$ for all $(a,b,c) \in \mathbf Q$, it will follow that $f$ has no root in $K$.

Answer 3

For good measure, here is a solution closer to what's in Artin's Algebra.

As Bruno noted, the discriminants are $-59,-31$. These two are not squares, so by Galois Theory for Cubics (16.8.5), the splitting fields $L_f,L_g$ both have degree $6$ over $\mathbb{Q}$, with Galois group $S_3$. If $g(x)$ had a root in $K$, then it splits completely in $K$ by the Splitting Theorem (16.3.2), hence $L_f \cong L_g$ since both are of degree $6$ over $\mathbb{Q}$. Since the Galois group is $S_3$, there should be one intermediate field of degree $2$ over $\mathbb{Q}$, but we have two, $\mathbb{Q}(\sqrt{-59})$ and $\mathbb{Q}(\sqrt{-31})$, a contradiction.