# Thread: generators for one algebra

1. ## generators for one algebra

Hello,

Let $\displaystyle I=<x^{2}-\beta y^{2}>$ be an ideal of $\displaystyle \mathbb{Q}[x,y]$ where $\displaystyle \beta\neq a^{2}$ for any $\displaystyle a\in \mathbb{Q}$. Find two generators for the algebra $\displaystyle \frac{\mathbb{Q}[x,y]}{I}$.

2. Originally Posted by Biscaim
Hello,

Let $\displaystyle I=<x^{2}-\beta y^{2}>$ be an ideal of $\displaystyle \mathbb{Q}[x,y]$ where $\displaystyle \beta\neq a^{2}$ for any $\displaystyle a\in \mathbb{Q}$. Find two generators for the algebra $\displaystyle \frac{\mathbb{Q}[x,y]}{I}$.

since $\displaystyle \sqrt{\beta} \notin \mathbb{Q},$ the polynomial $\displaystyle x^2 - \beta y^2$ is irreducible over $\displaystyle \mathbb{Q}[y][x]=\mathbb{Q}[x,y].$ thus: $\displaystyle \frac{\mathbb{Q}[x,y]}{I} = \frac{\mathbb{Q}[y][x]}{<x^2-\beta y>} \cong \mathbb{Q}[y][\sqrt{\beta}y]=\mathbb{Q}[y,\sqrt{\beta}y],$ where $\displaystyle \sqrt{\beta} \in \mathbb{C}-\mathbb{Q}.$

note that if $\displaystyle \beta = a^2,$ for some $\displaystyle 0 \neq a \in \mathbb{Q},$ then $\displaystyle x^2-\beta y^2=(x- ay)(x+ay)$ and thus by the Chinese remainder theorem: $\displaystyle \frac{\mathbb{Q}[x,y]}{I} \cong \mathbb{Q}[y] \times \mathbb{Q}[y].$