Math Genius: Dimension of affine algebraic set in $mathbb{A^6}$

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If $X=V(f_1, f_2, f_3) subseteq mathbb{A}^6$, with
$$f_1=x_1x_5-x_4x_2, qquad f_2=x_1x_6-x_4x_3, qquad f_3=x_2x_6-x_5x_3,
$$

how can I show $dim X=4$?
I was trying to find $operatorname{ht} I(X)$, since $dim X=6- operatorname{ht} I(X)$ or construct explicitely an isomorphism between $A(X)$ and a polynomial ring with $4$ variables, but I haven’t succeeded.

Expanding on Sasha’s comment above.

Consider the Segre embedding $sigma:mathbb{P}^1times mathbb{P}^2 to mathbb{P}^5$ given by
$$sigma:((a_0:a_1),(b_0:b_1:b_2)) longmapsto (a_0b_0:a_0b_1:a_0b_2:a_1b_0:a_1b_1:a_1b_2).$$
Letting $x_1,ldots,x_6$ be the coordinate functions on $mathbb{P}^5,$ we see that
$$sigma(mathbb{P}^1timesmathbb{P}^2)=V(x_1x_5-x_2x_4,,x_1x_6-x_3x_4,,x_2x_6-x_3x_5)subseteq mathbb{P}^5.$$
Hence your variety $X$ is the affine cone in $mathbb{A}^6$ over $sigma(mathbb{P}^1times mathbb{P}^2)$ and so we have
$$dim X = dim sigma(mathbb{P}^1times mathbb{P}^2) +1 = (1+2)+1 = 4.$$

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Math Genius: Dimension of affine algebraic set in $mathbb{A^6}$

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If $X=V(f_1, f_2, f_3) subseteq mathbb{A}^6$, with
$$f_1=x_1x_5-x_4x_2, qquad f_2=x_1x_6-x_4x_3, qquad f_3=x_2x_6-x_5x_3,
$$

how can I show $dim X=4$?
I was trying to find $operatorname{ht} I(X)$, since $dim X=6- operatorname{ht} I(X)$ or construct explicitely an isomorphism between $A(X)$ and a polynomial ring with $4$ variables, but I haven’t succeeded.

Expanding on Sasha’s comment above.

Consider the Segre embedding $sigma:mathbb{P}^1times mathbb{P}^2 to mathbb{P}^5$ given by
$$sigma:((a_0:a_1),(b_0:b_1:b_2)) longmapsto (a_0b_0:a_0b_1:a_0b_2:a_1b_0:a_1b_1:a_1b_2).$$
Letting $x_1,ldots,x_6$ be the coordinate functions on $mathbb{P}^5,$ we see that
$$sigma(mathbb{P}^1timesmathbb{P}^2)=V(x_1x_5-x_2x_4,,x_1x_6-x_3x_4,,x_2x_6-x_3x_5)subseteq mathbb{P}^5.$$
Hence your variety $X$ is the affine cone in $mathbb{A}^6$ over $sigma(mathbb{P}^1times mathbb{P}^2)$ and so we have
$$dim X = dim sigma(mathbb{P}^1times mathbb{P}^2) +1 = (1+2)+1 = 4.$$

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Math Genius: Factorization into prime ideals in dedekind domain $mathbb{C}[t]_{(t)}[x]/(x^3+x^2+t)$

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$R = mathbb{C}[t]_{(t)}[x]/(x^3+x^2+t)$ is a dedekind domain. Therefore every proper ideal $I$ can be written as a product of prime ideals.

I want to find the factorization of $I = (t+x^2-x)$ into prime ideals in $R$.
I tried to describe the prime ideals of $S = mathbb{C}[t]_{(t)}[x]$ the following way:

  • zero ideal $(0)$
  • $(f)$ where $f$ irreducible in $S$
  • $(p)$ where $p$ prime in $mathbb{C}[t]_{(t)}$
  • $(p,f)$ where $p$ prime in $mathbb{C}[t]_{(t)}$ and $f$ irreducible mod $p$

Then, the prime ideals in $R$ would correspond to those of $S$ that contain $(x^3+x^2+t)$.

Am I going in the right direction with this?
Also I’m not able to find the factorization, how do I have to do this?

I think you are on the right track. In fact, $I$ is a prime ideal.

In $R$, one sees that
$$
t + x^2 -x = (-x^3-x^2) + x^2 – x= -x^3-x = -x(x-i)(x+i).
$$

Thus any prime ideal of $P$ containing $I$ must contain one of the ideals $(t,x)R, (t -1 -i,x-i)R$, or $(t-1+i,x+i)R$.
Notice that the second and third cases cannot happen as $t-1-i, t-1+i$ are units in $R$.
Therefore, since $R$ is a Dedekind domain, $I = P^a$ for some $a$, where $P = (t, x)$ (it is straightforward to check $P$ is prime).

One can determine $a$ locally. In $R_P$, $x$ is a uniformizing parameter, and the images of $x-i,x+i$ will be units. Thus
$IR_P = xR_P$. Hence $I = P$.

Of course, we could’ve checked that $R/I$ is a field directly, but I believe this approach is more general.

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Math Genius: Clarification of open affine neighborhood at $p$

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From Eisenbud’s Commutative Algebra chapter 2:

enter image description here

If $Y$ is defined by one equation $f=0$, then isn’t $Y=mathcal Z(0)=mathbb A^r$?

The phrase “algebraic set defined by one equation $f=0$” stands for $mathcal Z(f)$, because $mathcal Z(f)$ is the set of the points $xi$ such that $f(xi)=0$.

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Math Genius: Example of a polynomial with a lesser degree than the minimal monic polynomial.

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Let $A$ and $B$ ($Asubset B$) be commutative rings with unity. Let $ain B$ be integral over $A$. Therefore, there exists a monic polynomial $pin A[x]$, such that $p(a)=0$.

By well-ordering principle, there exists a monic polynomial $p_0 in A[x]$, such that $p_0(a) = 0$ and
$$text{degree}(p_0) := min{text{degree}(g); gin A[x], g text{is monic and }g(a)=0}. $$

Question: Does anyone know an example of commutative rings $A$ and $B$ (with unity), such that there is a polynomial $gin A[x]$ (not necessarily monic) such that $g(a)=0$ and $text{degree}(g)<text{degree}(p_0)?$

If $A$ is a field then it is obvious that it is impossible. But in the general case, I am not being able to find a counterexample. Can anyone help me?

One can choose $A = Q[u]$ and $B = Q[u,v]/(uv, v^2+1)$.

The equation of integrality of $v$ over $A$ is $T^2+1 in A[T]$. If there exists a monic polynomial $f(T) in A[T]$ of degree $1$ such that $f(v) = 0$, it must be of the form $T-v$. Since $v notin A$, this is impossible. So, $deg p_0 = 2$.

For $g(T)$, one can take $g(T) = uT in A[T]$ which is of degree $1$.

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Math Genius: Zero divisors in annihilator of modules

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I faced a theorem

Let $R$ be a noetherian local ring, $M$ a finitely generated module of finite projective dimension.
Then if the annihilator of $M$ is not trivial, then it contains a nonzero zero divisor in $R$.

Or more precisely, in this form.

enter image description here

This is a part of Vasconcelos’ paper Ideals Generated by R-Sequences, he referred paper of Auslander and Buchsbaum Homological dimension in local ring. But there is no Proporsition 6.2 in the paper referred.

My question is, how to prove this proposition?

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Math Genius: Show that $phi^{-1}((y_1-a_1,ldots,y_n-a_n))=(x_1-b_1,ldots,x_m-b_m)$ for points in affine varieties $psi((a_1,ldots,a_n))=(b_1,ldots,b_m)$

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Let $A=k[x_1,ldots,x_m]$ and $B=k[y_1,ldots,y_n]$ over some algebraically closed field $k$. Let $psi : mathbb{A}^nrightarrow mathbb{A}^m$ be the morphism corresponding to a $k$-algebra homomorphism $phi: Arightarrow B$. Let $a=(a_1,ldots,a_n)in mathbb{A}^n$ and $psi(a)=(b_1,ldots,b_m)inmathbb{A}^m$. I want to show that $phi^{-1}((y_1-a_1,ldots,y_n-a_n))=(x_1-b_1,ldots,x_m-b_m)$.

I know that since $phi$ is a $k$-algebra homomorphism and $A$ and $B$ are finitely generated $k$-algebra, then the preimage of a maximal ideal is again a maximal ideal. I also know that by Hilberts Nullstellensatz, then a point $(a_1,ldots,a_n)$ in a affine variety corresponds to the maximal ideal $(a_1-p_1,ldots,x_n-a_n)$. And by how $psi$ is defined on the point $a$, then this seems to be natural that $phi^{-1}((y_1-a_1,ldots,y_n-a_n))=(x_1-b_1,ldots,x_m-b_m)$. However I was wondering if there were a more rigorously way of doing it or seeing it. My argument seems a little loose.

Your argument is the right idea. Let me try to give you some help.

A $k$-point in $Bbb A^n$ is the same as a map from $Bbb A^0to Bbb A^n$ picking out the point we care about. On the coordinate algebra side, this is represented by the map $k[y_1,cdots,y_n]to k$ given by $y_imapsto a_i$, and the preimage of the maximal ideal $(0)subset k$ is exactly the maximal ideal $(y_1-a_1,cdots,y_n-a_n)$.

Composing this with our map $Bbb A^nto Bbb A^m$, we get a map $Bbb A^0to Bbb A^m$ which corresponds to $k[x_1,cdots,x_m]to k$. The map $Bbb A^0to Bbb A^m$ picks out $(b_1,cdots,b_m)$, so we have that our map on coordinate rings $k[x_1,cdots,x_m]to k$ is given by $x_jmapsto b_j$, and the preimage of the maximal ideal $(0)subset k$ is exactly the maximal $(x_1-b_1,cdots,x_m-b_m)$.

Now look at the sequence of maps $k[x_1,cdots,x_m]to k[y_1,cdots,y_n]to k$. The preimage of $(0)$ along the first map is $(y_1-a_1,cdots,y_n-a_n)$, and the preimage of $(0)$ along the composite is $(x_1-b_1,cdots,x_m-b_m)$. But this means that the preimage of $(y_1-a_1,cdots,y_n-a_n)$ is $(x_1-b_1,cdots,x_m-b_m)$ by, for instance, the second-to-last line of this answer.

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Math Genius: Factorization into prime ideals in dedekind domain $mathbb{C}[t]_{(t)}[x]/(x^3+x^2+t)$

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$R = mathbb{C}[t]_{(t)}[x]/(x^3+x^2+t)$ is a dedekind domain. Therefore every proper ideal $I$ can be written as a product of prime ideals.

I want to find the factorization of $I = (t+x^2-x)$ into prime ideals in $R$.
I tried to describe the prime ideals of $S = mathbb{C}[t]_{(t)}[x]$ the following way:

  • zero ideal $(0)$
  • $(f)$ where $f$ irreducible in $S$
  • $(p)$ where $p$ prime in $mathbb{C}[t]_{(t)}$
  • $(p,f)$ where $p$ prime in $mathbb{C}[t]_{(t)}$ and $f$ irreducible mod $p$

Then, the prime ideals in $R$ would correspond to those of $S$ that contain $(x^3+x^2+t)$.

Am I going in the right direction with this?
Also I’m not able to find the factorization, how do I have to do this?

I think you are on the right track. In fact, $I$ is a prime ideal.

In $R$, one sees that
$$
t + x^2 -x = (-x^3-x^2) + x^2 – x= -x^3-x = -x(x-i)(x+i).
$$

Thus any prime ideal of $P$ containing $I$ must contain one of the ideals $(t,x)R, (t -1 -i,x-i)R$, or $(t-1+i,x+i)R$.
Notice that the second and third cases cannot happen as $t-1-i, t-1+i$ are units in $R$.
Therefore, since $R$ is a Dedekind domain, $I = P^a$ for some $a$, where $P = (t, x)$ (it is straightforward to check $P$ is prime).

One can determine $a$ locally. In $R_P$, $x$ is a uniformizing parameter, and the images of $x-i,x+i$ will be units. Thus
$IR_P = xR_P$. Hence $I = P$.

Of course, we could’ve checked that $R/I$ is a field directly, but I believe this approach is more general.

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Math Genius: Ideal of the twisted cubic

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The twisted cubic is the image of the morphism $phi : mathbb{P}^1 to mathbb{P}^3 , (x:y) mapsto (x^3:x^2 y:x y^2:y^3)$, it is given by $X = V(ad-bc,b^2-ac,c^2-bd)$. Now I would like to compute $I(X)$, which equals by the Nullstellensatz the radical of the ideal $I := (ad-bc,b^2-ac,c^2-bd) subseteq k[a,b,c,d]$. I think that that $I$ is already a radical ideal, even a prime ideal. Namely, I suspect that
$$phi^* : Q:=k[a,b,c,d]/I to k[s,t] , a mapsto s^3, b mapsto s^2 t , c mapsto s t^2 , d mapsto t^3$$
is an injection. If this is true: How can we prove that? I’ve already tried to find a $k$-basis of the quotient, but this turned out to be a big mess. Even the representation of the quotient as a monoid algebra doesn’t seem to help. Another idea is the following: A formal manipulation of generators and relations implies $Q_a cong k[a,b]_a$. Thus it suffices to prove that $Q to Q_a$ is injective, i.e. that $a$ is not a zero divisor.

Here is a purely algebraic proof that $I(X)=I$.
It is of course sufficient to prove that $I(X) subset I$ and for that it suffices to prove that every homogeneous polynomial $P(a,b,c,d)$
which vanishes on $X$ is in $I$.

Lemma
Any homogeneous polynomial $P(a,b,c,d)in k[a,b,c,d]$ can be written $$P(a,b,c,d)=R(a,d) +S(a,d)b+T(a,d)c+i(a,b,c,d) $$ for some polynomials $R,S,Tin k[a,d]$
and a polynomial $iin I$
The easy proof is by induction on the degree of $P$ and I’ll leave it to you.

Now back to our problem. If now that homogeneous $P$ is in $ I(X)$ , we write it as in the lemma and get by using that $P$ vanishes on $X$ that for all $(x:y)in mathbb P^1_k$ $$0=P(x^3,x^2y,xy^2,y^3)=R(x^3,y^3) +S(x^3,y^3)x^2y+T(x^3,y^3)xy^2+0 $$
By considering exponents modulo $3$ for $x$ and $y$, we see that no cancellation occurs, hence that $R=S=T=0$ and thus $P=iin I$ as required.

The ideal $I$ is prime if and only if its associated projective scheme $V_p(I)subset mathbb P^3_k$ is integral.
This in turn can be checked on the four standard affine open subschemes covering $mathbb P^3_k$.
For example in the affine open subscheme (isomorphic to $mathbb A^3_k$) $U_dsubset mathbb P^3_k$ corresponding to $d=1$, the scheme $V_p(I)cap U_d$ is defined by the ideal $(a-bc,b^2-ac,c^2-b)$ which is trivially prime in $k[a,b,c]$.
Three similar calculations will imply that indeed the original ideal $I$ is prime.

There are several algorithms that can compute whether your ideal $I$ is prime or not, most of the algorithms use Groebner basis. I gave your ideal to a maple package, namely PrimDecomp, obtainable from http://wwwb.math.rwth-aachen.de/~markus/ and it says that your ideal is prime.

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Math Genius: Question regarding p-basis in complete local ring

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Let $k$ be the residue field of characteristic $p>0$ of a complete local ring $R$ with maximal ideal $m$. A $p$-basis for $k$ is a subset $B subset k$ such that

(i) $k^p(B)=k$

(ii) all monomials of the type $b_1^{e_1}b_2^{e_2}…..b_r^{e_r}$ , $b_i in B$, $0leq e_i leq p-1$ are linearly independent.

Now what I can’t understand is how can we show existence of this $p$– basis of any field $k$ with $ch k=p>0$,
because as I think if $k$ is perfect field , then it may not have a $p$-basis. Any help will be appreciated, thanks in advanced

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