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

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
$$dim X = dim sigma(mathbb{P}^1times mathbb{P}^2) +1 = (1+2)+1 = 4.$$

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

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
$$dim X = dim sigma(mathbb{P}^1times mathbb{P}^2) +1 = (1+2)+1 = 4.$$

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

$$R = mathbb{C}[t]_{(t)}[x]/(x^3+x^2+t)$$ 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.

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

From Eisenbud’s Commutative Algebra chapter 2: 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 $$f(xi)=0$$.

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

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)

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 $$1$$.

## Math Genius: Zero divisors in annihilator of modules

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. 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?

## 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)\$

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.

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

$$R = mathbb{C}[t]_{(t)}[x]/(x^3+x^2+t)$$ 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.

## Math Genius: Ideal of the twisted cubic

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.

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

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