-1

Simple question ( Originally I proposed a more stronger question first, which seems to be false through communicating with user26857 below. So I re-edit our question so that only second question leaves. )

Let $f(x,y,z) := y^2z+yz^2-x^3+xz^2 \in \mathbb{Z}[x,y,z]$. Then are there homogeneous prime ideals $\mathfrak{p}_1, \mathfrak{p}_2 \subseteq \mathbb{Z}[x,y,z]$ with $\mathfrak{p}_1 , \mathfrak{p}_2 \nsupseteq (x,y,z)$ ( i.e., not necessary $\mathfrak{p}_1,\mathfrak{p}_2 \subsetneq (x,y,z)$ ) such that $$ I:=(f(x,y,z)) \subsetneq \mathfrak{p}_{1,+}:=\mathfrak{p}_1 \cap (x,y,z) \subsetneq \mathfrak{p}_{2,+}:=\mathfrak{p}_2\cap (x,y,z)$$ ?

Can we find such homogeneous prime ideals?

This question originates from following question that I posed but not yet answered : Why $\operatorname{Proj}\mathbb{Z}[x,y,z]/(y^2z+yz^2-x^3+xz^2)$ is fibered surface over $\operatorname{Spec}\mathbb{Z}$? ; Refer to the 'first attempt to the first question' in the link.

Can anyone helps?

Plantation
  • 2,417
  • 1
    There are no primes satisfying your requirement, let alone homogeneous since the height of (x,y,z) in the polynomial ring is 3. – user26857 Sep 25 '23 at 10:03
  • 1
    @user26857 isn't it 4 since we are in the polynomial ring over the integers? – Sergey Guminov Sep 25 '23 at 10:14
  • @user26857 : Uhm.. if the height of $(x,y,z)$ is 3, then it contradicts to the existence of chain of length 4 : $(0) \subsetneq I \subsetneq p_1 \subsetneq p_2 \subsetneq (x,y,z)$. But is it really true? I found associated theorem (Eisenbud's Commutative algebra, Corollary 10.4) : $(x_1,\dots,x_c) \subset k[x_1,\dots x_r]$ has codimension $c$. But is it limited only when k is a field? By the Principal ideal theorem, if $(x,y,z)$ is minimal among primes of $\mathbb{Z}[x,y,z]$ containing $x,y,z$, then $\operatorname{ht}(x,y,z) \le 3$. Correct argument? And is that true?.. – Plantation Sep 25 '23 at 10:27
  • 2
    @SergeyGuminov The Krull dimension of the ring is 4, indeed, but the height of (x,y,z) is 3. (Note that this is not a maximal ideal.) – user26857 Sep 25 '23 at 10:34
  • 1
    @Plantation Yes, your argument is correct. More precisely, the height of an ideal (prime or not) is at most the minimal number of generators of that ideal. – user26857 Sep 25 '23 at 10:35
  • @user26857 : O.K. I edit my question asking more weaker statement. It is directly connected to my linked question. Can you also see the question ? – Plantation Sep 25 '23 at 10:40
  • @user26857 : Oh, sorry. I edit quetsion more correctly reflecting my intention. Can you also see? – Plantation Sep 25 '23 at 11:00
  • 1
    Could you do something like $p_1 = (y^2z+yz^2-x^3+xz^2, x,z)$ and $p_2 = (y^2z+yz^2-x^3+xz^2, x, z,7)$ ? – Jenny Kenkel Sep 27 '23 at 14:43
  • O.K. Thank you. And they are primes!? How can we show? Checking primness is one of key in our question. Certainly, $p_2 \subsetneq (x,y,z)$. So it seems that its choice does not violate to our discussion in upper comments ( If our discussion is correct ). – Plantation Sep 28 '23 at 08:58
  • @Jenny kenkel : I am found out method. Perhaps. Let $f(x,y,z) := y^2z+yz^2-x^3+xz^2$. Then $(f,x,z) = (f) + (x,z)$ is true? If so, then by the second and third isomorphism theorem about rings (!) , we can show that $\mathbb{Z}[x,y,z]/(f,x,z) \cong \mathbb{Z}[y]$. So $(f,x,z)$ is prime ideal. True? The primeness of the $(f,x,z,7)$ can be proved similarly? Perhaps, $\mathbb{Z}[x,y,z]/(x,z,7) \cong \mathbb{Z}[y]$? – Plantation Sep 28 '23 at 10:17
  • 1
    @JennyKenkel : I think that your example is choosed by careful consideration. – Plantation Sep 28 '23 at 10:19
  • 1
    @Jenny Kenkel Correction : through direct calculation, I think that $\mathbb{Z}[x,y,z]/(x,z,7) \cong \mathbb{Z}[y]/(7)$, which is integral domain ( https://math.stackexchange.com/questions/2167051/prime-p-in-r-remains-prime-in-rx-gausss-lemma ) – Plantation Sep 28 '23 at 11:32
  • @Jenny Kenkel : I wonder if $\mathfrak{p}{1,+}:=\mathfrak{p}_1 \cap (x,y,z) \subsetneq \mathfrak{p}{2,+}:=\mathfrak{p}_2\cap (x,y,z)$ is satisfied for your examples. – Plantation Sep 29 '23 at 12:13
  • 1
    Ah, you're right, it isn't. I guess you want $p_2=(y^2z+yz^2-x^3+xz^2,x,z,y)$ instead – Jenny Kenkel Sep 29 '23 at 13:55
  • @Jenny Kenkel I think that your $p_2$ doesn't satisfy condition $p_2 \supsetneq (x,y,z)$ in our question. T.T. Can we think any other object? Anyway your trial was helpful to understand about testing primeness of ideal. – Plantation Sep 30 '23 at 04:37
  • @Jenny Kenkel : I think that I find suitable such $p_2$. I leave that as a answer. Can you see? I think that this answer may be helpful to you :) – Plantation Oct 04 '23 at 08:44

1 Answers1

0

O.K.

Let $f(x,y,z) := y^2z+yz^2-x^3+xz^2 \in \mathbb{Z}[x,y,z]$.

Let $\mathfrak{p}_1 := (f, x,z)$ and $\mathfrak{p}_2:= (f,x,z,7y,7)$ (ideals in $\mathbb{Z}[x,y,z]$.) Then $\mathfrak{p}_1 , \mathfrak{p}_2 \nsupseteq (x,y,z)$ (True?) and such that $$ I:=(f) \subsetneq \mathfrak{p}_{1,+}:=\mathfrak{p}_1 \cap (x,y,z) \subsetneq \mathfrak{p}_{2,+}:=\mathfrak{p}_2\cap (x,y,z).$$

( $\mathfrak{p}_{1,+} \subsetneq \mathfrak{p}_{2,+}$ since $7y$ is not contained in $\mathfrak{p}_{1,+}$ ).

It remains to show that each $\mathfrak{p}_1 , \mathfrak{p}_2$ are prime ideals in $\mathbb{Z}[x,y,z]$. But by using the second and third isomorphism theorems about ring (!) well we can show that

  • $\mathbb{Z}[x,y,z]/\mathfrak{p}_1 \cong \mathbb{Z}[y]$
  • $\mathbb{Z}[x,y,z]/\mathfrak{p}_2 \cong \mathbb{Z}[y]/(7)$

which are integral domains! ( C.f. Prime $p\in R$ remains prime in $R[x]$ (Gauss's Lemma)) ( As an exercise, calculation of this by yourself is recommended. Still, if anyone is interested, I will upload detailed calculation.)

So we are done.

Plantation
  • 2,417