Finding maximal ideals in an artinian ring.

Asked Aug 17 '21 at 03:56

Active Aug 17 '21 at 03:56

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Consider the ring $R:=\mathbb{C}[x,y]/\langle x^2-y^3, 4y^2-5x^2y+x^4\rangle$. I want to do the following:

Show that $R$ is artinian and calculate the maximal ideals $\mathfrak{m}$ of $R$. How many ideals there are?
Calculate the dimension of $R$ and $R_\mathfrak{m}$ as complex vector spaces.

I think it could be useful the following facts:

$R$ is noetherian (but I am not sure why the Krull dimension of $R$ is $0$)
$(x^2-y^3)$ is prime ideal of $\mathbb{C}[x,y]$. Actually, there is an isomorphims between $\mathbb{C}[t]$ and $\mathbb{C}[x,y]/\langle x^2-y^3\rangle$ such that $\overline{x}\mapsto t^3$ and $\overline{y}\mapsto t^2$.
$R$ has finitely many maximal ideals.

So, I am not certain how to calculate explicitly such ideals and dimensions. I aprreciate your ideas.

asked Aug 17 '21 at 03:56

Andrés Felipe

To prove its Krull dimension is $0$, you might be able to use that $\mathbb C[x,y]/(x^2-y^3)\cong\mathbb C[t]$ has Krull dimension $1$. – Kenta S Aug 17 '21 at 03:58
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I think you can use the relation $y^3-x^2=0$ to plug into $4y^2+x^4-5x^2y$ – Aug 17 '21 at 04:06
Like $4^2+^4−5^2 =y^2(y-2)(y+2)(y-1)(y+1)$ – Aug 17 '21 at 04:07
Let $x/y=t$, you have $t^2=x^2/y^2=y$ and $t^3=x$ – Aug 17 '21 at 04:09
Then you will get something like $\mathbb{C}[t]/(t^4(t^2-2)(t^2+2)(t^2-1)(t^2+1))$. Then the rest is obvious. – Aug 17 '21 at 04:10
2

The rings $\Bbb C[t]$ and $\Bbb C[x,y]/\langle x^2-y^3\rangle$ are not isomorphic. Indeed the first is normal while the second is not. – leoli1 Aug 17 '21 at 07:45
Related – Viktor Vaughn Aug 17 '21 at 16:06
Thank you for your comments. With these I can conclude that the $\mathbb{C}$-dimension of $A$ is 12 and there are nine maximal ideals. ¿Is it right? – Andrés Felipe Oct 12 '21 at 03:41

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