# Dimension of Factor Ring

• Mar 16th 2011, 03:59 PM
hurle32
Dimension of Factor Ring
Find the dimension of $\displaystyle \mathbb{R}[x,y]/(2xy,x+y+1).$
• Mar 16th 2011, 10:58 PM
NonCommAlg
if by "dimension" you mean dimension as a vector space over $\displaystyle \mathbb{R}$, the answer is $\displaystyle 2$. this is easy to see: put $\displaystyle x+y+1=u$ and $\displaystyle 2xy=v$ and let $\displaystyle I$ be the ideal of $\displaystyle \mathbb{R}[x,y]$ which is generated by $\displaystyle u$ and $\displaystyle v$. show that $\displaystyle x^2+x \in I$ and conclude that $\displaystyle \{1+I, x+I\}$ is an $\displaystyle \mathbb{R}$-basis for your ring.
• Mar 16th 2011, 11:16 PM
hurle32
I get everything up to the last conclusion: why does $\displaystyle x^2 + x \in I$ imply that $\displaystyle \{1+I,x+I\}$ is a basis?

Thanks!
• Mar 16th 2011, 11:35 PM
NonCommAlg
$\displaystyle y+I=-x-1 + I$ and so every element of your ring is in the form $\displaystyle p(x)+I$, for some $\displaystyle p(x) \in \mathbb{R}[x].$ also, since $\displaystyle x^2 + x \in I$, we have $\displaystyle p(x)+I=ax + b + I$, for some $\displaystyle a,b \in \mathbb{R}.$ so $\displaystyle \{1+I, x + I\}$ generates your ring. it is obvious that the set is $\displaystyle \mathbb{R}$-linearly independent.