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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!

Sera Gunn
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Quiet_waters
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1 Answers1

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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.

and I think it is $⟨f⟩$, is it?

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.

Sera Gunn
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  • Thank you! I'll study the details, but already understand what you mean, many thanks for this detailed answer. – Quiet_waters Nov 29 '19 at 18:23
  • Hi, please, could you explain a little bit more this part? "Conversely, we know that dim[x1,x2]/(f)>0 since if (a1,a2) is any zero of f, then (f)⊂(x1−a1,x2−a2)." I understood that (f)⊂(x1−a1,x2−a2). However, why this show that dim[x1,x2]/(f)>0? If for instance (f)=(x1−a1,x2−a2), so dim[x1,x2]/(f)=0 (because is a field)... Or I am wrong? Many thanks! – Quiet_waters Dec 05 '19 at 22:01
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    Because $(f) \subset (x_1 - a_1, x_2 - a_2)$ is a chain of prime ideals containing $(f)$. When I write $\subset$ I mean strict subset. So you may be wondering why $(f) \ne (x_1 - a_1, x_2 - a_2)$. One form of Hilbert's Nullstellensatz says that if $k$ is algebraically closed, then the only maximal ideals are of the form $(x_1 - a_1, x_2 - a_2)$. For general fields, Krull's principal ideal theorem says that a principal ideal has height $1$, but $(0) \subset (x_1 - a_1) \subset (x_1 - a_1, x_2 - a_2)$ so the latter ideal has height $2$. There may be an elementary way of seeing this as well. – Sera Gunn Dec 05 '19 at 22:05
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    @Na'omi If you believe that $f$ has more than one zero, then you can see that $(f) \subset (x_1 - a_1, x_2 - a_2)$ for any zero $(a_1, a_2)$. But now it should be easy to see that $(x_1 - a_1, x_2 - a_2) \subseteq (x_1 - a_1', x_2 - a_2')$ if and only if $(a_1,a_2) = (a_1',a_2')$. – Sera Gunn Dec 05 '19 at 22:07
  • Thank you so much, I think I understood what you say. However, it may seen easy, but I could not see yet that $f$ always have more than one root, could you help me again? Many thanks! – Quiet_waters Dec 05 '19 at 22:45
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    Honestly, I'm not sure what the proof looks like without appealing to dimension @Na'omi Unless maybe you plug in a certain a to get $\deg f(a,x_2) \ge 2$ but I don't think that always works. – Sera Gunn Dec 05 '19 at 23:27
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    Hi, I hope I found something interesting: https://math.stackexchange.com/questions/286352/polynomial-with-infinitely-many-zeros. Many thanks! – Quiet_waters Dec 06 '19 at 00:25
  • The fact that some authors use as results what other authors use as definition and vice-versa makes something difficult. But now I think the question is closed. If you want, you can put the link above as an edit of your answer, so it will be complete...! Thank you so much for the help. Best regards. – Quiet_waters Dec 06 '19 at 00:45
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    One certainly does not need to appeal to Krull's Height Theorem to prove that the ideal $(X_{1}-a_{1}, X_{2} - A_{2})$ is not principal. Indeed, if this ideal were generated by some $f$, then $f$ would necessarily be a divisor of both $X_{1}-a_{1}$ and $X_{2}-a_{2}$, hence a unit. – Alex Wertheim Dec 06 '19 at 01:58
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    @AlexWertheim Thank you, I knew there was something simple I was missing. – Sera Gunn Dec 06 '19 at 02:01
  • @TrevorGunn, sorry by new comment. I was wondering. The link I post consider $k$ algebrically closed. This was not an initial condition. However, I was wondering this is necessary, what do you think? Otherwise, for example, $ f (x, y) = x ^ 2 + 1 \in \mathbb {R} [x, y] $ has no roots. Thus, $ Z (f) = \emptyset $ and there is a convention that the Krull dimension for the empty set is $ -1 $ (for example, 'Algebraic Approach To Differential Equations', by Le Dung Trang). Thank you for the attention. – Quiet_waters Dec 06 '19 at 14:03
  • @AlexWertheim,thank you for the addendum! – Quiet_waters Dec 06 '19 at 14:04
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    Yes, dimension doesn't work as nicely if the field isn't algebraically closed for the reason you point out. You'll also notice that $\mathbb{R}[x,y]/(x^2 + 1) \cong \mathbb{C}[y]$ which is 1-dimensional. @Na'omi – Sera Gunn Dec 06 '19 at 14:11