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Suppose I have a $\mathbb{Q}$-algebra R, and I have two elements x and y in R. I happen to know that x is a scalar multiple of y. Is there a way to figure out what the scalar is?

In other words, I want to identify the $\mathbb{Q}$-span of $y$ with 1-dimensional vector space, choose an isomorphism of this vector space with $\mathbb{Q}$ (sending y to 1), and see where x goes.

The general context is that R is a finite-dimensional graded $\mathbb{Q}$-algebra (given as a quotient of a polynomial ring), and the top degree piece has dimension 1. I have an isomorphism of this top degree piece with $\mathbb{Q}$ (defined by sending a certain element to 1), and I want to be able to compute what it does to other elements.

Suppose I have a $\mathbb{Q}$-algebra R, and I have two elements x and y in R. I happen to know that x is a scalar multiple of y. Is there a way to figure out what the scalar is?

In other words, I want to identify the $\mathbb{Q}$-span of $y$ y with 1-dimensional vector space, choose an isomorphism of this vector space with $\mathbb{Q}$ (sending y to 1), and see where x goes.

The general context is that R is a finite-dimensional graded $\mathbb{Q}$-algebra (given as a quotient of a polynomial ring), and the top degree piece has dimension 1. I have an isomorphism of this top degree piece with $\mathbb{Q}$ (defined by sending a certain element to 1), and I want to be able to compute what it does to other elements.

Suppose I have a $\mathbb{Q}$-algebra R, and I have two elements x and y in R. I happen to know that x is a scalar multiple of y. Is there a way to figure out what the scalar is?

In other words, I want to identify the $\mathbb{Q}$-span of y with 1-dimensional vector space, choose an isomorphism of this vector space with $\mathbb{Q}$ (sending y to 1), and see where x goes.

The general context is that R is a finite-dimensional graded $\mathbb{Q}$-algebra (given as a quotient of a polynomial ring), and the top degree piece has dimension 1. I have an isomorphism of this top degree piece with $\mathbb{Q}$ (defined by sending a certain element to 1), and I want to be able to compute what it does to other elements.

An example of such a ring: let $A = \mathbb{Q}[x_1, x_2, x_3, x_4]$. Let $I_1 = (x_1x_3, x_2x_4)$. Let $I_2 = (x_1 + x_3, x_2 + x_4)$. Let $R = A/(I_1 + I_2)$.

(Note that $R$ is the Stanley-Reisner ring of simplicial complex (a triangulation of the circle) modulo a linear system of parameters. So it is graded by degree, and as the simplicial complex is a circle, the degree 2 is a 1-dimensional vector space.)

Let $y = x_1 x_2$, and let $x = (x_1 + x_3)(x_2 + x_4)$. x_4)(x_2 + x_3)$. Then x is a scalar multiple of y (as they are both in degree 2), and I would like to know what the scalar is.

(In this case, x = 4y.)

Suppose I have a $\mathbb{Q}$-algebra R, and I have two elements x and y in R. I happen to know that x is a scalar multiple of y. Is there a way to figure out what the scalar is?

In other words, I want to identify the $\mathbb{Q}$-span of y with 1-dimensional vector space, choose an isomorphism of this vector space with $\mathbb{Q}$ (sending y to 1), and see where x goes.

The general context is that R is a finite-dimensional graded $\mathbb{Q}$-algebra (given as a quotient of a polynomial ring), and the top degree piece has dimension 1. I have an isomorphism of this top degree piece with $\mathbb{Q}$ (defined by sending a certain element to 1), and I want to be able to compute what it does to other elements.

An example of such a ring: let $A = \mathbb{Q}[x_1, x_2, x_3, x_4]$. Let $I_1 = (x_1x_3, x_2x_4)$. Let $I_2 = (x_1 + x_3, x_2 + x_4)$. Let $R = A/(I_1 + I_2)$.

(Note that $R$ is the Stanley-Reisner ring of simplicial complex (a triangulation of the circle) modulo a linear system of parameters. So it is graded by degree, and as the simplicial complex is a circle, the degree 2 is a 1-dimensional vector space.)

Let $y = x_1 x_2$, and let $x = (x_1 + x_4)(x_2 + x_3)$. Then x is a scalar multiple of y (as they are both in degree 2), and I would like to know what the scalar is.

(In this case, x = 4y.)2y.)

Later edit: fixed typo