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Maximal ideal (x,y) - and then primary ideal (x,y)^n

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Example (2) on page 682 of Dummit and Foote reads as follows: (see attached)

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

(2) For any field k, the ideal (x) in k[x,y] is primary since it is a prime ideal.

For any [TEX] n \ge 1 [/TEX], the ideal [TEX] (x,y)^n [/TEX] is primary

since it is a power of the maximal ideal (x,y)

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

My first problem with this example is as follows:

How can we demonstrate the the ideal (x,y) in k[x,y] is maximal



Then my second problem with the example is as follows:

How do we rigorously demonstrate that the ideal [TEX] (x,y)^n [/TEX] is primary.

D&F say that this is because it is the power of a maximal ideal - but where have they developed that theorem/result?

The closest result they have to that is the following part of Proposition 19 (top of page 682 - see attachment)

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

Proposition 19. Let R be a commutative ring with 1

... ...

(5) Suppose M is a maximal ideal and Q is an ideal with [TEX] M^n \subseteq Q \subseteq M [/TEX]

for some [TEX] n \ge 1[/TEX].

Then Q is a primary ideal with rad Q = M

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

Now if my suspicions are correct and Proposition 19 is being used, then can someone explain (preferably demonstrate formally and rigorously)

how part (5) of 19 demonstrates that the ideal [TEX] (x,y)^n [/TEX] is primary on the basis of being a power of a maximal ideal.

Would appreciate some help

Peter
 
Last edited:

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Example (2) on page 682 of Dummit and Foote reads as follows: (see attached)

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

(2) For any field k, the ideal (x) in k[x,y] is primary since it is a prime ideal.

For any [TEX] n \ge 1 [/TEX], the ideal [TEX] (x,y)^n [/TEX] is primary

since it is a power of the maximal ideal (x,y)

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

My first problem with this example is as follows:

How can we demonstrate the the ideal (x,y) in k[x,y] is maximal



Then my second problem with the example is as follows:

How do we rigorously demonstrate that the ideal [TEX] (x,y)^n [/TEX] is primary.

D&F say that this is because it is the power of a maximal ideal - but where have they developed that theorem/result?

The closest result they have to that is the following part of Proposition 19 (top of page 682 - see attachment)

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

Proposition 19. Let R be a commutative ring with 1

... ...

(5) Suppose M is a maximal ideal and Q is an ideal with [TEX] M^n \subseteq Q \subseteq M [/TEX]

for some [TEX] n \ge 1[/TEX].

Then Q is a primary ideal with rad Q = M

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

Now if my suspicions are correct and Proposition 19 is being used, then can someone explain (preferably demonstrate formally and rigorously)

how part (5) of 19 demonstrates that the ideal [TEX] (x,y)^n [/TEX] is primary on the basis of being a power of a maximal ideal.

Would appreciate some help

Peter
I have been doing some reflecting and reading around the two issues/problems mentioned in my post above.

First problem/issue was as follows:

"My first problem with this example is as follows:

How can we demonstrate the the ideal (x,y) in k[x,y] is maximal"

In the excellent book "Ideals, Varieties and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative Algebra" by David Cox, John Little and Donal O'Shea we find the following theorem (and its proof) on pages 201-202.

Proposition 9. If k is any field, an ideal [TEX] I \subseteq k[x_1, x_2, ... ... , x_n] [/TEX] of the form

[TEX] I = (x_1 - a_1, x_2 - a_2, ... ... x_n - a_n) [/TEX] where [TEX] a_1, a_2, ... ... , a_n \in k [/TEX]

is maximal.


Now (x, y) is of the form mentioned in Cox et al Proposition 9 since [TEX] (x,y) = (x-0, y-0) [/TEX] and so by Cox et al Proposition 9, (x,y) is maximal

Can someone confirm that this is correct.

Now reflecting on my second problem/issue.

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
With regards to your first dilemma, showing that $(x,y)$ is maximal....try to think about which polynomials are NOT in this ideal. It should hopefully be obvious that any element of $(x,y)$ is of the form:

$f(x,y) = xp(x,y) + yq(x,y)$ for certain elements $p,q \in k[x,y]$.

What does this tell you about the "constant term"? Equivalently, what does:

$k[x,y]/(x,y)$ look like?

Now, what happens if you add any polynomial containing a non-zero constant term to $(x,y)$ and create an ideal containing such a polynomial AND all of $(x,y)$? What does this tell you about the maximality of $(x,y)$?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
With regards to your first dilemma, showing that $(x,y)$ is maximal....try to think about which polynomials are NOT in this ideal. It should hopefully be obvious that any element of $(x,y)$ is of the form:

$f(x,y) = xp(x,y) + yq(x,y)$ for certain elements $p,q \in k[x,y]$.

What does this tell you about the "constant term"? Equivalently, what does:

$k[x,y]/(x,y)$ look like?

Now, what happens if you add any polynomial containing a non-zero constant term to $(x,y)$ and create an ideal containing such a polynomial AND all of $(x,y)$? What does this tell you about the maximality of $(x,y)$?


Hi Deveno,

First, to respond to your comment:

" It should hopefully be obvious that any element of $(x,y)$ is of the form:

$f(x,y) = xp(x,y) + yq(x,y)$ for certain elements $p,q \in k[x,y]$."


Well, the ideal generated by \(\displaystyle (x,y) = (f_1, f_2) \)

where \(\displaystyle f_1(x,y) = x \) and \(\displaystyle f_2(x,y) = y \)

is the set of all finite sums of \(\displaystyle f_1 \) and \(\displaystyle f_2 \).

So ...

\(\displaystyle (x,y) = (f_1, f_2) = \{ f_1p + f_2q \ | \ p,q \in k[x.y] \)

\(\displaystyle = \{ xp(x,y) + yq(x,y) \ | \ p,q \in k[x.y] \)

So, as you say, any element f of (x,y) is of the form:

\(\displaystyle f(x,y) = xp(x,y) + yq(x,y) \) for elements \(\displaystyle p,q \in k[x.y] \)

...

You then ask about the constant term in the elements f - clearly the constant term is 0.

Now thinking about the nature of k[x,y]/(x,y)

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Hi Deveno,

First, to respond to your comment:

" It should hopefully be obvious that any element of $(x,y)$ is of the form:

$f(x,y) = xp(x,y) + yq(x,y)$ for certain elements $p,q \in k[x,y]$."


Well, the ideal generated by \(\displaystyle (x,y) = (f_1, f_2) \)

where \(\displaystyle f_1(x,y) = x \) and \(\displaystyle f_2(x,y) = y \)

is the set of all finite sums of \(\displaystyle f_1 \) and \(\displaystyle f_2 \).

So ...

\(\displaystyle (x,y) = (f_1, f_2) = \{ f_1p + f_2q \ | \ p,q \in k[x.y] \)

\(\displaystyle = \{ xp(x,y) + yq(x,y) \ | \ p,q \in k[x.y] \)

So, as you say, any element f of (x,y) is of the form:

\(\displaystyle f(x,y) = xp(x,y) + yq(x,y) \) for elements \(\displaystyle p,q \in k[x.y] \)

...

You then ask about the constant term in the elements f - clearly the constant term is 0.

Now thinking about the nature of k[x,y]/(x,y)

Peter

Hi Deveno,

I am having considerable problems framing a sensible answer to your question:

"what does [FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main])[/FONT] look like?"

Can you help further in characterizing the elements of [FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main])[/FONT]?

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Well, loosely speaking, "modding out" x and y amounts to setting x = y = 0, so what's left?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Well, loosely speaking, "modding out" x and y amounts to setting x = y = 0, so what's left?
Well, in that case, you get the constant term of the polynomial that you "mod out"

So then, if you add a polynomial with a constant term to (x,y) you would generate polynomials with any given constant term as the polynomial that you added is multiplied by all elements of k[x,y] - so I think you would generate all the elements of k[x,y] - which would mean (x,y) is maximal.

But ...would it actually do this ... if the constant term was 4 ... could you get all possible constant terms by multiplying by other polynomials ,,, well k is a field ... so I think you would.

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Well, yeah, it's a maximal ideal....adding anything to it and then taking all the other stuff you need to force an ideal eventually means you get a constant polynomial in it, and since constant polynomials are units, we must have the entire ring.

For example, suppose we want the smallest ideal containing the set $\{4 + xy,x,y\}$.

Clearly, since $4 + xy$ is in our ideal, and so is $xy$, their difference must be in such an ideal, which means $4$ is in the ideal.

Since (provided $k$ is not of characteristic 2) $4 \neq 0$, it has an inverse, which is again "some element" of $k$, so:

$4^{-1}4 = 1$ is again in our ideal.

But surely, if we have 1 in our ideal, we're kinda screwed, we have to take EVERYTHING.

In other words, $k[x,y]/(x,y)$ looks "just like $k$" (the polynomial we get from multiplying two polynomials and "forgetting the non-constant terms" is the same polynomial we get from "forgetting the non-constant terms FIRST", and then multiplying:

$(4 + xy + y^2)(2 + x + xy) = 8 + 4x + 2y^2 + 6xy + x^2y + xy^2 + x^2y^2 + xy^3 \mapsto 8$

$(4 + xy + y^2)(2 + x + xy) \mapsto (4)(2) = 8$).

If you were to think of these 2-variable polynomials as surfaces in 3-dimensional space, "modding out" x and y just leaves the z-axis, which is isomorphic to the underlying field.

I mention this because in more advanced mathematics (which I never undertook a full study of) these "geometric" interpretations of ideals become very important.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Well, yeah, it's a maximal ideal....adding anything to it and then taking all the other stuff you need to force an ideal eventually means you get a constant polynomial in it, and since constant polynomials are units, we must have the entire ring.

For example, suppose we want the smallest ideal containing the set $\{4 + xy,x,y\}$.

Clearly, since $4 + xy$ is in our ideal, and so is $xy$, their difference must be in such an ideal, which means $4$ is in the ideal.

Since (provided $k$ is not of characteristic 2) $4 \neq 0$, it has an inverse, which is again "some element" of $k$, so:

$4^{-1}4 = 1$ is again in our ideal.

But surely, if we have 1 in our ideal, we're kinda screwed, we have to take EVERYTHING.

In other words, $k[x,y]/(x,y)$ looks "just like $k$" (the polynomial we get from multiplying two polynomials and "forgetting the non-constant terms" is the same polynomial we get from "forgetting the non-constant terms FIRST", and then multiplying:

$(4 + xy + y^2)(2 + x + xy) = 8 + 4x + 2y^2 + 6xy + x^2y + xy^2 + x^2y^2 + xy^3 \mapsto 8$

$(4 + xy + y^2)(2 + x + xy) \mapsto (4)(2) = 8$).

If you were to think of these 2-variable polynomials as surfaces in 3-dimensional space, "modding out" x and y just leaves the z-axis, which is isomorphic to the underlying field.

I mention this because in more advanced mathematics (which I never undertook a full study of) these "geometric" interpretations of ideals become very important.
Thanks Deveno.

I really wish to understand your post fully, but I fear I must be missing something, since after showing that adding the polynomial 4 + xy to (x,y) we, as you say, get everything - so (x,y) is maximal in k[x,y].

But then you say

"In other words, \(\displaystyle k[x,y]/(x,y) \) looks "just like \(\displaystyle k \)" ... "

You seem to be saying this follows from your analysis above .... ??? ... but how does this follow from what you said above it?

Just one further question that your posts have raised is how does your process of "modding out" link conceptually with the idea of equivalence classes ... so, in particular, how do we know in the case of k[x,y]/(x,y) that taking a polynomial and putting x = y = 0 gives us the requisite equivalence class/coset?

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
OK.

Let $I = (x,y)$.

Take any polynomial $f(x,y)$.

Explicitly, let:

$f(x,y) = c_0 + c_1x + c_2y + c_3x^2 + c_4xy + c_5y^2 + \cdots + c_rx^my^n$

Consider the polynomial $g(x,y) = f(x,y) - c_0 = c_1x + c_2y + c_3x^2 + c_4xy + c_5y^2 + \cdots + c_rx^my^n$.

We can re-write this as:

$g(x,y) = x(c_1 + c_3x + c_4y + \cdots c_rx^{m-1}y^n) + y(c_2 + c_5y + \cdots + c_sy^t)$

for some positive integer $t \leq m+n$.

Clearly, $g(x,y) \in I$, so by the very definition of ideal cosets:

$g(x,y) + I = c_0 + I$.

Note that $c_0 = f(0,0)$. This is the one-to-one correspondence between the cosets:

$f(x,y) + I \leftrightarrow f(0,0)$

Two polynomials with the same constant term are going to have a difference of terms involving either x,y or both, and hence are going to belong to the same coset.

So the cosets $c_0 + I$ for $c_0 \in k$ are a complete enumeration of all the cosets. Moreover, if two polynomials have DIFFERENT constant terms, they will lie in DIFFERENT cosets of $I = (x,y)$.

That is, the map:

$\phi_{(0,0)}: k[x,y] \to k$ given by:

$\phi_{(0,0)}(f(x,y)) = f(0,0)$ is a surjective ring homomorphism from $k[x,y]$ to $k$, with kernel $(x,y)$. This establishes that:

$k[x,y]/(x,y) \cong k$

Think back to the first ring congruence you were introduced to: the integers mod $n$. Yes, one CAN think of these formally as the (infinite) sets:

$k + n\Bbb Z = \{m \in \Bbb Z: m = k + nt, t \in \Bbb Z\}$

But it is much SIMPLER to regard these as just "cyclical numbers": we do ORDINARY integer arithmetic, but we stipulate that $n = 0$ (so we only need to keep track of 0 and the first $n-1$ integers after that...the rest are superfluous, we can "reduce them mod $n$" to one of our first $n$ integers).

One of the purposes of ideals is to "get rid of stuff which makes things complicated", revealing a similar but simpler structure often with much more desirable properties. For example, when we set the polynomial $x^2 + 1$ equal to 0 in $\Bbb R[x]$, instead of a vast ring which has many undesirable properties (such as not being algebraically complete, and also the polynomial $x$ has no inverse, making division a dicey affair, indeed) we get a much "nicer" structure, a field, which inherits many of the ring properties we started out with, and follows many of the same rules of algebraic manipulation (in particular, the DISTRIBUTIVE law holds, so we can treat the root of $x^2 + 1$ formally as just a symbol "$i$", and use the rule:

$(a + bi)(c + di) = ac + bci + adi + bdi^2 = (ac - bd) + (ad + bc)i$

since if $i^2 + 1 = 0$, then $i^2 = i^2 + 1 - 1 = 0 - 1 = -1$).

"Modding out" $x^2 + 1$ effectively knocks all our possible "polynomials in $i$" down to degree 1 or less:

$a = a + 0i,\ a \in \Bbb R$ <--polynomials of degree 0
$a + bi,\ a,b \in \Bbb R$ <--polynomials of degree 1

with "coset algebra" we can effectively IGNORE the ideal, because of the rules:

$(x + I) + (y + I) = (x + y) + I$
$(x + I)(y + I) = xy + I$

Often, if one wishes to emphasize this one might write:

$[x]$ instead of $x + I$, so that these rules become:

$[x] + [y] = [x+y]$
$[x]\cdot[y] = [xy]$

but even the brackets can become baggage that gets in the way, and they are often omitted (as long as it is clear what CONTEXT we are operating in).

One rarely sees complex numbers written as:

$a + bx + (x^2 + 1)$ it's just too inefficient.

Think about this, for a second: what are the two important algebraic properties of the number 0?

1) $ 0 + 0 = 0$ (additive identity forces this, right?)
2) $ x \cdot 0 = 0$ (this follows from the ring axioms, try to prove it holds in ANY ring).

Ideals just "generalize" this to a set that has "zero-like properties". In fact, in the quotient ring, the ideal becomes "the new zero".

The Fundamental Isomorphism Theorem (for rings) isn't just there "for show". It tells us that quotient rings preserve sums and products from our original ring (though things that were "different" in the original ring may get "lumped together" in the quotient). We should expect this, because we are replacing EQUALITY with EQUIVALENCE.

This is typical of abstract thought in general: in reality, every bit of everything is unique (the old saying goes: "you can't step in the same river twice"). But a lot of what makes something unique is irrelevant to our quest for knowledge: if I want to predict the motion of a billiard ball, I don't need to know what color it is.

The power of abstraction is that we can deduce what rules apply to all members of a given class, instead of having to separately re-derive algebraic rules for integers, and real numbers, and polynomials and functions defined on an open set of the complex plane, we can investigate rings, and apply principles that are quite general to different instances as they occur. This allows great economy of effort, and often affords great insight into things that are at first mysterious.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
OK.

Let $I = (x,y)$.

Take any polynomial $f(x,y)$.

Explicitly, let:

$f(x,y) = c_0 + c_1x + c_2y + c_3x^2 + c_4xy + c_5y^2 + \cdots + c_rx^my^n$

Consider the polynomial $g(x,y) = f(x,y) - c_0 = c_1x + c_2y + c_3x^2 + c_4xy + c_5y^2 + \cdots + c_rx^my^n$.

We can re-write this as:

$g(x,y) = x(c_1 + c_3x + c_4y + \cdots c_rx^{m-1}y^n) + y(c_2 + c_5y + \cdots + c_sy^t)$

for some positive integer $t \leq m+n$.

Clearly, $g(x,y) \in I$, so by the very definition of ideal cosets:

$g(x,y) + I = c_0 + I$.

Note that $c_0 = f(0,0)$. This is the one-to-one correspondence between the cosets:

$f(x,y) + I \leftrightarrow f(0,0)$

Two polynomials with the same constant term are going to have a difference of terms involving either x,y or both, and hence are going to belong to the same coset.

So the cosets $c_0 + I$ for $c_0 \in k$ are a complete enumeration of all the cosets. Moreover, if two polynomials have DIFFERENT constant terms, they will lie in DIFFERENT cosets of $I = (x,y)$.

That is, the map:

$\phi_{(0,0)}: k[x,y] \to k$ given by:

$\phi_{(0,0)}(f(x,y)) = f(0,0)$ is a surjective ring homomorphism from $k[x,y]$ to $k$, with kernel $(x,y)$. This establishes that:

$k[x,y]/(x,y) \cong k$

Think back to the first ring congruence you were introduced to: the integers mod $n$. Yes, one CAN think of these formally as the (infinite) sets:

$k + n\Bbb Z = \{m \in \Bbb Z: m = k + nt, t \in \Bbb Z\}$

But it is much SIMPLER to regard these as just "cyclical numbers": we do ORDINARY integer arithmetic, but we stipulate that $n = 0$ (so we only need to keep track of 0 and the first $n-1$ integers after that...the rest are superfluous, we can "reduce them mod $n$" to one of our first $n$ integers).

One of the purposes of ideals is to "get rid of stuff which makes things complicated", revealing a similar but simpler structure often with much more desirable properties. For example, when we set the polynomial $x^2 + 1$ equal to 0 in $\Bbb R[x]$, instead of a vast ring which has many undesirable properties (such as not being algebraically complete, and also the polynomial $x$ has no inverse, making division a dicey affair, indeed) we get a much "nicer" structure, a field, which inherits many of the ring properties we started out with, and follows many of the same rules of algebraic manipulation (in particular, the DISTRIBUTIVE law holds, so we can treat the root of $x^2 + 1$ formally as just a symbol "$i$", and use the rule:

$(a + bi)(c + di) = ac + bci + adi + bdi^2 = (ac - bd) + (ad + bc)i$

since if $i^2 + 1 = 0$, then $i^2 = i^2 + 1 - 1 = 0 - 1 = -1$).

"Modding out" $x^2 + 1$ effectively knocks all our possible "polynomials in $i$" down to degree 1 or less:

$a = a + 0i,\ a \in \Bbb R$ <--polynomials of degree 0
$a + bi,\ a,b \in \Bbb R$ <--polynomials of degree 1

with "coset algebra" we can effectively IGNORE the ideal, because of the rules:

$(x + I) + (y + I) = (x + y) + I$
$(x + I)(y + I) = xy + I$

Often, if one wishes to emphasize this one might write:

$[x]$ instead of $x + I$, so that these rules become:

$[x] + [y] = [x+y]$
$[x]\cdot[y] = [xy]$

but even the brackets can become baggage that gets in the way, and they are often omitted (as long as it is clear what CONTEXT we are operating in).

One rarely sees complex numbers written as:

$a + bx + (x^2 + 1)$ it's just too inefficient.

Think about this, for a second: what are the two important algebraic properties of the number 0?

1) $ 0 + 0 = 0$ (additive identity forces this, right?)
2) $ x \cdot 0 = 0$ (this follows from the ring axioms, try to prove it holds in ANY ring).

Ideals just "generalize" this to a set that has "zero-like properties". In fact, in the quotient ring, the ideal becomes "the new zero".

The Fundamental Isomorphism Theorem (for rings) isn't just there "for show". It tells us that quotient rings preserve sums and products from our original ring (though things that were "different" in the original ring may get "lumped together" in the quotient). We should expect this, because we are replacing EQUALITY with EQUIVALENCE.

This is typical of abstract thought in general: in reality, every bit of everything is unique (the old saying goes: "you can't step in the same river twice"). But a lot of what makes something unique is irrelevant to our quest for knowledge: if I want to predict the motion of a billiard ball, I don't need to know what color it is.

The power of abstraction is that we can deduce what rules apply to all members of a given class, instead of having to separately re-derive algebraic rules for integers, and real numbers, and polynomials and functions defined on an open set of the complex plane, we can investigate rings, and apply principles that are quite general to different instances as they occur. This allows great economy of effort, and often affords great insight into things that are at first mysterious.

Deveno,

Thank you for this thoughtful and extremely helpful post.

I am working through it carefully now.

Thanks again.

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
OK.

Let $I = (x,y)$.

Take any polynomial $f(x,y)$.

Explicitly, let:

$f(x,y) = c_0 + c_1x + c_2y + c_3x^2 + c_4xy + c_5y^2 + \cdots + c_rx^my^n$

Consider the polynomial $g(x,y) = f(x,y) - c_0 = c_1x + c_2y + c_3x^2 + c_4xy + c_5y^2 + \cdots + c_rx^my^n$.

We can re-write this as:

$g(x,y) = x(c_1 + c_3x + c_4y + \cdots c_rx^{m-1}y^n) + y(c_2 + c_5y + \cdots + c_sy^t)$

for some positive integer $t \leq m+n$.

Clearly, $g(x,y) \in I$, so by the very definition of ideal cosets:

$g(x,y) + I = c_0 + I$.

Note that $c_0 = f(0,0)$. This is the one-to-one correspondence between the cosets:

$f(x,y) + I \leftrightarrow f(0,0)$

Two polynomials with the same constant term are going to have a difference of terms involving either x,y or both, and hence are going to belong to the same coset.

So the cosets $c_0 + I$ for $c_0 \in k$ are a complete enumeration of all the cosets. Moreover, if two polynomials have DIFFERENT constant terms, they will lie in DIFFERENT cosets of $I = (x,y)$.

That is, the map:

$\phi_{(0,0)}: k[x,y] \to k$ given by:

$\phi_{(0,0)}(f(x,y)) = f(0,0)$ is a surjective ring homomorphism from $k[x,y]$ to $k$, with kernel $(x,y)$. This establishes that:

$k[x,y]/(x,y) \cong k$

Think back to the first ring congruence you were introduced to: the integers mod $n$. Yes, one CAN think of these formally as the (infinite) sets:

$k + n\Bbb Z = \{m \in \Bbb Z: m = k + nt, t \in \Bbb Z\}$

But it is much SIMPLER to regard these as just "cyclical numbers": we do ORDINARY integer arithmetic, but we stipulate that $n = 0$ (so we only need to keep track of 0 and the first $n-1$ integers after that...the rest are superfluous, we can "reduce them mod $n$" to one of our first $n$ integers).

One of the purposes of ideals is to "get rid of stuff which makes things complicated", revealing a similar but simpler structure often with much more desirable properties. For example, when we set the polynomial $x^2 + 1$ equal to 0 in $\Bbb R[x]$, instead of a vast ring which has many undesirable properties (such as not being algebraically complete, and also the polynomial $x$ has no inverse, making division a dicey affair, indeed) we get a much "nicer" structure, a field, which inherits many of the ring properties we started out with, and follows many of the same rules of algebraic manipulation (in particular, the DISTRIBUTIVE law holds, so we can treat the root of $x^2 + 1$ formally as just a symbol "$i$", and use the rule:

$(a + bi)(c + di) = ac + bci + adi + bdi^2 = (ac - bd) + (ad + bc)i$

since if $i^2 + 1 = 0$, then $i^2 = i^2 + 1 - 1 = 0 - 1 = -1$).

"Modding out" $x^2 + 1$ effectively knocks all our possible "polynomials in $i$" down to degree 1 or less:

$a = a + 0i,\ a \in \Bbb R$ <--polynomials of degree 0
$a + bi,\ a,b \in \Bbb R$ <--polynomials of degree 1

with "coset algebra" we can effectively IGNORE the ideal, because of the rules:

$(x + I) + (y + I) = (x + y) + I$
$(x + I)(y + I) = xy + I$

Often, if one wishes to emphasize this one might write:

$[x]$ instead of $x + I$, so that these rules become:

$[x] + [y] = [x+y]$
$[x]\cdot[y] = [xy]$

but even the brackets can become baggage that gets in the way, and they are often omitted (as long as it is clear what CONTEXT we are operating in).

One rarely sees complex numbers written as:

$a + bx + (x^2 + 1)$ it's just too inefficient.

Think about this, for a second: what are the two important algebraic properties of the number 0?

1) $ 0 + 0 = 0$ (additive identity forces this, right?)
2) $ x \cdot 0 = 0$ (this follows from the ring axioms, try to prove it holds in ANY ring).

Ideals just "generalize" this to a set that has "zero-like properties". In fact, in the quotient ring, the ideal becomes "the new zero".

The Fundamental Isomorphism Theorem (for rings) isn't just there "for show". It tells us that quotient rings preserve sums and products from our original ring (though things that were "different" in the original ring may get "lumped together" in the quotient). We should expect this, because we are replacing EQUALITY with EQUIVALENCE.

This is typical of abstract thought in general: in reality, every bit of everything is unique (the old saying goes: "you can't step in the same river twice"). But a lot of what makes something unique is irrelevant to our quest for knowledge: if I want to predict the motion of a billiard ball, I don't need to know what color it is.

The power of abstraction is that we can deduce what rules apply to all members of a given class, instead of having to separately re-derive algebraic rules for integers, and real numbers, and polynomials and functions defined on an open set of the complex plane, we can investigate rings, and apply principles that are quite general to different instances as they occur. This allows great economy of effort, and often affords great insight into things that are at first mysterious.
You write:

"We can re-write this as:

$g(x,y) = x(c_1 + c_3x + c_4y + \cdots c_rx^{m-1}y^n) + y(c_2 + c_5y + \cdots + c_sy^t)$

for some positive integer $t \leq m+n$."

and then you write:

"Clearly, $g(x,y) \in I$, so by the very definition of ideal cosets:"

I am taking it that you mean that \(\displaystyle g(x,y) \) is of the form \(\displaystyle g(x,y) = f_1(x,y)g_1(x,y) + f_2(x,y)g_2(x,y) \) where \(\displaystyle f_1(x,y) = x \) and \(\displaystyle f_2(x,y) = y \)

Then you write:

"$g(x,y) + I = c_0 + I$."

I do not understand how you get this ... can you explain?

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
What is the criterion for equality of two cosets in a quotient ring?

It is NOT:

$a + I = b + I \iff a = b$ (that is: if $a - b = 0$)

but rather:

$a + I = b + I \iff a - b \in I$.

Rings are, first and foremost, abelian groups under addition. This group structure dominates their algebraic character (for example: ideals, that is kernels of ring-homomorphisms are pre-images of the ADDITIVE identity, instead of, say, the unity of a ring).

It looks like I may have made a typo: it should read-

$f(x,y) + I = c_0 + I$ (since $g(x,y) = f(x,y) - c_0 \in I$).

I apologize for that.

But the larger point I am trying to make is this:

When you are reading Dummit and Foote (or any other algebra textbook), you have to assimilate the basic concepts into larger chunks, at some point. To proceed theorem-by-theorem without any real confidence in what a theorem is saying in practical terms is to defeat the purpose of learning: understanding.

By the time you are studying polynomial ideal properties, you should have complete confidence about "how" abelian groups tick.

I see this kind of "conceptual stumbling block" most frequently in linear algebra. Most people have been using properties of abelian groups all their lives, they just don't KNOW it. What mathematicians call a "field" they think of as a "number-system". For most people, addition and multiplication are learned so early, and in the context of such "concrete" things, they feel as though what addition and multiplication ARE is bound up intrinsically with these "number-thingies".

Such people take easy enough to the thought of vectors as $n$-tuples, and then get VERY confused when dealing with polynomial vector spaces. In truth, an abelian group is a simpler structure than a field (only ONE operation to consider).

Scalar multiplication (which is where the trouble usually sets in) is an abelian group endomorphism with certain other properties. Of course, this isn't going to make any sense if one has never heard of an "endomorphism".

Now, one can (because of the well-behaved nature of the direct product in abelian groups) "bypass" this, and pretty much treat all finite-dimensional vector spaces as some $F^n$ (typically with $F = \Bbb C$ or $F = \Bbb R$). But if a student ever studies quotient spaces, or direct sums of vector spaces, or null spaces and column spaces, he/she often gets very confused, because he/she has no experience in the abelian group concepts underpinning these structures.

For example, the rank-nullity theorem, an almost all-powerful tool in linear algebra, is a direct consequence of the Fundamental Isomorphism Theorem for Abelian Groups. Vector Spaces are all abelian groups, just as rings are. This abelian group-ness pervades their character (in rather a nice way).

To recap...it seems to me as if you need to spend some time really thinking about groups, and abelian groups in particular. Because it's hindering how well you can assimilate some of these more advanced topics. Get a bigger picture, mathematics isn't all about the gears of the machinery.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
What is the criterion for equality of two cosets in a quotient ring?

It is NOT:

$a + I = b + I \iff a = b$ (that is: if $a - b = 0$)

but rather:

$a + I = b + I \iff a - b \in I$.

Rings are, first and foremost, abelian groups under addition. This group structure dominates their algebraic character (for example: ideals, that is kernels of ring-homomorphisms are pre-images of the ADDITIVE identity, instead of, say, the unity of a ring).

It looks like I may have made a typo: it should read-

$f(x,y) + I = c_0 + I$ (since $g(x,y) = f(x,y) - c_0 \in I$).

I apologize for that.

But the larger point I am trying to make is this:

When you are reading Dummit and Foote (or any other algebra textbook), you have to assimilate the basic concepts into larger chunks, at some point. To proceed theorem-by-theorem without any real confidence in what a theorem is saying in practical terms is to defeat the purpose of learning: understanding.

By the time you are studying polynomial ideal properties, you should have complete confidence about "how" abelian groups tick.

I see this kind of "conceptual stumbling block" most frequently in linear algebra. Most people have been using properties of abelian groups all their lives, they just don't KNOW it. What mathematicians call a "field" they think of as a "number-system". For most people, addition and multiplication are learned so early, and in the context of such "concrete" things, they feel as though what addition and multiplication ARE is bound up intrinsically with these "number-thingies".

Such people take easy enough to the thought of vectors as $n$-tuples, and then get VERY confused when dealing with polynomial vector spaces. In truth, an abelian group is a simpler structure than a field (only ONE operation to consider).

Scalar multiplication (which is where the trouble usually sets in) is an abelian group endomorphism with certain other properties. Of course, this isn't going to make any sense if one has never heard of an "endomorphism".

Now, one can (because of the well-behaved nature of the direct product in abelian groups) "bypass" this, and pretty much treat all finite-dimensional vector spaces as some $F^n$ (typically with $F = \Bbb C$ or $F = \Bbb R$). But if a student ever studies quotient spaces, or direct sums of vector spaces, or null spaces and column spaces, he/she often gets very confused, because he/she has no experience in the abelian group concepts underpinning these structures.

For example, the rank-nullity theorem, an almost all-powerful tool in linear algebra, is a direct consequence of the Fundamental Isomorphism Theorem for Abelian Groups. Vector Spaces are all abelian groups, just as rings are. This abelian group-ness pervades their character (in rather a nice way).

To recap...it seems to me as if you need to spend some time really thinking about groups, and abelian groups in particular. Because it's hindering how well you can assimilate some of these more advanced topics. Get a bigger picture, mathematics isn't all about the gears of the machinery.
Thanks Deveno, I was revising quotient or factor rings! but will now go back and revise quotient groups? I will also work through the basic theory of abelian groups.

Thank you for your guidance ... It has been enormously helpful and gives me confidence to go on towards my goal of understanding commutative algebra and algebraic geometry.

Time to go back to group theory for some revision ...

Peter