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I wonder if somebody can provide enlightenment on my question on the polynomial below.

The polynomial has the property that if you pick up one root, the other two can be obtained by successively acting certain fractional linear transformation on it. I wonder whether this phenomenon is ubiquitous (at least to a certain family of polynomials over rational numbers, say) or sporadic.

This fact is remarked by Shanks (an excerpt is attached below), but he gives no explanation as to why that is the case nor how he obtained the result. Serre refers to this polynomial in one of his papers below, but he says nothing about the (fractional linear) group action.

In this article, Shanks says:

The cubic equation $$(a) \qquad x^3 =ax^2 + (a + 3)x + 1$$ has the discriminant $$(b) \qquad D = (a^2 + 3a + 9)^2,$$ and if $a^2 + 3a + 9$ is prime, (b) is obviously also the discriminant of the field $\mathbb{Q}(\rho)$ where $\rho$ is a root of (a). One may easily verify that the other two roots of (a) are $\rho_{2} = -l/(l+\rho)$ and $\rho_{3} = - 1/(1 +\rho_{2})$.

Serre refers to this polynomial, saying "c'est même là une extension universelle" in Théorème 3 here.

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Welcome to MO!...Regarding your question, I am not comfortable with the phrase "but unfortunately he said he had no answer". In my case, I mention private correspondences only upon the agreement of the other party – Unknown Nov 21 '10 at 14:05
These have come to be known as "Shanks' simplest cubics," and there is considerable literature available, not to mention generalizations to "simplest quartics," "simplest quintics," etc. See Math Reviews. – Gerry Myerson Nov 21 '10 at 23:03

This is true of every cubic polynomial with Galois group $C_3$ (in particular, in characteristic not equal to $2$ it is true of every separable irreducible cubic polynomial with discriminant a square). Let $f$ be such a cubic polynomial over a base field $k$ and let $L$ be the splitting field, hence $[L : k] = 3$. Let $r_1, r_2, r_3$ be the roots of $f$ in $L$. Then there is some nontrivial linear dependence among the elements $1, r_1, r_2, r_1 r_2$, hence a fractional linear transformation with coefficients in $k$ such that $r_2 = \frac{a r_1 + b}{c r_1 + d}$. This relation respects the action of the Galois group, so if we choose $r_1, r_2$ so that the Galois group acts by $r_1 \mapsto r_2, r_2 \mapsto r_3, r_3 \mapsto r_1$ we get $r_3 = \frac{a r_2 + b}{c r_2 + d}$ as desired.

On the other hand, this sufficient condition is necessary, since it implies that $[L : k] = 3$.

(Full disclosure: this was an exercise I did recently.)

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Nice solution, Qiaochu. – Todd Trimble Nov 21 '10 at 14:01

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