I am dealing with the following question:
Le be $f(x_1,x_2)\in k[x_1,x_2]$ non-constant. Prove that $\dim Z(f)=1.$
I know that $\dim Z(f)=\dim \dfrac{k[x_1,x_2]}{I(Z(f))}$, where $I(Z(f))$ is the set of zeros of $Z(f)$ (and I think it is $\langle f\rangle $, is it?).
So, I wonder that the dimension is $1$ once it cannot be $2$ (because of there is more than one root, so dimension is lower) and cannot be $0$, because of there is some point that is not root.
This is very unformal and I'd like to get some clue to formalize an answer.
Thanks in advance!
You really only want to talk about the dimension of $I$ when $R/I$ is a domain. That is, only dimensions of irreducible varieties. For example, if $I$ is the ideal corresponding to the line $x = 0$ union the plane $y = z = 0$ then the total variety has one dimension $1$ part (the line) and one dimension $2$ part (the plane). Of course, when you have a principal ideal $(f)$ and $f = g_1 \cdots g_k$ where each $g_i$ is irreducible, then $\operatorname{codim} (g_i) = 1$ by what I will show next. Therefore every irreducible component of $Z(f)$ has codimension $1$.
So let's take $f$ to be irreducible. Then $I(Z(f)) = (f)$ is a prime ideal and chains of (prime) ideals of $k[x_1, x_2]/(f)$ correspond to chains of (prime) ideals containing $f$. So if we have a chain of length $2$ in $k[x_1, x_2]/(f)$:
$$ \mathfrak{p}_0/(f) \subset \mathfrak{p}_1/(f) \subset \mathfrak{p}_3/(f), $$
we get a chain of length at least $3$ in $k[x_1, x_2]$:
$$ (0) \subset (f) \subseteq \mathfrak{p}_0 \subset \mathfrak{p}_1 \subset \mathfrak{p}_2 $$
(possibly $\mathfrak{p}_0 = (f)$). Since we know that $\dim [x_1, x_2] = 2$, this can't happen.
Conversely, we know that $\dim [x_1, x_2]/(f) > 0$ since if $(a_1, a_2)$ is any zero of $f$, then
$$ (f) \subset (x_1 - a_1, x_2 - a_2). $$
Also note that this must be a proper subset because $(x_1 - a_1, x_2 - a_2)$ cannot be a principal ideal. Otherwise $f \mid x_1 - a_1$ and $f \mid x_2 - a_2$ which is only possible if $f \in k$.
You can work this out using polynomial division. First divide by $x_1 - a_1$ until you get a remainder which is a polynomial in $x_2$ only. Then divide by $x_2 - a_2$ until you get a remainder which is constant. Then the remainder must be $0$ since it equals $f(a_1,a_2)$.
I recommend Eisenbud's book Commutative Algebra (Section II: Dimension Theory) for more on Krull dimension.
If $f = g_1^{i_1} \cdots g_k^{i_k}$ is the irreducible factorization of $f$ (where $g_i$ are irreducible and distinct), then $$I(Z(f)) = (g_1 \cdots g_k).$$ You can maybe see why this should be the case, because $Z(f) = Z(g_1\cdots g_k)$. To prove this, you use Hilbert's Nullstellensatz.