find ker of map from poly ring to poly ring,related to segre

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find ker of map from poly ring to poly ring,related to segre

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#1 h Mar 4, 2007, 3:09 PM • 4 Y

Y by dantx5, Adventure10, Mango247, and 1 other user

Hello,

Let $k$ be a field
let $R=k[Z_{i, j};i=0\cdots m,j=0\cdots n ]$
Let $S= k[x_{0},\cdots,x_{m},y_{0},\cdots,y_{n}]$

Let $\phi$ be the unique ring morphism from $R$ to $S$ , mapping $Z_{i,j}$ onto $x_{i}y_{j}$ .

Show that the kernel of $\phi$ is the ideal $W$ generated by all elements of type $Z_{a, b}Z_{c,d}-Z_{a,c}Z_{b,d}$ .
[ Moderator edit: Typo! This should be $Z_{a, b}Z_{c,d}-Z_{a,d}Z_{c,b}$ .]

Now the ideal I described is obviously a subset of the kernel, but how do I see all of the kernel is generated?

Thanks,
fredbel6

This post has been edited 3 times. Last edited by levans, Apr 6, 2015, 4:19 PM

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#2 h Mar 4, 2007, 9:41 PM • 2 Y

Y by Adventure10, Mango247

Maybe I should add that the point is that I need to prove that the generated ideal is a prime ideal (hence implying that the image of the Segre embedding is irreducible)
This should work because the quotient would then be isomorpic with a subring of S and thus be a domain.

But I'm still stuck.

(My book says it's just straightforward

)

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#3 h Mar 5, 2007, 2:36 PM • 1 Y

Y by Adventure10

Please let me know if you guys think this question is too easy, because I'm really trying hard to solve it.

I have come up with an alternative problem : since the ultimate goal is proving that the Segre variety is irreducible, it might be enough just to prove that $W$ is a prime ideal.

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-oo-

#4 h Mar 5, 2007, 5:17 PM • 1 Y

Y by Adventure10

keep cool. ;-) it doesn't seem to be easy.

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#5 h Mar 5, 2007, 5:26 PM • 1 Y

Y by Adventure10

-oo- wrote:

keep cool. ;-) it doesn't seem to be easy.

Is that good or bad news?

Maybe I should just say what the original problem was : consider the sets of projective points over an algebraically close field $k$ in $\mathcal{P}^{n m+n+m}$ of form $(a_{0}b_{0},\cdots,a_{0}b_{n},\cdots a_{m}b_{n})$ (not all zero of course)

Well prove that this set of points is not only a projective variety, but also an irreducible one.

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lofar

#6 h Mar 5, 2007, 10:07 PM • 2 Y

Y by Adventure10, Mango247

Polynomial set $\{Z_{a,b}Z_{c,d}-Z_{a,c}Z_{b,d}\}$ is a standard (Groebner) basis of the kernel. In particular it generates kernel.

You may refer to "An Introduction to Gröbner Bases" by Adams and Loustanau. It is a good easy-to-read book devoted to standard bases.

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-oo-

#7 h Mar 5, 2007, 11:51 PM • 2 Y

Y by Adventure10, Mango247

ok fredbel, I've also worked hard on it *g*. somewhat disappointing that lofar indicates that there's a more abstract and elegant approarch, which is unknown to me, but the following works with elementary means:

first, we may assume $m \leq n$ by symmetry. in the case $m=0$ , one easily sees that $\phi$ has trivial kernel, and that $W_{m,n}=0$ by definition, which says

$W_{m,n}= < Z_{ab}Z_{cd}-Z_{ac}Z_{bd}: 0 \leq a,c \leq m , 0 \leq b,d \leq \min(m,n) >$

now the key idea is the canonical isomorphism

$k[Z_{ij}; i=0...m, j=0...n] \cong k[Z_{ij}; i=0...m-1, j=0...n] [Z_{m0},...,Z_{mn}]$

and then induction by $m$ . let $f$ be an element of $\ker(\phi)$ , and write

$f = \sum_{\nu \in \mathbb{N}^{\{0,...,n\}}}a_{\nu}Z_{m0}^{\nu_{0}}... Z_{mn}^{\nu_{n}}\text{~~where~~}a_{\nu}\in k[Z_{ij}; i=0...m-1, j=0...n]$

now we apply $\phi$ . remark that $\phi(a_{\nu}) \in k[x_{0},... , x_{m-1}, y_{0},...y_{n}]$ , so that

$0 = \phi(f) = \sum_{\nu \in \mathbb{N}^{\{0,...,n\}}}\phi(a_{\nu}) x_{m}^{\nu_{0}+..+\nu_{n}}y_{0}^{\nu_{0}}... y_{n}^{\nu_{n}}\in k[x_{m}][x_{0},... , x_{m-1}, y_{0},...y_{n}]$

clearly implies $\phi(a_{\nu}) y_{0}^{\nu_{0}}... y_{n}^{\nu_{n}}=0$ oh, I just see that this is not clear :-(
and thus $\phi(a_{\nu})=0$ for all $\nu$ . by the induction hypothesis, it follows $a_{\nu}\subseteq W_{m-1,n}\subseteq W_{m,n}$ and hence $f \in W_{m,n}$ .

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buddha

#8 h Mar 6, 2007, 7:45 PM • 2 Y

Y by Adventure10, Mango247

Assume $f \in ker(\phi)$ and $f$ is an irreducible element in $\mathbb{k}[Z_{i}j]$ .now write $f = Z_{ij}Z_{mn}h+g;{(i,j)}\neq{(m,n)};g$ not divisible by $Z_{ij},Z_{mn}$ .Now as $\phi(f)=0$ ,and polynomial ring is a U.F.D we get $x_{i}x_{m}y_{j}y_{n}$ divides $-\phi(g)$ ,so $f = (Z_{ij}Z_{mn}-Z_{in}Z_{mj})l+q$ wid $q \in ker(\phi)$ . now proceeding wid a similar argument on $q$ u can arrive at the rquired result.
what do u think fred?.

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#9 h May 7, 2007, 2:47 PM • 2 Y

Y by Adventure10, Mango247

buddha wrote:

we get $x_{i}x_{m}y_{j}y_{n}$ divides $-\phi(g)$ ,so $f = (Z_{ij}Z_{mn}-Z_{in}Z_{mj})l+q$ wid $q \in ker(\phi)$ .

Can you explain the "so"? How do you proceed? to obtain that?

Quote:

now proceeding wid a similar argument on $q$ u can arrive at the rquired result.
what do u think fred?.

Hmm, can you explain how you proceed with a similar argument? I mean, we cannot afford to keep going in circles, what makes $q$ any different from $f$ ?

I'm sorry, I still don't get it.

Mean book.

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#10 h Sep 28, 2011, 2:11 AM • 2 Y

Y by Adventure10, Mango247

[EDIT: The content of the post below has been reposted with more detail and better notations at https://math.stackexchange.com/questions/353846/hartshorne-problem-1-2-14-on-segre-embedding .]

Notice a typo in the problem statement: The ideal $W$ should be generated by all elements of the type $Z_{a,b}Z_{c,d}-Z_{a,d}Z_{c,b}$ , not $Z_{a, b}Z_{c,d}-Z_{a,c}Z_{b,d}$ .

Also note that $k$ needs not be a field, but can be any commutative ring with $1$ .

A week ago I asked Bjorn Poonen for a solution of this problem (in response to him giving Hartshorne Chapter I Exercise 2.14 as homework). He came up with a very nice proof, which I am going to sketch:

Let $M=\left\{0,1,...,m\right\}$ and $N=\left\{0,1,...,n\right\}$ . We order the set $M\times N$ lexicographically.

We must show that $\mathrm{Ker}\phi = W$ .

We notice that $\mathrm{Ker}\phi\supseteq W$ is very easy to prove (in fact, a trivial computation shows that $\mathrm{Ker}\phi$ contains $Z_{a,b}Z_{c,d}-Z_{a,d}Z_{c,b}$ for all $a$ , $b$ , $c$ , $d$ ).

Every $k$ -tuple $\left(\left(a_1,b_1\right),\left(a_2,b_2\right),...,\left(a_k,b_k\right)\right)\in \left(M\times N\right)^k$ and every permutation $\sigma\in S_k$ satisfy

(1) $Z_{a_1,b_1}Z_{a_2,b_2}...Z_{a_k,b_k} \equiv Z_{a_1,b_{\sigma 1}}Z_{a_2,b_{\sigma 2}}...Z_{a_k,b_{\sigma k}} \mod W$ .

(In fact, this is obvious from the definition of $W$ when $\sigma$ is a transposition, and hence, by induction, it also holds for every permutation $\sigma$ because every permutation is a composition of transpositions.)

Now, let $T$ be the $k$ -submodule of $k\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$ generated by all products of the form $Z_{a_1,c_1}Z_{a_2,c_2}...Z_{a_k,c_k}$ with $k$ being a positive integer and $\left(\left(a_1,c_1\right),\left(a_2,c_2\right),...,\left(a_k,c_k\right)\right)\in \left(M\times N\right)^k$ being a $k$ -tuple satisfying $a_1\leq a_2\leq ...\leq a_k$ and $c_1\leq c_2\leq ...\leq c_k$ . It is easy to see that the map $\phi\mid_T:T\to k\left[x_0,x_1,...,x_m,y_0,y_1,...,y_n\right]$ is injective. (In fact,

$\left(\phi\mid_T\right)\left(Z_{a_1,c_1}Z_{a_2,c_2}...Z_{a_k,c_k}\right) = \phi\left(Z_{a_1,c_1}Z_{a_2,c_2}...Z_{a_k,c_k}\right) = x_{a_1}y_{c_1}x_{a_2}y_{c_2}...x_{a_k}y_{c_k}$
$=x_{a_1}x_{a_2}...x_{a_k}y_{c_1}y_{c_2}...y_{c_k}$

is a monomial from which we can recover the $k$ -tuple $\left(a_1,a_2,...,a_k\right)$ up to order and the $k$ -tuple $\left(c_1,c_2,...,c_k\right)$ up to order; but since the order of each of these two $k$ -tuples is predetermined by the condition that $a_1\leq a_2\leq ...\leq a_k$ and $c_1\leq c_2\leq ...\leq c_k$ , we can therefore recover these two $k$ -tuples completely; hence, the map $\phi\mid_T$ sends distinct monomials to distinct monomials, and thus is injective.)

Next we are going to show that:

(2) every monomial in $k\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$ lies in $T+W$ .

Proof of (2). Let $\mu$ be any monomial in $k\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$ . Then, $\mu=Z_{a_1,b_1}Z_{a_2,b_2}...Z_{a_k,b_k}$ for some $k$ and some $k$ -tuple $\left(\left(a_1,b_1\right),\left(a_2,b_2\right),...,\left(a_k,b_k\right)\right)\in \left(M\times N\right)^k$ such that $\left(a_1,b_1\right)\leq\left(a_2,b_2\right)\leq ...\leq \left(a_k,b_k\right)$ . Consider such a $k$ and such a $k$ -tuple $\left(\left(a_1,b_1\right),\left(a_2,b_2\right),...,\left(a_k,b_k\right)\right)$ . Since $\left(a_1,b_1\right)\leq\left(a_2,b_2\right)\leq ...\leq \left(a_k,b_k\right)$ , we have $a_1\leq a_2\leq ...\leq a_k$ (since our order is lexicographic). Clearly there exists a permutation $\sigma\in S_k$ such that $b_{\sigma 1}\leq b_{\sigma 2}\leq ... \leq b_{\sigma k}$ . Consider such a $\sigma$ . Let $c_i=b_{\sigma i}$ for every $i\in\left\{1,2,...,k\right\}$ . Then,

$\mu=Z_{a_1,b_1}Z_{a_2,b_2}...Z_{a_k,b_k} \equiv Z_{a_1,b_{\sigma 1}}Z_{a_2,b_{\sigma 2}}...Z_{a_k,b_{\sigma k}}$ (by (1))
$= Z_{a_1,c_1} Z_{a_2,c_2} ... Z_{a_k,c_k} \mod W$ (since $b_{\sigma i}=c_i$ for every $i\in\left\{1,2,...,k\right\}$ ).

But since $a_1\leq a_2\leq ...\leq a_k$ and $c_1\leq c_2\leq ...\leq c_k$ (the latter is a rewriting of $b_{\sigma 1}\leq b_{\sigma 2}\leq ... \leq b_{\sigma k}$ ), we have $Z_{a_1,c_1} Z_{a_2,c_2} ... Z_{a_k,c_k}\in T$ (by the definition of $T$ ), so this rewrites as follows:

$\mu\equiv \left(\text{an element of }T\right)\mod W$ .

In other words, $\mu\in T+W$ . Since this holds for every monomial $\mu$ in $k\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$ , this proves (2).

Since the monomials in $k\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$ generate the $k$ -module $k\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$ , and since $T+W$ is a submodule of this $k$ -module, we obtain $k\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]=T+W$ from (2).

Now there is a very easy algebraic lemma: If $A$ and $B$ are two submodules of a $k$ -module $C$ such that $C=A+B$ , and if $\psi$ is a $k$ -module map from $C$ to another $k$ -module $D$ such that $\psi\mid_A$ is injective and $\mathrm{Ker}\psi \supseteq B$ , then $\mathrm{Ker}\psi = B$ .

Applying this to $C=k\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$ , $A=T$ , $B=W$ and $\psi=\phi$ , we conclude that $\mathrm{Ker}\phi = W$ , qed.

This post has been edited 1 time. Last edited by darij grinberg, Nov 19, 2018, 7:46 AM

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