Checking if a ring is Artinian.

Question

I am reviewing some commutative algebra notes for an upcoming course, and I found an exercise I wanted to solve, but I found out I absolutely have no idea how.

Let $K$ be a field (I think $\operatorname{char}K\neq 2$, but the exercise doesn't outright state it), find whether $$R = k[x,y]/(2x^2 + 3y^2 - 11, x^2 - y^2 - 3)$$ and $$S = k[x,y]/(x^2 - xy + x, xy - y^2 + y)$$ are artinian or not.

I mean, they are of course noetherian, should I check if its Krull dimension is $0$, but I don't know how to check it. Could someone please help me? Thank you!

Answer 1

Note that the meaning for Krull dimension of a ring $A$ equal to $0$ is that every prime ideal in $A$ is also maximal.

Here as you have realized, $$R=k[x,y]/(2x^2+3y^2-11,x^2-y^2-3)=k[x,y]/(x^2-4,y^2-1).$$

Let $\mathfrak{p}$ be a prime ideal in $R$.

Recall that a prime (resp. maximal) ideal in the quotient ring $A/I$ are precisely of the form $P/I$, where $P$ is a prime (resp. maximal) ideal of $A$ containing $I$.

Therefore, $$\mathfrak{p}=P/(x^2-4,y^2-1),$$ where $P$ is a prime ideal of $k[x,y]$ containing $(x^2-4,y^2-1)$. Here we have trivial factorizations $$x^2-4=(x-2)(x+2)\quad\text{and}\quad y^2-1=(y-1)(y+1).$$ Because $P$ is prime, at least one of $x-2$ and $x+2$ is in $P$, and similarly, at least one of $y-1$ and $y+1$ belongs to $P$.

Every ideal in $k[x_1,\ldots,x_n]$ of the form $(x_1-a_1,\ldots,x_n-a_n)$ for $a_1,\ldots,a_n\in k$ is maximal.

By displaying all cases, you shall see that $P$ is indeed a maximal ideal in $k[x,y]$, hence $\mathfrak{p}=P/(x^2-4,y^2-1)$ is also a maximal ideal in $R$, as desired.

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Similarly, for $S$, observe that $$x^2-xy+x=x(x-y+1)\quad\text{and}\quad xy-y^2+y=y(x-y+1).$$ If $P$ is a prime ideal in $k[x,y]$ containing $(x^2-xy+x,xy-y^2+y)$, at least one of $x$ and $x-y+1$, and also at least one of $y$ and $x-y+1$, lies in $P$. However, there is one case that $$P=(x-y+1),$$ which is not maximal because $P\subsetneq(x,y-1)$ where $(x,y-1)$ is maximal in $k[x,y]$. Therefore, $S$ is not Artinian.