Dimension of quotient rings and zero divisors

Question

In a previous question, I asked about the correctness of a method to compute the Krull dimension of quotient rings which works well if the ring in question is of the form $A/(x_1,\ldots,x_n)$, where for each $i$, $x_i$ is not a zero divisor in $A/(x_1,\ldots,x_{i-1})$.

However, I am now confronted with the problem to compute the Krull dimension of the following ring: $$A=\mathbb{C}[x,y]/(xy-x,y^2-y)$$ in which the required property doesn't hold. Now, in this special case it is easy to show that $\dim A\le 1$ and that we have the chain of primes in $A$ composed of the image of the chain: $$(y-1)\subseteq(y-1,x)$$ and thus $\dim A = 1$, but I am wondering: is there a fast way to find out the dimension of a ring of the form written above if we do not require the condition of non-zero-divisibility of the $x_i$?

Answer 1

EDIT: Since there was a typo in the original question, the answer before did not address the right ring.

There are various ways to compute the dimension in your specific example. Given a quotient ring $A/(x_1,...,x_n)$, one can think of the intersection of the various hypersurfaces $V(x_i) \subseteq \text{Spec}(A)$. In your case, for the ring $\mathbb{C}[x,y]/(xy - x, y^2 - y)$, this is the intersection of two hypersurfaces: the first being the union of the $x$-axis and the line $y = 1$, and the second being the union of the lines $y = 0$ and $y = 1$. Their intersection consists of the line $y = 1$ and the origin, and has dimension $1$.

Alternatively, $xy - x$ and $y^2 - y$ have a common factor of $y - 1$. Thus the ideal $I := (xy - x, y^2 - y)$ is a product of two ideals, $(x,y)(y - 1)$, which is also the intersection $(x,y) \cap (y - 1)$. This yields $\dim \mathbb{C}[x,y]/I = \max \{\dim \mathbb{C}[x,y]/(x,y), \dim \mathbb{C}[x,y]/(y-1) \} = 1$ (in general, if $I = I_1...I_n$ or $I = I_1 \cap \ldots \cap I_n$, then $\dim R/I = \max\{\dim R/I_i\}$).

Finally, one always has $\dim R/I \le \dim R - \text{ht}(I)$, but if $R$ is a finitely generated domain over a field (which $\mathbb{C}[x,y]$ certainly is), then equality holds, so in these cases one can think about computing heights of ideals rather than dimensions of quotients. Krull's altitude theorem always gives an upper bound on height, and for rings of small dimension one can often find the correct value by experimenting.

Answer 2

Just to emphasize the geometric aspect of user115654's answer:

For simple systems of equations like these, you can just "plot the graph'' (more precisely, compute the associated variety).

You have $xy = x,$ so $x = 0$ or $y = 1$, and also $y^2 - y$, so $y = 0$ or $y = 1$.

So the variety has two components: the line $y = 1$, and the point $x = y = 0$. Thus the dimension of the associated ring is the max. of the dimension of these two components, which is $1$.