TITLE: Find all possible rational values of the parameter of a parametric cubic such that it is reducible
QUESTION [0 upvotes]: Description:  Given the following parametric cubic polynomials ${E}^{3}
- 15\, {\beta}_{\pm}\, {E}^{2} - 3 \left({71\, {\beta}_{\pm}^{2} + 352\, {\beta}_{\mp}}\right) E
+ 135\, {\beta}_{\pm} \left({5\, {\beta}_{\pm}^{2} - 32\, {\beta}_{\mp}}\right)$ where ${\beta}_{\pm} = 1 \pm \eta$, how do I find all possible rational values of $\eta$ such that this polynomial is reducible.
Background:  I have found 61 rational values of $\eta$ that factor these polynomials into a rational linear $E - p/q$ for integer $p$ and $q$ relatively prime and $q \ne 0$ times an irreducible quadratic of the form ${E}^{2} + a\, E + b$.  I do not know if there are an infinite number of such values and if not do I have the complete set.

REPLY [1 votes]: I just reopened the question so I can write an answer, although my comment should have sufficed but I may not be getting my message across.
Anyway, let's use some sane variables ($x=\beta_{\pm},y = E$), so the polynomial is
$$y^3-15xy^2-3(71x^2+352x)y+135x(5x^2-32x)$$
Let's now replace $y$ by $xz$ and get
$$(z^3 - 15z^2 - 213z + 675)x^3 + (-1056z - 4320)x^2.$$
So you can pick any rational $z$ that you like, compute
$$x = (1056z + 4320)/(z^3 - 15z^2 - 213z + 675)$$
and if you replace the value of $x=\beta_{\pm}$ by what you computed, the resulting cubic in $E$ has a factor $E - xz$.


TITLE: The variety associated to a polynomial ring with a particular grading
QUESTION [0 upvotes]: We impose various grading on an algebra or a module to understand its homological properties. So I made up the following problem and I would like to understand its correspondence as a variety.
Suppose $k[x_1,x_2, x_3, x_4]$ is a polynomial ring over $k=\bar{k}$.
Let $R = k[x_1, x_2]$ and consider the graded module
$$
S = R \oplus (R x_1 x_3 \oplus R x_4) \oplus (R x_1 x_3 x_ 4 \oplus R x_4^2 \oplus R x_1^2 x_3^2) \oplus \ldots .
$$
What is $Proj(S)$?
Any comment or feedback is appreciated.
$$
$$
Note that  $R x_1 x_3 = \{  a x_1 x_3: a \in R\} = \{   b x_3: b \in R x_1\}$.
$$
$$

REPLY [1 votes]: If I interpret well, the ring $S$ coincide with $k[x_1, x_2, x_1 x_3, x_4]$ with grading $\deg x_1 = \deg x_2 = 0$ and $\deg x_1 x_3 = \deg x_4 = 1$. It is quite clear that $x_1 x_3$ is trascendent over $k[x_1, x_2, x_4]$, hence $S$ is a polynomial ring over $R = k[x_1, x_2]$ with two indeterminates, therefore
$$
\mathrm{Proj}(S) = \mathbb{P}^1_R = \mathbb{P}^1_k \times_k \mathrm{Spec}(R) = \mathbb{P}^1_k \times_k \mathbb{A}^2_k.
$$


TITLE: Show that $\angle AQP = \angle PQC$
QUESTION [3 upvotes]: Given a triangle $\triangle ABC$ whose incenter is $I$, its $A-$exincenter is $O$, the midpoint of $AC$ is $M$, $P = OM \cap BC$ and $Q = AI \cap (ABC) \neq A$. Prove that $\angle AQP = \angle PQC$.

It is easy to see that $QB= QC=QO = QI$, so I thought about the spiral similarity between $AC$ and $IO$, but it is a weird one because its centers would be either points $C$ or $A$? Anyways, I couldn't find a simple solution neither find some harmonic pencil.
the great problem seems to be that point $M$

REPLY [1 votes]: Let $IP$ meet $AC$ at $X$ and let $AO$ cuts $BC$ at $A'$.
Then we have \begin{align} (A,C;X,M)&= (PA,PC;PX,PM) \\ &= (PA, PA'; PI, PO) \\ &= (A,A'; I,O) \\ &=-1
\end{align}
Since $M$ is the midpoint of $AC$ it means $X=\infty$, so $IP||AC$.
But now it is easy to see that $BQPI$ ic cyclic and by simle angle chase we get the conclusion.