Quotient of Polynomial Ring is Artinian

Question

I start with the polynomial ring $R = \mathbb{C}[x,y]$ and the ideal $I=(x^2 + ax, y^2 + by, xy + bx, xy +ay)$ for some $a,b \in \mathbb{C}^*$, $a\neq b$. I would like to prove that $R/I$ is artinian. I know that that $R$ and thus $R/I$ are noetherian, so I am left to show that $\dim(R/I)=0$.

Any ideas on how this can be done (rather elementarily)?

score 2 · Accepted Answer · answered Mar 07 '20 at 18:51

2

Even easier than doing "Noetherian + $0$ dimensional":

$R/I$ is spanned as a $\mathbb C$ vector space by the set $\{1+I, x+I, y+I\}$, so it is a finite dimensional $\mathbb C$ algebra, hence Artinian.

answered Mar 07 '20 at 18:51

rschwieb

153,510

1

How does one prove that ${1+I,x+I,y+I}$ is a basis, in fact, isn't ${1+I,x+I}$ already a basis? – EinStone Mar 08 '20 at 09:53
@EinStone It is, but I didn’t want to waste time proving anything was a basis. Since it is a generating set, that’s already enough, and it’s easy to see it is a generating set. – rschwieb Mar 08 '20 at 13:22

score 1 · Answer 2 · answered Mar 07 '20 at 16:36

1

Find all the solutions! If it is finite, then the ring has dimension 0.

answered Mar 07 '20 at 16:36

RghtHndSd

7,715

1

Why does finitely many solutions imply dim = 0? – EinStone Mar 07 '20 at 16:38

Quotient of Polynomial Ring is Artinian

2 Answers2