MHB Elementary Algebraic Geometry - D&F Section 15.1 - Exercise 15

[Math Amateur]

Dummit and Foote (D&F), Ch15, Section 15.1, Exercise 15 reads as follows:

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If k = \mathbb{F}_2 and V = \{ (0,0), (1,1) \} \subset \mathbb{A}^2,

show that \mathcal{I} (V) is the product ideal m_1m_2

where m_1 = (x,y) and m_2 = (x -1, y-1).

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I am having trouble getting started on this problem.

One issue/problem I have is - what is the exact nature of m_1, m_2 and m_1m_2. What (explicitly) are the nature of the elements of these ideals.

I would appreciate some help and guidance.

Peter
Note: D&F define \mathcal{I} (V) as follows:

$$ \mathcal{I} (V) = \{ f \in k(x_1, x_2, ... , x_n) \ | \ f(a_1, a_2, ... , a_n) = 0 \ \ \forall \ \ (a_1, a_2, ... , a_n) \in V \} $$

 

[Math Amateur]

Dummit and Foote (D&F), Ch15, Section 15.1, Exercise 15 reads as follows:

----------------------------------------------------------------------------------------------------

If k = \mathbb{F}_2 and V = \{ (0,0), (1,1) \} \subset \mathbb{A}^2,

show that \mathcal{I} (V) is the product ideal m_1m_2

where m_1 = (x,y) and m_2 = (x -1, y-1).

------------------------------------------------------------------------------------------------------

I am having trouble getting started on this problem.

One issue/problem I have is - what is the exact nature of m_1, m_2 and m_1m_2. What (explicitly) are the nature of the elements of these ideals.

I would appreciate some help and guidance.

Peter
Note: D&F define \mathcal{I} (V) as follows:

$$ \mathcal{I} (V) = \{ f \in k(x_1, x_2, ... , x_n) \ | \ f(a_1, a_2, ... , a_n) = 0 \ \ \forall \ \ (a_1, a_2, ... , a_n) \in V \} $$

================================================================
================================================================

I am still working on this problem, but have made little progress. I have clarified some of the elements of the problem as follows:

$$ m_1 = (x,y) $$ and $$ m_2 = (x-1, y-1) $$The product $$ m_1m_2 $$ = the set of all finite sums of the form ab with $$ a \in m_1 $$ and $$ b \in m_2 $$

$$ m_1 = \{f_1x + f_2y \ | \ f_1, f_2 \in \mathbb{F}_2 (x,y) \} $$

$$ m_2 = \{g_1 (x-1) + g_2 (y - 1) \ | \ g_1, g_2 \in \mathbb{F}_2 (x,y) \} $$

$$ V = \mathcal{Z} (S) = \{ (0,0), (1,1) \} \subset \mathbb{A}^2 $$

i.e. V is the set of functions f such that f = 0 on V$$ \mathcal{I}(V) = \mathcal{I} ( \{ (0,0), (1,1) \} ) $$

$$ = \{ f \in \mathbb{F}_2 (x,y) \ | \ f(0,0) = f(1,1) = 0 \}$$
... ... so the above gives the elements of the exercise, but now ... how to solve ?

Would appreciate help and guidance.

Peter

 

[Turgul]

In reading your questions it is clear that a big picture understanding of what is going on has been elusive. I don't have a copy of Dummit & Foote on hand to remember what exactly they say, but let's see if we can't shed some light on the situation.

In algebraic geometry, we are interested in studying "algebraic sets" as objects of geometric interest. What is an algebraic set? It is the zero set of a collection of polynomials. For example all of the points $(x,y)$ satisfying $y=x^2$ form an algebraic set because they are the solutions of $y-x^2=0$. Another example of an algebraic set is $\{(0,0),(1,1)\}$ because it is the common zeroes of the polynomials $y-x=0$ and $y-x^2=0$, that is both of these points are zeroes of both polynomials at the same time, and no other points have this property.

There are a couple of things to note above:

First, in the above examples all points were ordered pairs and all polynomials were in two variables. That tacitly assumes that I'm living in an ambient two dimensional space (the space of all possible ordered pairs $(x,y)$ with no restrictions on $x$ or $y$ at all). Note that in this space, the equation $x=0$ cuts a line of points $(0,y)$ where $y$ can be anything. If we were living in one dimensional space, then $x=0$ would cut out precisely the point $x=0$. This means we should pay at least some attention to where we are looking for zeroes of our polynomials. At some point, we would like to realize that "one-space" lives inside of "two-space" lives in "three-space" and so on, but some care must be taken as to what we mean when we say that, and it's a more technical point to be puzzled later.

Second, polynomials have different solutions over different rings. Consider $x^2+1=0$. Over the real numbers, this has no zeroes at all! Over the complex numbers, we have $x = i$ and $x = -i$, and over the field of two elements, we have the single solution $x=1$. So if we are looking for solutions to polynomials, we have to pick a ring to look in! For practical purposes, we often restrict our attention to fields to make our lives easier (in practice, we actually often further assume that the field is algebraically closed, but there is something to be learned about how that matters, so we'll avoid that assumption for now), which is what D&F certainly do and we will as well, for the time being.
For now, let's restrict our attention to algebraic sets sitting inside of two-dimensional space over a fixed field $k$. Algebraic sets are just special kinds of sets, we can ask if they have nice properties in terms of typical set operations. Consider the polynomials $y-x=0$ and $y-x^2=0$. One cuts out a line, the other a parabola. If I intersect the two shapes, I get the two points from before. That is to say, if I require that points satisfy both at the same time, I get the intersection of the shapes. Since algebraic sets are allowed to be simultaneous zero sets of multiple polynomials (we don't even require finitely many!), this operation is allowed.

On the other hand, what happens if we look at $(y-x)(y-x^2)=0$, the zero set of the product of the two polynomials? Since we live over a field, if the product of two things is zero, either one or the other must be zero, so if $(a,b) \in k^2$ with $(b-a)(b-a^2)=0$, it must be that $b-a = 0$ or $b-a^2=0$. Hence the zero set of $(y-x)(y-x^2)=0$ is contained in the union of the line and the parabola. On the other hand, any point on either the line or the parabola will be a zero of the product. So by looking at the zeroes of the product polynomial, we get the union of the first two!

What's more is that if $A$ is an algebraic set defined by $f_1(x,y)=0,f_2(x,y)=0,\ldots$
and $B$ is defined by $g_1(x,y)=0,g_2(x,y)=0,\ldots$, the the intersection $A \cap B$ is defined by $f_1(x,y)=0,g_1(x,y)=0,f_2(x,y)=0,g_2(x,y)=0, \ldots$ the common zeroes of all the polynomials defining each. On the other hand, the union $A \cup B$ is defined by $f_1(x,y)g_1(x,y)=0,f_1(x,y)g_2(x,y)=0,f_2(x,y)g_1(x,y)=0,f_2(x,y)g_2(x,y)=0, \ldots$ the products of every polynomial defining $A$ by each polynomial defining $B$. You should convince yourself of this.

Now, with many polynomials running around, this gets very complicated quickly, so it is natural to wonder if there is a better way of managing the collections of polynomials. Fortunately, there is. It turns out that the ideal of $k[x,y]$ generated by all of the polynomials defining an algebraic set defines the same algebraic set! At first glance this just seems to have made things worse. After all, if we defined a set using two polynomials, the ideal generated by them has infinitely many polynomials! The great simplification comes when we realize that all that really matters are the generators of the ideal, so we can pick our favorite generators for the ideal and do all the computations with them. Instead of looking at the zero set of $xy,x^2y,xy^2,x^3y^4$, we can just look at the zero set of $xy$ since it generates the same ideal as the other four polynomials!
I suspect this gets overwhelming, so instead of rambling on, I will leave you with the following question:

The first thing to consider before you attack your problem is why is the union of algebraic sets given by the common zeroes of the products of the defining polynomials of each? That is, instead of considering $m_1$ and $m_2$, consider just the pairs of polynomials $x,y$ and $x-1, y-1$ defining the two points $(0,0)$ and $(1,1)$ seperately. Why are the two points together defined by the four polynomials $xy,x(y-1),(x-1)y,(x-1)(y-1)$?

Your problem is just a supped up version of this in terms of ideals instead of polynomials.

 

[Math Amateur]

In reading your questions it is clear that a big picture understanding of what is going on has been elusive. I don't have a copy of Dummit & Foote on hand to remember what exactly they say, but let's see if we can't shed some light on the situation.

In algebraic geometry, we are interested in studying "algebraic sets" as objects of geometric interest. What is an algebraic set? It is the zero set of a collection of polynomials. For example all of the points $(x,y)$ satisfying $y=x^2$ form an algebraic set because they are the solutions of $y-x^2=0$. Another example of an algebraic set is $\{(0,0),(1,1)\}$ because it is the common zeroes of the polynomials $y-x=0$ and $y-x^2=0$, that is both of these points are zeroes of both polynomials at the same time, and no other points have this property.

There are a couple of things to note above:

First, in the above examples all points were ordered pairs and all polynomials were in two variables. That tacitly assumes that I'm living in an ambient two dimensional space (the space of all possible ordered pairs $(x,y)$ with no restrictions on $x$ or $y$ at all). Note that in this space, the equation $x=0$ cuts a line of points $(0,y)$ where $y$ can be anything. If we were living in one dimensional space, then $x=0$ would cut out precisely the point $x=0$. This means we should pay at least some attention to where we are looking for zeroes of our polynomials. At some point, we would like to realize that "one-space" lives inside of "two-space" lives in "three-space" and so on, but some care must be taken as to what we mean when we say that, and it's a more technical point to be puzzled later.

Second, polynomials have different solutions over different rings. Consider $x^2+1=0$. Over the real numbers, this has no zeroes at all! Over the complex numbers, we have $x = i$ and $x = -i$, and over the field of two elements, we have the single solution $x=1$. So if we are looking for solutions to polynomials, we have to pick a ring to look in! For practical purposes, we often restrict our attention to fields to make our lives easier (in practice, we actually often further assume that the field is algebraically closed, but there is something to be learned about how that matters, so we'll avoid that assumption for now), which is what D&F certainly do and we will as well, for the time being.
For now, let's restrict our attention to algebraic sets sitting inside of two-dimensional space over a fixed field $k$. Algebraic sets are just special kinds of sets, we can ask if they have nice properties in terms of typical set operations. Consider the polynomials $y-x=0$ and $y-x^2=0$. One cuts out a line, the other a parabola. If I intersect the two shapes, I get the two points from before. That is to say, if I require that points satisfy both at the same time, I get the intersection of the shapes. Since algebraic sets are allowed to be simultaneous zero sets of multiple polynomials (we don't even require finitely many!), this operation is allowed.

On the other hand, what happens if we look at $(y-x)(y-x^2)=0$, the zero set of the product of the two polynomials? Since we live over a field, if the product of two things is zero, either one or the other must be zero, so if $(a,b) \in k^2$ with $(b-a)(b-a^2)=0$, it must be that $b-a = 0$ or $b-a^2=0$. Hence the zero set of $(y-x)(y-x^2)=0$ is contained in the union of the line and the parabola. On the other hand, any point on either the line or the parabola will be a zero of the product. So by looking at the zeroes of the product polynomial, we get the union of the first two!

What's more is that if $A$ is an algebraic set defined by $f_1(x,y)=0,f_2(x,y)=0,\ldots$
and $B$ is defined by $g_1(x,y)=0,g_2(x,y)=0,\ldots$, the the intersection $A \cap B$ is defined by $f_1(x,y)=0,g_1(x,y)=0,f_2(x,y)=0,g_2(x,y)=0, \ldots$ the common zeroes of all the polynomials defining each. On the other hand, the union $A \cup B$ is defined by $f_1(x,y)g_1(x,y)=0,f_1(x,y)g_2(x,y)=0,f_2(x,y)g_1(x,y)=0,f_2(x,y)g_2(x,y)=0, \ldots$ the products of every polynomial defining $A$ by each polynomial defining $B$. You should convince yourself of this.

Now, with many polynomials running around, this gets very complicated quickly, so it is natural to wonder if there is a better way of managing the collections of polynomials. Fortunately, there is. It turns out that the ideal of $k[x,y]$ generated by all of the polynomials defining an algebraic set defines the same algebraic set! At first glance this just seems to have made things worse. After all, if we defined a set using two polynomials, the ideal generated by them has infinitely many polynomials! The great simplification comes when we realize that all that really matters are the generators of the ideal, so we can pick our favorite generators for the ideal and do all the computations with them. Instead of looking at the zero set of $xy,x^2y,xy^2,x^3y^4$, we can just look at the zero set of $xy$ since it generates the same ideal as the other four polynomials!
I suspect this gets overwhelming, so instead of rambling on, I will leave you with the following question:

The first thing to consider before you attack your problem is why is the union of algebraic sets given by the common zeroes of the products of the defining polynomials of each? That is, instead of considering $m_1$ and $m_2$, consider just the pairs of polynomials $x,y$ and $x-1, y-1$ defining the two points $(0,0)$ and $(1,1)$ seperately. Why are the two points together defined by the four polynomials $xy,x(y-1),(x-1)y,(x-1)(y-1)$?

Your problem is just a supped up version of this in terms of ideals instead of polynomials.

Thanks so much for your post Turgil.

I will work through it carefully now.

Peter

 

[Math Amateur]

In reading your questions it is clear that a big picture understanding of what is going on has been elusive. I don't have a copy of Dummit & Foote on hand to remember what exactly they say, but let's see if we can't shed some light on the situation.

In algebraic geometry, we are interested in studying "algebraic sets" as objects of geometric interest. What is an algebraic set? It is the zero set of a collection of polynomials. For example all of the points $(x,y)$ satisfying $y=x^2$ form an algebraic set because they are the solutions of $y-x^2=0$. Another example of an algebraic set is $\{(0,0),(1,1)\}$ because it is the common zeroes of the polynomials $y-x=0$ and $y-x^2=0$, that is both of these points are zeroes of both polynomials at the same time, and no other points have this property.

There are a couple of things to note above:

First, in the above examples all points were ordered pairs and all polynomials were in two variables. That tacitly assumes that I'm living in an ambient two dimensional space (the space of all possible ordered pairs $(x,y)$ with no restrictions on $x$ or $y$ at all). Note that in this space, the equation $x=0$ cuts a line of points $(0,y)$ where $y$ can be anything. If we were living in one dimensional space, then $x=0$ would cut out precisely the point $x=0$. This means we should pay at least some attention to where we are looking for zeroes of our polynomials. At some point, we would like to realize that "one-space" lives inside of "two-space" lives in "three-space" and so on, but some care must be taken as to what we mean when we say that, and it's a more technical point to be puzzled later.

Second, polynomials have different solutions over different rings. Consider $x^2+1=0$. Over the real numbers, this has no zeroes at all! Over the complex numbers, we have $x = i$ and $x = -i$, and over the field of two elements, we have the single solution $x=1$. So if we are looking for solutions to polynomials, we have to pick a ring to look in! For practical purposes, we often restrict our attention to fields to make our lives easier (in practice, we actually often further assume that the field is algebraically closed, but there is something to be learned about how that matters, so we'll avoid that assumption for now), which is what D&F certainly do and we will as well, for the time being.
For now, let's restrict our attention to algebraic sets sitting inside of two-dimensional space over a fixed field $k$. Algebraic sets are just special kinds of sets, we can ask if they have nice properties in terms of typical set operations. Consider the polynomials $y-x=0$ and $y-x^2=0$. One cuts out a line, the other a parabola. If I intersect the two shapes, I get the two points from before. That is to say, if I require that points satisfy both at the same time, I get the intersection of the shapes. Since algebraic sets are allowed to be simultaneous zero sets of multiple polynomials (we don't even require finitely many!), this operation is allowed.

On the other hand, what happens if we look at $(y-x)(y-x^2)=0$, the zero set of the product of the two polynomials? Since we live over a field, if the product of two things is zero, either one or the other must be zero, so if $(a,b) \in k^2$ with $(b-a)(b-a^2)=0$, it must be that $b-a = 0$ or $b-a^2=0$. Hence the zero set of $(y-x)(y-x^2)=0$ is contained in the union of the line and the parabola. On the other hand, any point on either the line or the parabola will be a zero of the product. So by looking at the zeroes of the product polynomial, we get the union of the first two!

What's more is that if $A$ is an algebraic set defined by $f_1(x,y)=0,f_2(x,y)=0,\ldots$
and $B$ is defined by $g_1(x,y)=0,g_2(x,y)=0,\ldots$, the the intersection $A \cap B$ is defined by $f_1(x,y)=0,g_1(x,y)=0,f_2(x,y)=0,g_2(x,y)=0, \ldots$ the common zeroes of all the polynomials defining each. On the other hand, the union $A \cup B$ is defined by $f_1(x,y)g_1(x,y)=0,f_1(x,y)g_2(x,y)=0,f_2(x,y)g_1(x,y)=0,f_2(x,y)g_2(x,y)=0, \ldots$ the products of every polynomial defining $A$ by each polynomial defining $B$. You should convince yourself of this.

Now, with many polynomials running around, this gets very complicated quickly, so it is natural to wonder if there is a better way of managing the collections of polynomials. Fortunately, there is. It turns out that the ideal of $k[x,y]$ generated by all of the polynomials defining an algebraic set defines the same algebraic set! At first glance this just seems to have made things worse. After all, if we defined a set using two polynomials, the ideal generated by them has infinitely many polynomials! The great simplification comes when we realize that all that really matters are the generators of the ideal, so we can pick our favorite generators for the ideal and do all the computations with them. Instead of looking at the zero set of $xy,x^2y,xy^2,x^3y^4$, we can just look at the zero set of $xy$ since it generates the same ideal as the other four polynomials!
I suspect this gets overwhelming, so instead of rambling on, I will leave you with the following question:

The first thing to consider before you attack your problem is why is the union of algebraic sets given by the common zeroes of the products of the defining polynomials of each? That is, instead of considering $m_1$ and $m_2$, consider just the pairs of polynomials $x,y$ and $x-1, y-1$ defining the two points $(0,0)$ and $(1,1)$ seperately. Why are the two points together defined by the four polynomials $xy,x(y-1),(x-1)y,(x-1)(y-1)$?

Your problem is just a supped up version of this in terms of ideals instead of polynomials.

Thanks Turgil, you certainly have helped my understanding of affine algebraic sets, but I still have a way to go in trying to get an understanding of the basics of algebraic geometry ,,,

In this post, I will try to address your question:

" ... ... why is the union of algebraic sets given by the common zeroes of the products of the defining polynomials of each?"

Since the exercise deals in 2 dimensions let us consider your question in $$ k[x_1, x_2] $$.

In Dummit and Foote's terminology (see attachment)

In $$ k[x_1, x_2] $$ we have

$$ V_1 = \mathcal{Z} (S) $$

$$ = \mathcal{Z} ( \{ f_1, f_2, ... \ ... f_s \} )$$

$$ = \{ (a_1, a_2) \in \mathbb{A}^2 \ | \ f_i (a_1, a_2) = 0 \ \forall \ f_i \in S \} $$

and

$$ V_2 = \mathcal{Z} (T) $$

$$ = \mathcal{Z} ( \{ g_1, g_2, ... \ ... g_s \} )$$

$$ = \{ (a_1, a_2) \in \mathbb{A}^2 \ | \ g_i (a_1, a_2) = 0 \ \forall \ g_i \in T \} $$

Then for $$ V_1 \cup V_2 $$ we have

$$ f_i (a_1, a_2) \times g_j (a_1, a_2) = 0 \ \forall \ i, j $$ and $$ \forall \ (a_1, a_2) \in V_1 \cup V_2 $$ since if $$ (a_1, a_2) \in V_1 $$ then $$ f_i (a_1, a_2) = 0 $$ and if $$ (a_1, a_2) \in V_2 $$ then $$ g_j (a_1, a_2) = 0 $$.

===============================================================

I am still puzzling a bit over your next suggestions since to apply the above re affine algebraic sets $$ V_1, V_2 $$ I would be looking for two affine algebraic sets (again in D&F's terminology - see attachment) , say, $$ V_1 = \mathcal{Z} (S_1) , V_2 = \mathcal{Z} (S_2) $$ where $$ S_1 = \{ f_1, f_2, ... \ ... f_{s_1} \} , S_2 = \{ g_1, g_2, ... \ ... g_{s_2} \} $$.

But when we come to $$ V= \{ (0,0) , (1,1) \} $$, although it looks like we could express this as $$ V = V_1 \cup V_2 $$ where $$ V_1 = \{ (0,0) \} $$ and $$ V_2 = \{ (1,1) \} $$ it does not appear intuitive to do this as the polynomial $$ f_1 = x-y $$ defines an affine algebraic set $$ V_1 \cup V_2 = \{ (0,0) , (1,1) \} $$ as does $$ f_2 = y - x^2 $$ as do some other polynomials like $$ f_3 = y^2 - x $$ and $$ f_4 = x^2 - y $$.

However, I am still thinking over this ...

================================================================

You advise me to consider the four polynomials $$ g_1 = xy, g_2 = x(y-1), g_3 = (x-1)y , g_4 = (x-1)(y-1) $$

Inspection reveals that in $$ \mathbb{F}_2 $$:

$$ g_1 = xy = 0 $$ at the points $$ (0,0), (0,1), (1,0) $$

$$ g_2 = x(y-1) = 0 $$ at the points $$ (0,0), (0,1), (1,1) $$

$$ g_3 = (x-1)y = 0 $$ at the points $$ (0,0), (1,0), (1,1) $$

$$ g_4 = (x-1)(y-1) = 0 $$ at the points $$ (0,1), (1,0), (1,1) $$

Thus in D&F's terminology (see attachment)

$$ V_1 = \mathcal{Z} (g_1) = \{ (0,0), (0,1), (1,0) \} $$

$$ V_2 = \mathcal{Z} (g_2) = \{ (0,0), (0,1), (1,1) \} $$

$$ V_3 = \mathcal{Z} (g_3) = \{ (0,0), (1,0), (1,1) \} $$

$$ V_4 = \mathcal{Z} (g_4) = \{ (0,1), (1,0), (1,1) \} $$

SO just seeing how $$ g_1 = xy, g_2 = x(y-1), g_3 = (x-1)y , g_4 = (x-1)(y-1) $$ might define $$ \{ (0,0), (1,1) \} $$

Any comments on the above?

Thanks again for your post, it was most helpful!

Peter

 

[Math Amateur]

In reading your questions it is clear that a big picture understanding of what is going on has been elusive. I don't have a copy of Dummit & Foote on hand to remember what exactly they say, but let's see if we can't shed some light on the situation.

In algebraic geometry, we are interested in studying "algebraic sets" as objects of geometric interest. What is an algebraic set? It is the zero set of a collection of polynomials. For example all of the points $(x,y)$ satisfying $y=x^2$ form an algebraic set because they are the solutions of $y-x^2=0$. Another example of an algebraic set is $\{(0,0),(1,1)\}$ because it is the common zeroes of the polynomials $y-x=0$ and $y-x^2=0$, that is both of these points are zeroes of both polynomials at the same time, and no other points have this property.

There are a couple of things to note above:

First, in the above examples all points were ordered pairs and all polynomials were in two variables. That tacitly assumes that I'm living in an ambient two dimensional space (the space of all possible ordered pairs $(x,y)$ with no restrictions on $x$ or $y$ at all). Note that in this space, the equation $x=0$ cuts a line of points $(0,y)$ where $y$ can be anything. If we were living in one dimensional space, then $x=0$ would cut out precisely the point $x=0$. This means we should pay at least some attention to where we are looking for zeroes of our polynomials. At some point, we would like to realize that "one-space" lives inside of "two-space" lives in "three-space" and so on, but some care must be taken as to what we mean when we say that, and it's a more technical point to be puzzled later.

Second, polynomials have different solutions over different rings. Consider $x^2+1=0$. Over the real numbers, this has no zeroes at all! Over the complex numbers, we have $x = i$ and $x = -i$, and over the field of two elements, we have the single solution $x=1$. So if we are looking for solutions to polynomials, we have to pick a ring to look in! For practical purposes, we often restrict our attention to fields to make our lives easier (in practice, we actually often further assume that the field is algebraically closed, but there is something to be learned about how that matters, so we'll avoid that assumption for now), which is what D&F certainly do and we will as well, for the time being.
For now, let's restrict our attention to algebraic sets sitting inside of two-dimensional space over a fixed field $k$. Algebraic sets are just special kinds of sets, we can ask if they have nice properties in terms of typical set operations. Consider the polynomials $y-x=0$ and $y-x^2=0$. One cuts out a line, the other a parabola. If I intersect the two shapes, I get the two points from before. That is to say, if I require that points satisfy both at the same time, I get the intersection of the shapes. Since algebraic sets are allowed to be simultaneous zero sets of multiple polynomials (we don't even require finitely many!), this operation is allowed.

On the other hand, what happens if we look at $(y-x)(y-x^2)=0$, the zero set of the product of the two polynomials? Since we live over a field, if the product of two things is zero, either one or the other must be zero, so if $(a,b) \in k^2$ with $(b-a)(b-a^2)=0$, it must be that $b-a = 0$ or $b-a^2=0$. Hence the zero set of $(y-x)(y-x^2)=0$ is contained in the union of the line and the parabola. On the other hand, any point on either the line or the parabola will be a zero of the product. So by looking at the zeroes of the product polynomial, we get the union of the first two!

What's more is that if $A$ is an algebraic set defined by $f_1(x,y)=0,f_2(x,y)=0,\ldots$
and $B$ is defined by $g_1(x,y)=0,g_2(x,y)=0,\ldots$, the the intersection $A \cap B$ is defined by $f_1(x,y)=0,g_1(x,y)=0,f_2(x,y)=0,g_2(x,y)=0, \ldots$ the common zeroes of all the polynomials defining each. On the other hand, the union $A \cup B$ is defined by $f_1(x,y)g_1(x,y)=0,f_1(x,y)g_2(x,y)=0,f_2(x,y)g_1(x,y)=0,f_2(x,y)g_2(x,y)=0, \ldots$ the products of every polynomial defining $A$ by each polynomial defining $B$. You should convince yourself of this.

Now, with many polynomials running around, this gets very complicated quickly, so it is natural to wonder if there is a better way of managing the collections of polynomials. Fortunately, there is. It turns out that the ideal of $k[x,y]$ generated by all of the polynomials defining an algebraic set defines the same algebraic set! At first glance this just seems to have made things worse. After all, if we defined a set using two polynomials, the ideal generated by them has infinitely many polynomials! The great simplification comes when we realize that all that really matters are the generators of the ideal, so we can pick our favorite generators for the ideal and do all the computations with them. Instead of looking at the zero set of $xy,x^2y,xy^2,x^3y^4$, we can just look at the zero set of $xy$ since it generates the same ideal as the other four polynomials!
I suspect this gets overwhelming, so instead of rambling on, I will leave you with the following question:

The first thing to consider before you attack your problem is why is the union of algebraic sets given by the common zeroes of the products of the defining polynomials of each? That is, instead of considering $m_1$ and $m_2$, consider just the pairs of polynomials $x,y$ and $x-1, y-1$ defining the two points $(0,0)$ and $(1,1)$ seperately. Why are the two points together defined by the four polynomials $xy,x(y-1),(x-1)y,(x-1)(y-1)$?

Your problem is just a supped up version of this in terms of ideals instead of polynomials.

Turgul,

In your post you suggest ... "instead of considering $m_1$ and $m_2$, consider just the pairs of polynomials $x,y$ and $x-1, y-1$ defining the two points $(0,0)$ and $(1,1)$ separately. "

So to follow your post ...

Consider the polynomials $$ f_1 = x, f_2 = y $$ ... ...

In D&F's terminology:

$$ \mathcal{Z} (S) = \mathcal{Z} (x,y) $$

$$ \{ (a_1, a_2) \in \mathbb{F}_2 \ | \ f_i (a_1, a_2) = 0 , \ \forall \ f_i \in S \} $$

$$ \{ (a_1, a_2) \in \mathbb{F}_2 \ | \ x= 0, \ y = 0 \} $$

$$ = \{ (0,0) \} $$

since $$ (0,0) $$ is the only point in $$ \mathbb{F}_2 $$ where the polynomials $$ f_1, f_2 $$ are both simultaneously zero.Now, in a similar fashion, consider the polynomials $$ g_1 = x-1, g_2 = y-1 $$ ... ...

In D&F's terminology:

$$ \mathcal{Z} (T) = \mathcal{Z} (x-1, y-1) $$

$$ \{ (a_1, a_2) \in \mathbb{F}_2 \ | \ g_i (a_1, a_2) = 0 , \ \forall \ g_i \in S \} $$

$$ \{ (a_1, a_2) \in \mathbb{F}_2 \ | \ x-1 = 0, \ y-1 = 0 \} $$

$$ = \{ (1,1) \} $$

since $$ (1,1) $$ is the only point in $$ \mathbb{F}_2 $$ where the polynomials $$ f_1, f_2 $$ are both simultaneously zero.Thus the set $$ V = V_1 \cup V_2 = \{ (0,0), (1,1) \} $$

where $$ V_1 = \mathcal{Z} (S) = \mathcal{Z} (x,y) $$

and

$$ V_2 = \mathcal{Z} (T) = \mathcal{Z} (x-1, y-1) $$

Now I need to consider your question:

Why are the two points together defined by the four polynomials $xy,x(y-1),(x-1)y,(x-1)(y-1)$?

I am puzzling over this, particularly as xy = 0 at points such as (1,0) and (0,1) as well as (0,0).

Can you help?

Peter

 

[Math Amateur]

In reading your questions it is clear that a big picture understanding of what is going on has been elusive. I don't have a copy of Dummit & Foote on hand to remember what exactly they say, but let's see if we can't shed some light on the situation.

In algebraic geometry, we are interested in studying "algebraic sets" as objects of geometric interest. What is an algebraic set? It is the zero set of a collection of polynomials. For example all of the points $(x,y)$ satisfying $y=x^2$ form an algebraic set because they are the solutions of $y-x^2=0$. Another example of an algebraic set is $\{(0,0),(1,1)\}$ because it is the common zeroes of the polynomials $y-x=0$ and $y-x^2=0$, that is both of these points are zeroes of both polynomials at the same time, and no other points have this property.

There are a couple of things to note above:

First, in the above examples all points were ordered pairs and all polynomials were in two variables. That tacitly assumes that I'm living in an ambient two dimensional space (the space of all possible ordered pairs $(x,y)$ with no restrictions on $x$ or $y$ at all). Note that in this space, the equation $x=0$ cuts a line of points $(0,y)$ where $y$ can be anything. If we were living in one dimensional space, then $x=0$ would cut out precisely the point $x=0$. This means we should pay at least some attention to where we are looking for zeroes of our polynomials. At some point, we would like to realize that "one-space" lives inside of "two-space" lives in "three-space" and so on, but some care must be taken as to what we mean when we say that, and it's a more technical point to be puzzled later.

Second, polynomials have different solutions over different rings. Consider $x^2+1=0$. Over the real numbers, this has no zeroes at all! Over the complex numbers, we have $x = i$ and $x = -i$, and over the field of two elements, we have the single solution $x=1$. So if we are looking for solutions to polynomials, we have to pick a ring to look in! For practical purposes, we often restrict our attention to fields to make our lives easier (in practice, we actually often further assume that the field is algebraically closed, but there is something to be learned about how that matters, so we'll avoid that assumption for now), which is what D&F certainly do and we will as well, for the time being.
For now, let's restrict our attention to algebraic sets sitting inside of two-dimensional space over a fixed field $k$. Algebraic sets are just special kinds of sets, we can ask if they have nice properties in terms of typical set operations. Consider the polynomials $y-x=0$ and $y-x^2=0$. One cuts out a line, the other a parabola. If I intersect the two shapes, I get the two points from before. That is to say, if I require that points satisfy both at the same time, I get the intersection of the shapes. Since algebraic sets are allowed to be simultaneous zero sets of multiple polynomials (we don't even require finitely many!), this operation is allowed.

On the other hand, what happens if we look at $(y-x)(y-x^2)=0$, the zero set of the product of the two polynomials? Since we live over a field, if the product of two things is zero, either one or the other must be zero, so if $(a,b) \in k^2$ with $(b-a)(b-a^2)=0$, it must be that $b-a = 0$ or $b-a^2=0$. Hence the zero set of $(y-x)(y-x^2)=0$ is contained in the union of the line and the parabola. On the other hand, any point on either the line or the parabola will be a zero of the product. So by looking at the zeroes of the product polynomial, we get the union of the first two!

What's more is that if $A$ is an algebraic set defined by $f_1(x,y)=0,f_2(x,y)=0,\ldots$
and $B$ is defined by $g_1(x,y)=0,g_2(x,y)=0,\ldots$, the the intersection $A \cap B$ is defined by $f_1(x,y)=0,g_1(x,y)=0,f_2(x,y)=0,g_2(x,y)=0, \ldots$ the common zeroes of all the polynomials defining each. On the other hand, the union $A \cup B$ is defined by $f_1(x,y)g_1(x,y)=0,f_1(x,y)g_2(x,y)=0,f_2(x,y)g_1(x,y)=0,f_2(x,y)g_2(x,y)=0, \ldots$ the products of every polynomial defining $A$ by each polynomial defining $B$. You should convince yourself of this.

Now, with many polynomials running around, this gets very complicated quickly, so it is natural to wonder if there is a better way of managing the collections of polynomials. Fortunately, there is. It turns out that the ideal of $k[x,y]$ generated by all of the polynomials defining an algebraic set defines the same algebraic set! At first glance this just seems to have made things worse. After all, if we defined a set using two polynomials, the ideal generated by them has infinitely many polynomials! The great simplification comes when we realize that all that really matters are the generators of the ideal, so we can pick our favorite generators for the ideal and do all the computations with them. Instead of looking at the zero set of $xy,x^2y,xy^2,x^3y^4$, we can just look at the zero set of $xy$ since it generates the same ideal as the other four polynomials!
I suspect this gets overwhelming, so instead of rambling on, I will leave you with the following question:

The first thing to consider before you attack your problem is why is the union of algebraic sets given by the common zeroes of the products of the defining polynomials of each? That is, instead of considering $m_1$ and $m_2$, consider just the pairs of polynomials $x,y$ and $x-1, y-1$ defining the two points $(0,0)$ and $(1,1)$ seperately. Why are the two points together defined by the four polynomials $xy,x(y-1),(x-1)y,(x-1)(y-1)$?

Your problem is just a supped up version of this in terms of ideals instead of polynomials.

Hi again, Turgul,

Here is an attempt to define a set of polynomial equations that define the affine algebraic set $$ V = \mathcal{Z} (S) = \{ (0,0), (1,1) \} $$

Consider the polynomials $$ f_1 = x, \ f_2 = y $$.

So for these polynomials $$ V_1 = \mathcal{Z} (S_1) = \mathcal{Z}(x, y) = \{(0,0) \}$$

Note that for $$ \mathcal{Z} (S_1) $$ that f_1 = 0 , f_2 = 0 have to be both satisfied simultaneously.

Now, similarly, consider the polynomials $$ g_1 = x - 1, \ g_2 = y - 1 $$.

So for these polynomials $$ V_2 = \mathcal{Z} (S_2) = \mathcal{Z}(x - 1, y - 1) = \{(1,1) \}$$

Note that for $$ \mathcal{Z} (S_1) $$ that g_1 = 0 , g_2 = 0 have to be both satisfied simultaneously.Now consider $$ V = V_1 \cup V_2 = \{ (0,0), (1,1) \} $$

where $$ V = \{ f_ig_j \} $$

Thus we must now consider the products $$ f_1g_1, f_1g_2, f_2g_1, f_2,g_2 $$ where we require all these products to be zero simultaneously.

so we have:

x(x - 1) = 0 at (0, 0), (0,1), (1,1), (1,0)

x(y - 1) = 0 at (0, 0), (0,1), (1,1)

y(x - 1) = 0 at (0, 0), (1,0), (1,1)

y(y -1) = 0 at (0, 0), (0,1), (1,1), (1,0)By inspection, all the above products are zero simultaneously at the points of $$ V = V_1 \cup V_2 = \{ (0,0), (1,1) \} $$

Can you confirm that the above analysis is correct?

If it is, I now have to connect the analysis with the required ideals ...

Note that I cannot see how you generated the set of polynomials in your question as follows:

"Why are the two points together defined by the four polynomials [FONT=MathJax_Math]x[FONT=MathJax_Math]y[FONT=MathJax_Main],[FONT=MathJax_Math]x[FONT=MathJax_Main]([FONT=MathJax_Math]y[FONT=MathJax_Main]−[FONT=MathJax_Main]1[FONT=MathJax_Main])[FONT=MathJax_Main],[FONT=MathJax_Main]([FONT=MathJax_Math]x[FONT=MathJax_Main]−[FONT=MathJax_Main]1[FONT=MathJax_Main])[FONT=MathJax_Math]y[FONT=MathJax_Main],[FONT=MathJax_Main]([FONT=MathJax_Math]x[FONT=MathJax_Main]−[FONT=MathJax_Main]1[FONT=MathJax_Main])[FONT=MathJax_Main]([FONT=MathJax_Math]y[FONT=MathJax_Main]−[FONT=MathJax_Main]1[FONT=MathJax_Main])?

Why [FONT=MathJax_Math]x[FONT=MathJax_Math]y[FONT=MathJax_Main],[FONT=MathJax_Math]x[FONT=MathJax_Main]([FONT=MathJax_Math]y[FONT=MathJax_Main]−[FONT=MathJax_Main]1[FONT=MathJax_Main])[FONT=MathJax_Main],[FONT=MathJax_Main]([FONT=MathJax_Math]x[FONT=MathJax_Main]−[FONT=MathJax_Main]1[FONT=MathJax_Main])[FONT=MathJax_Math]y[FONT=MathJax_Main],[FONT=MathJax_Main]([FONT=MathJax_Math]x[FONT=MathJax_Main]−[FONT=MathJax_Main]1[FONT=MathJax_Main])[FONT=MathJax_Main]([FONT=MathJax_Math]y[FONT=MathJax_Main]−[FONT=MathJax_Main]1[FONT=MathJax_Main])? How did you arrive at these polynomials

Can you help?

Peter

 

[Turgul]

My apologies, the polynomials I left at the end are incorrect, which is why you've been having difficulty. Your corrections are exactly right. The points $(0,0)$ and $(1,1)$ are cut out by $x,y$ and $x-1,y-1$ respectfully, so we should be looking at the common zeroes of the products $x(x-1),x(y-1),y(x-1),y(y-1)$.

One thing to note is that the above polynomials work over ANY field, not just $\mathbb{F}_2$. You are absolutely correct that, over $\mathbb{F}_2$, we could just look at the zero sets of $x-y$ or $y-x^2$ or many other polynomials. For reasons that are not at all clear at the moment, this is somewhat "pathological" behavior which is exhibited over finite fields.

Notice that if we were looking over an infinite field, it would be much more difficult to cut out just two points of 2d space with a single polynomial (sometimes still possible, but you've got to be more clever and you can't in general). Generally speaking, if we are living in n-dimensional space, looking at the zero set of a single polynomial should give us some sort of (n-1)-dimensional space (think that the parabola $y-x^2=0$ is a curve in 2-space, hence a 1d subset). It just so happens that talking about dimensions geometrically over finite sets of points ends up being a bit silly. Dummit and Foote use this example because it's computationally very easy to do things over $\mathbb{F}_2$, but they are trying to illustrate how to make things work more generally; that's why they (and I) use the polynomials they do and not others.

The far more interesting observation from this problem is that algebraic sets (there is a perfectly good reason they use the word affine, but until you start looking at things that aren't affine, I'll omit the adjective) behave nicely under intersections and unions. More precisely, we can get the union of algebraic sets by looking at the appropriate products of the polynomials defining each, and we can get intersections just by looking at the common zeroes of all involved polynomials. It is worth noting that this means that algebraic sets are closed under finite union (finite products of polynomials still give polynomials) and arbitrary intersection (looking at common zeroes of any number of polynomials is still looking at the common zeroes of a collection of polynomials). Furthermore these observations hold over any fixed choice of field.

The other important observation is that not only do these operations work for generating sets of polynomials (your $S_1,S_2$), but they also hold for the ideals generated by $S_1$ and $S_2$, and these ideals give you the same algebraic set. That is $Z(<f_1, \ldots, f_s>) = Z(f_1, \ldots, f_s)$.

To answer part of your original question about what $m_1m_2$ looks like, let's recall that $m_1 = <x,y>$ is any sum of polynomials divisible by either $x$ or $y$. So it contains $x,x^2,xy,y^3,x^2+y,3x+7y^{30}$ and so forth. Similarly, $m_2$ is all possible sums of polynomials divisible by $x-1$ or $y-1$.
The product $m_1m_2$ is all possible sums of products $fg$ of polynomials $f \in m_1$ with polynomials of $g \in m_2$. It contains $x(x-1),x^2(y-1)^3,\big( x^2+xy^7 \big) \big(5(x-1)+(y-1)^2 \big)+x(x-1)$ and so on.
In particular, it is generated by the products of the generators: $m_1m_2 = <x(x-1),x(y-1),y(x-1),y(y-1)>$.

Hopefully this clears up some confusion and gives you something to think about. Again, you should try to see why solving my *fixed* question and the D&F problem are essentially the same thing.

 

[Math Amateur]

My apologies, the polynomials I left at the end are incorrect, which is why you've been having difficulty. Your corrections are exactly right. The points $(0,0)$ and $(1,1)$ are cut out by $x,y$ and $x-1,y-1$ respectfully, so we should be looking at the common zeroes of the products $x(x-1),x(y-1),y(x-1),y(y-1)$.

One thing to note is that the above polynomials work over ANY field, not just $\mathbb{F}_2$. You are absolutely correct that, over $\mathbb{F}_2$, we could just look at the zero sets of $x-y$ or $y-x^2$ or many other polynomials. For reasons that are not at all clear at the moment, this is somewhat "pathological" behavior which is exhibited over finite fields.

Notice that if we were looking over an infinite field, it would be much more difficult to cut out just two points of 2d space with a single polynomial (sometimes still possible, but you've got to be more clever and you can't in general). Generally speaking, if we are living in n-dimensional space, looking at the zero set of a single polynomial should give us some sort of (n-1)-dimensional space (think that the parabola $y-x^2=0$ is a curve in 2-space, hence a 1d subset). It just so happens that talking about dimensions geometrically over finite sets of points ends up being a bit silly. Dummit and Foote use this example because it's computationally very easy to do things over $\mathbb{F}_2$, but they are trying to illustrate how to make things work more generally; that's why they (and I) use the polynomials they do and not others.

The far more interesting observation from this problem is that algebraic sets (there is a perfectly good reason they use the word affine, but until you start looking at things that aren't affine, I'll omit the adjective) behave nicely under intersections and unions. More precisely, we can get the union of algebraic sets by looking at the appropriate products of the polynomials defining each, and we can get intersections just by looking at the common zeroes of all involved polynomials. It is worth noting that this means that algebraic sets are closed under finite union (finite products of polynomials still give polynomials) and arbitrary intersection (looking at common zeroes of any number of polynomials is still looking at the common zeroes of a collection of polynomials). Furthermore these observations hold over any fixed choice of field.

The other important observation is that not only do these operations work for generating sets of polynomials (your $S_1,S_2$), but they also hold for the ideals generated by $S_1$ and $S_2$, and these ideals give you the same algebraic set. That is $Z(<f_1, \ldots, f_s>) = Z(f_1, \ldots, f_s)$.

To answer part of your original question about what $m_1m_2$ looks like, let's recall that $m_1 = <x,y>$ is any sum of polynomials divisible by either $x$ or $y$. So it contains $x,x^2,xy,y^3,x^2+y,3x+7y^{30}$ and so forth. Similarly, $m_2$ is all possible sums of polynomials divisible by $x-1$ or $y-1$.
The product $m_1m_2$ is all possible sums of products $fg$ of polynomials $f \in m_1$ with polynomials of $g \in m_2$. It contains $x(x-1),x^2(y-1)^3,\big( x^2+xy^7 \big) \big(5(x-1)+(y-1)^2 \big)+x(x-1)$ and so on.
In particular, it is generated by the products of the generators: $m_1m_2 = <x(x-1),x(y-1),y(x-1),y(y-1)>$.

Hopefully this clears up some confusion and gives you something to think about. Again, you should try to see why solving my *fixed* question and the D&F problem are essentially the same thing.

Thanks so much, Turgil, your posts have really helped me begin on the road to algebraic geometry ... Just a very humble beginning, but anyway ... THANKS!

What books/online notes or lectures. Would you suggest as giving an introduction (elementary preferably) to algebraic geometry.

Peter

 

[Turgul]

Unfortunately, while there is a vast literature on algebraic geometry, almost all of it is completely incomprehensible to the beginner. From what I can tell of your general knowledge, your current effort to work through D&F is one of the better things you could be doing. As a supplement, you might consider Fulton's book on curves:

http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf

which is roughly at a similar level to D&F. Other "beginner" books tend to appeal more to a geometric intuition. Perhaps the most popular is Reid's Undergraduate Algebraic Geometry, which is probably worth getting a hold of as well, though he jumps straight into projective geometry before looking at the affine picture. Electronic copies don't seem hard to get, but since I don't know their legality, I won't post a link here.

Given my impression of what you know, I would probably start with D&F, Fulton, and Reid.

If you are comfortable with complex analysis and/or some basic differential geometry, you might consider Miranda's Algebraic Curves and Riemann Surfaces, but this is a very different perspective and won't provide much insight into what you are currently looking at until you are over 150 pages in.

And finally, it is worth noting that for the last four centuries, the major users of algebraic geometry have been number theorists, so some of the most readable commentary about algebraic geometry for the non-geometer appear interspersed throughout books on number theory. If you are not interested in number theory, then you might skip this, but if you are you should consider picking up a copy of Ireland & Rosen's A Classical Introduction to Modern Number Theory. Overall it is one of my favorite math books and it does happen to house a nice little chapter on algebraic geometry over finite fields.

Ultimately, one of the best resources you can have is other people. Absent that, simply organizing your thoughts enough to write down what exactly you are confused about can be surprisingly helpful. So you should continue to formulate questions and ask them to places like this. Hopefully you'll get helpful feedback, but simply asking the question can help direct your own efforts too.