Math Help - Polynomial Rigs _ Dummit and Foote Chapter 9

1. Polynomial Rigs _ Dummit and Foote Chapter 9

I am reading Dummit and Foote Chapter 9 on Polynomial Rings and am trying to get a good understanding of Propostion 2 (see Attachment - page 296) which states:

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Let I be an ideal of the ring R and let (I) = I[x] denote the ideal of R[x] generated by I (set of polynomials with co-efficients in I). Then

R[x]/I[x] $\cong$ (R/I) [x]

In particular, if I is a prime ideal of R then (I) is a prime ideal of R[x]

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I decided to generate a simple example using R= $\mathbb{Z}$ and I = 2 $\mathbb{Z}$

Then $\mathbb{Z}$ = { ..., -2, -1, 0, 1, 2, 3, ... } and 2 $\mathbb{Z}$ = { ..., -4, -2, 0, 2, 4, 6 .... }

also $\mathbb{Z}$/2 $\mathbb{Z}$ = { $\overline{0}, \overline{1}$ }

Then it appears to me that R[x] = $\mathbb{Z}$ [x] is the set of all polynomials with integer coefficients and I[x] is the set of polynomials with even integers as coefficients

Now how do I formally express R[x], I[x] and $\mathbb{Z}$/2 $\mathbb{Z}$ formally and algebraicly?? Is my text above OK?

It seem that $\mathbb{Z}$ [x] /2 $\mathbb{Z}$ [x] would have two elements - one which was all the polynomials with even co-efficients and one which contains all the polynomials with odd integer co-efficients - but again - how do I express this in formal algebraic symbolism

Further, returning to Propostion 2 above

R/I = $\mathbb{Z}$/2 $\mathbb{Z}$ = { $\overline{0}, \overline{1}$ } and so (R/I)[x] appears to have two elements - one the set of polynomials with coefficients in $\overline{0}$ and one with coefficients in [TEX]\overline{1}TEX] which seems to be correct.

Is my example and reasoning correct? Would appreciate an assurance from someone that all is OK.

How would my exampole ber expressed in more formal algebraic symbolism?

Note: I was also somewhat thrown by D&F's use of the symbolism (I) for I[x]. Previously (see attachment on Properties of Ideals, D&F ch 7 page 251) the symbol (I) was used to denote the smallest idea of R containing I which D&F point out is the set of all finite sums of elements of the form ra with r $\in$ R and a $\in$ I ie sums such as $r_1 a_1 + r_2 a_2 + ... + r_n a_n$. Isn't the use of (I) for I[x] somewhat inconsistent with the use of the symbolism just described.

Peter

2. Re: Polynomial Rigs _ Dummit and Foote Chapter 9

First of all, yes your examples are correct.

$R[x] = \{a_nx^n+a_{n-1}x^{n-1}+...+a_0 | a_i \in R, n \in \mathbb{N} \}$
Same for I[x]

As for $\frac{\mathbb{Z}}{2\mathbb{Z}} = \{0 + 2\mathbb{Z}, 1 + 2\mathbb{Z} \}$

one which was all the polynomials with even co-efficients and one which contains all the polynomials with odd integer co-efficients - but again - how do I express this in formal algebraic symbolism
Since most of this is merely representation, i think if you are clear, you can write it anyway.

Like,
$(\frac{\mathbb{Z}}{2\mathbb{Z}})[X] = \{ \{a_0+a_1x+...+a_nx^n | a_i \in 0 + 2\mathbb{Z} \}, \{a_0+a_1x+...+a_nx^n | a_i \in 1 + 2\mathbb{Z} \} \}$

OR

$(\frac{\mathbb{Z}}{2\mathbb{Z}})[X] = \{ \{a_0+a_1x+...+a_nx^n | a_i = 2k \text{ for some } k \in \mathbb{N}\}, \{a_0+a_1x+...+a_nx^n | a_i = 2m + 1 \text{ for some } m \in \mathbb{N} \} \}$

3. Re: Polynomial Rigs _ Dummit and Foote Chapter 9

let's look at Z[x]/(2Z)[x] and (Z/2Z)[x] separately.

the first consists of cosets of the form:

p(x) + (2Z)[x]

for example, we have the cosets 4x3+1 + (2Z)[x] and 2x2+4x+3 + (2Z)[x]. note that these are actually the SAME coset, since:

4x3+1 - (2x2+4x+3) = 4x3-2x2-4x-2, which is in (2Z)[x].

the second consists of polynomials in (Z/2Z)[x], which have coefficients in the ring (Z/2Z) = {[0],[1]} (and a common abuse of notation is just to call these 0 and 1). for example, x+1, x2+1, and x3+x+1 are all elements of

if we want to show that these two rings are isomorphic, we can (and probably SHOULD) display an isomorphism:

let's try this one:

we know we have a (ring) homomorphism k-->k (mod 2) of Z onto Z/(2Z). let's call this φ.

so now, let's define a map ψ:Z[x]-->(Z/2Z)[x] by:

ψ(a0+a1x+...+anxn) = φ(a0) + φ(a1)x +....φ(an)xn.

i leave it to you to verify ψ is actually a homomorphism (it's easy). essentially, we're just replacing aj with aj (mod 2), so for example, in our polynomials above:

ψ(4x3+1) = φ(4)x3 + φ(1) = 0x3 + 1 = 1,

ψ(2x2+4x+3) = φ(2)x2 + φ(4)x + φ(3) = 0x2 + 0x + 1 = 1.

if we can show ker(ψ) = (2Z)[x], we're done (by the first isomorphism theorem for rings)!

but if ψ(p(x)) = 0, we must have φ(aj) = 0, for every coefficient aj of p, thus p is in (2Z)[x], so ker(ψ) is in (2Z)[x].

on the other hand if p(x) is in (2Z)[x] = (ker(φ))[x], then ψ(p(x)) = 0 + 0x +....+0xn = 0, so (2Z)[x] is contained in ker(ψ).

(the same proof works, by the way, for R and the ideal I).

thus Z[x]/(2Z)[x] is isomorphic to (Z/2Z)[x].

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(Z/2Z)[x] DOES NOT HAVE TWO ELEMENTS!!!!! it has infinitely many. all of them have "1's" as coefficients, but x and x2 are NOT the same polynomial in (Z/2z)[x]. x+1 and x are also distinct polynomials. in fact, for any degree n, we have 2n distinct polynomials of degree n. for example, we have, of degree 2:

x2
x2+1
x2+x
x2+x+1

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a word about the notation (I). note that this is the smallest ideal of R[x] containing I. since (I) is an ideal of R[x], it must contain all monomials of the form axj, with a in I. thus it must contain all sums of these (since ideals are closed under addition), and thus I[x]. hence I[x] is contained in (I). on the other hand, since I[x] is an ideal of R[x] containing I (as the polynomials a*1, for a in I), we must have (I) in I[x]. hence (I) = I[x].

4. Re: Polynomial Rigs _ Dummit and Foote Chapter 9

Originally Posted by Deveno
let's look at Z[x]/(2Z)[x] and (Z/2Z)[x] separately.

the first consists of cosets of the form:

p(x) + (2Z)[x]

for example, we have the cosets 4x3+1 + (2Z)[x] and 2x2+4x+3 + (2Z)[x]. note that these are actually the SAME coset, since:

4x3+1 - (2x2+4x+3) = 4x3-2x2-4x-2, which is in (2Z)[x].

the second consists of polynomials in (Z/2Z)[x], which have coefficients in the ring (Z/2Z) = {[0],[1]} (and a common abuse of notation is just to call these 0 and 1). for example, x+1, x2+1, and x3+x+1 are all elements of

if we want to show that these two rings are isomorphic, we can (and probably SHOULD) display an isomorphism:

let's try this one:

we know we have a (ring) homomorphism k-->k (mod 2) of Z onto Z/(2Z). let's call this φ.

so now, let's define a map ψ:Z[x]-->(Z/2Z)[x] by:

ψ(a0+a1x+...+anxn) = φ(a0) + φ(a1)x +....φ(an)xn.

i leave it to you to verify ψ is actually a homomorphism (it's easy). essentially, we're just replacing aj with aj (mod 2), so for example, in our polynomials above:

ψ(4x3+1) = φ(4)x3 + φ(1) = 0x3 + 1 = 1,

ψ(2x2+4x+3) = φ(2)x2 + φ(4)x + φ(3) = 0x2 + 0x + 1 = 1.

if we can show ker(ψ) = (2Z)[x], we're done (by the first isomorphism theorem for rings)!

but if ψ(p(x)) = 0, we must have φ(aj) = 0, for every coefficient aj of p, thus p is in (2Z)[x], so ker(ψ) is in (2Z)[x].

on the other hand if p(x) is in (2Z)[x] = (ker(φ))[x], then ψ(p(x)) = 0 + 0x +....+0xn = 0, so (2Z)[x] is contained in ker(ψ).

(the same proof works, by the way, for R and the ideal I).

thus Z[x]/(2Z)[x] is isomorphic to (Z/2Z)[x].

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(Z/2Z)[x] DOES NOT HAVE TWO ELEMENTS!!!!! it has infinitely many. all of them have "1's" as coefficients, but x and x2 are NOT the same polynomial in (Z/2z)[x]. x+1 and x are also distinct polynomials. in fact, for any degree n, we have 2n distinct polynomials of degree n. for example, we have, of degree 2:

x2
x2+1
x2+x
x2+x+1

**************************

a word about the notation (I). note that this is the smallest ideal of R[x] containing I. since (I) is an ideal of R[x], it must contain all monomials of the form axj, with a in I. thus it must contain all sums of these (since ideals are closed under addition), and thus I[x]. hence I[x] is contained in (I). on the other hand, since I[x] is an ideal of R[x] containing I (as the polynomials a*1, for a in I), we must have (I) in I[x]. hence (I) = I[x].
Yes sir, probably should have said 2 cosets.

5. Re: Polynomial Rigs _ Dummit and Foote Chapter 9

Originally Posted by jakncoke
Yes sir, probably should have said 2 cosets.
no, not even two cosets. infinitely many.

6. Re: Polynomial Rigs _ Dummit and Foote Chapter 9

Originally Posted by Deveno
no, not even two cosets. infinitely many.
Yes sir, right again, i'm gonna go to bed.

7. Re: Polynomial Rigs _ Dummit and Foote Chapter 9

A most helpful post - still working through the ideas in this post!

Thank you!!!

Peter

8. Re: Polynomial Rings _ Dummit and Foote Chapter 9

In the first part of your post above, you invite us to consider Z[x]/(2Z)[x].

This quotient ring of polynomials, you point out, consists of cosets of the form

p(x) + (2Z)[x]

and you give as an example the cosets of $4x^3 + 1 + (2Z)[x]$ and $2x^3 + 4x + 3 + (2Z)[x]$

and you point out that these are actually the same coset since $(4x^3 + 1) - (2x^3 + 4x + 3) \in (2Z)[x]$

Thus you seem to be using the following rule:

Two polynomials r(x) and s(x) belong to the same coset if $r(x) - S(x) \in (2Z)[x]$

Is this correct? Presumably you can generalise this from (2Z)[x] to the general ring R[x]

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To try to get a better understanding of your post and the above in general I turned to Hungerford's Abstract Algebra Ch 5 Congruence in F[x] and Congruence-Class Arithmetic - see attachment for pages 119-121.

I tried to fit or reconcile your definitions and process above with Hungerford's definitions at the bottom on p119 and page 120 plus his two illustrative examples of congrience modulo $(x^2 + 1)$ in R[x] and congruence modulo $(x^2 = x + 1)$ in $Z_2[x]$, but was unable to formally do so.

Can you please show how your analysis fits with the approach of Hungerford. I know they deal with the same structures so they should be in harmony but I cannot formally reconcile the two approaches. Can you show how this works?

By the Hungerford deals with F[x] where F is a field. DOes it work if F is simply a ring?

Peter

9. Re: Polynomial Rigs _ Dummit and Foote Chapter 9

in ANY ring (polynomial or not) R, if I is an ideal of R, we can form the factor ring R/I.

if R is a commutative ring, then we have to ways to "factor by I":

R[x]/I[x] (form the polynomial rings _[x] first, then factor the "big" polynomial ring by the "smaller" one).

(R/I)[x] (factor R by I first, THEN form a polynomial ring).

it turns out that this is "two paths to the same destination".

in general, for a factor ring R/I, we have a+I = b+I iff a-b in I (this is the abelian group version of aH = bH iff ab-1 in H). it turns out that for a ring, the additive structure "controls" how we form the cosets (ideals/kernels are the pre-images of the additive identity of the image).

the same rule holds for the factor ring F[x]/I, where I is an ideal of F[x], for a field F: f(x) + I = g(x) + I iff f(x) - g(x) is in I. however, F[x] where F is a field, has a property general polynomial rings R[x] do not: every ideal is principal (this is the corresponding idea to "cyclic group" for ideals: a principal ideal is generated by a single element, so every element in it is some multiple of the generator.

this is usually indicated in F[x] by <(p(x)> or (p(x)). so if f(x) - g(x) is in (p(x)), this means that f(x) - g(x) = k(x)p(x), for some polynomial k(x), or more pithily: p(x) DIVIDES f(x) - g(x).

to illustrate what goes "wrong" in a ring R[x], consider the ideal <2,x> in Z[x] (the ring of polynomials with integer coefficients). what do the elements of <2,x> look like?

they are the sum of a polynomial with all even coefficients and a polynomial with no constant term. a shorter description is: all polynomials with even constant term. so

2+x
3x + x2 <--0 constant terms (and thus a factor of x).
4 + 2x + x3 are all examples

now there is NO polynomial p(x) in Z[x] such that <2,x> = <p(x)>. to see this, note that such a polynomial would have to divide 2, and so must be 1,-1,2, or -2 (the only integer divisors of 2).

but 2 does NOT divide x (there is no polynomial k(x) in Z[x] with 2k(x) = x), and neither does -2. but <-1> = <1> = Z[x], and <2,x> is NOT equal to the entire ring (for example 1+x is NOT in <2,x> since it does not have even constant term).

a "plain old commutative ring" doesn't, in general, "have enough divisors" for the gcd of two polynomials to be a "maximal" generator. so the gcd "generates too much". when the ring is a field, this problem goes away, the "extra structure" we get with a field (namely, "more units") gives "more structure" to its corresponding polynomial field.

the example hungerford gives of R[x]/<x2+1> is an extremely important one.

here, our ideal is <x2+1>, consisting of all real polynomials that contain x2+1 as a factor. the elements of R[x]/<x2+1> are cosets of <x2+1>. i will write this ideal as I, to underscore the similarity with R/I for a general ring R (R[x] is a ring, after all).

as with ANY factor ring, we declare f(x) CONGRUENT to g(x) if they lie in the same coset of I = <x2+1> (this is by direct analogy with the factor ring Z/nZ = Z/(n) where we declare a = b (mod n) if a - b is in (n) = nZ, that is, if a and b differ by a multiple of n. the congruence class [a] = a (mod n) is actually the coset a + nZ = a + (n)).

as is the case with groups, a ring congruence creates a partition (or equivalence) on the ring. in this case, we seek to find "simple representatives" of the cosets (equivalence classes) f(x) + I.

it turns out that we can represent ANY coset as the coset of a LINEAR polynomial ax +b + I. rather than go about PROVING this, i will just show how it's done for a polynomial of degree 5.

say we have x5 + 3x + 1 in R[x]. then x5 + 3x + 1 = x3(x2 + 1) - x3 + 3x + 1 (take a minute, and work through the algebra on this).

thus x5 + 3x + 1 - (x2 + 3x + 1) = x3(x2 + 1), and the RHS is in I = <x2 + 1>, so

x5 + 3x + 1 = -x3 + 3x + 1 (mod x2+1), that is: x5 + 3x + 1 + I = -x3 + 3x + 1 + I.

so now we've reduced our "representative" down to a polynomial of degree 3. we can keep going:

-x3 + 3x + 1 = -x(x2 + 1) + x + 3x + 1 = -x(x2 + 1) + 4x + 1 so:

-x3 + 3x + 1 - (4x + 1) is in I, so -x3 + 3x + 1 = 4x + 1 (mod I).

if you are paying attention, you'll see i am just doing "polynomial long division" one step at a time. basically i am "dividing" x5 + 3x + 1 by x2 + 1 until i get a remainder of degree less than deg(x2 + 1) = 2.

so x5+ 3x+ 1 + I = 4x + 1 + I, 4x + 1 is the linear polynomial i promised i would produce.

now here is where it gets fun:

what do we get if we multiply:

(a+bx + I)(c+dx + I), after we "reduce mod x2 + 1"? (you 'll see later why i wrote the constant term first) let's do it:

(a+bx + I)(c+dx + I) = ac + (ad+bc)x + bdx2 + I. this is a quadratic, so we can reduce by one more degree. what we're going to do is add and subtract bd, so that we have a factor bd(x2 + 1).

ac + (ad+bc)x + bdx2 + I = ad - bc + (ad+bc)x + bd(x2 + 1) + I. but since bd(x2 + 1) is IN I, bd(x2 + 1) + I = I, so:

compare this with the product of complex numbers:

(a+bi)(c+di) = ac-bd + (ad+bc)i...see any similarities?

in fact, and this is MOST important: it turns out that the coset x+I is a ROOT of the polynomial x2+1 in R[x]/I:

(x+I)2 + (1+I) = (x2+I ) + (1+I) = x2+1 + I = I = 0+I (recall that "I" is the 0-element (additive identity) of the ring R[x]/I).

in other words, a+bx + I = a(1 + I) + b(x + I) acts "just like" the complex number a+bi with "1+I" playing the role of the real number 1, and "x+I" playing the role of the imaginary number i (which is a root of x2+1).

so one might suspect that R[x]/I is, in fact, a field isomorphic to the complex numbers, and it turns out this is true.

10. Re: Polynomial Rigs _ Dummit and Foote Chapter 9

Thank you for that extensive post - I will now happily work through it!

Peter

11. Re: Polynomial Rigs _ Dummit and Foote Chapter 9

You write:

"now there is NO polynomial p(x) in Z[x] such that <2,x> = <p(x)>. to see this, note that such a polynomial would have to divide 2, and so must be 1,-1,2, or -2 (the only integer divisors of 2)."

I am struggling to see why in the above <2,x> = <p(x)> implies that such a polynomial p(x) in Z[x] would have to divide 2

Can you formally show why this is the case

Peter