Let R = Z/4Z

(1) Show that 1 + 2x is a unit of R[x], and, further,

(2) show that every unit of R[x] has the form 1 + 2f(x) for some f(x) in R[x]

Would appreciate some help with this problem.

Peter

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- Dec 25th 2012, 02:30 PMBernhardPolynomial Rings -- Z/4Z
Let R = Z/4Z

(1) Show that 1 + 2x is a unit of R[x], and, further,

(2) show that every unit of R[x] has the form 1 + 2f(x) for some f(x) in R[x]

Would appreciate some help with this problem.

Peter - Dec 25th 2012, 07:06 PMDevenoRe: Polynomial Rings -- Z/4Z
for (1):

(1 + 2x)(1 - 2x) = 1 - 4x^{2}= 1 (since 4 = 0 (mod 4)).

for (2):

consider the homomorphism:

φ:(Z/4Z)[x]-->(Z/2Z)[x] given by:

a_{0}+ a_{1}x +...+ a_{n}x^{n}--> [a_{0}]+[a_{1}]x+...+[a_{n}]x^{n}, where [a_{k}] = a_{k}(mod 2).

if g(x) is a unit in (Z/4Z)[x], then g(x)k(x) = 1, whence φ(g(x))φ(k(x)) = φ(1) = 1, hence φ(g(x)) = φ(k(x)) = 1 (since the only units of (Z/2Z)[x] are the units of Z/2Z, that is: 1).

this means that φ(g(x) - 1) = φ(g(x)) - φ(1) = 1 - 1 = 0, that is g(x) - 1 is in ker(φ).

since the elements of ker(φ) are precisely the polynomials in (Z/4Z)[x] with coefficients of 0 or 2, g(x) - 1 = 2(f(x)), for some polynomial f(x).

hence g(x) = 1 + 2(f(x)). - Dec 25th 2012, 08:04 PMBernhardRe: Polynomial Rings -- Z/4Z
In your post you write:

"since the only units of (Z/2Z)[x] are the units of Z/2Z, that is: 1"

Is this relationship general for Z/nZ?

What is the general situation regarding units in a ring (or field) R and the units in R[x]

Can you clarify the situation?

Peter - Dec 25th 2012, 08:30 PMBernhardRe: Polynomial Rings -- Z/4Z
I would be most interested in understanding your problem solving strategy in this case

What caused you to use/invoke the homomorhism that you chose ... Which then turned out to prove the required result!!

Peter - Dec 25th 2012, 09:22 PMDevenoRe: Polynomial Rings -- Z/4Z
no this result does not, in general, hold for Z/nZ. it does however, hold for Z/pZ, where p is a prime, since Z/pZ is a field.

in general, (Z/nZ)[x] will contain 0 divisors (in addition to the ones that Z/nZ already has) for composite n.

however, every composite n has a prime factor, eh? - Dec 25th 2012, 09:27 PMBernhardRe: Polynomial Rings -- Z/4Z
So you are saying that if F is a field then the only units of F[x] are the units of F - and of course these will be 'constants' or polynomials of degree zero. Correct?

- Dec 25th 2012, 10:02 PMDevenoRe: Polynomial Rings -- Z/4Z
exactly. that is a theorem stated in most algebra texts.

- Dec 26th 2012, 02:29 PMBernhardRe: Polynomial Rings -- Z/4Z
In the above you seem to be saying that when dealing with Z/nZ[x] where n is composite - one should take one of the prime factors, say p, of n and then define a homomorphism of Z/nZ[x] onto Z/pZ[x]

In Z/pZ[x] we have that the only units of Z/pZ[x] are the units of Z/pZ - and we use this fact somehow ... possibly arguing back to Z/nZ[x] via the kernal of the homomorphism

Is that approximately correct? - Dec 26th 2012, 05:11 PMDevenoRe: Polynomial Rings -- Z/4Z
yes. it is a fact as well that for any ring R, the set of zero-divisors contains a prime ideal, say P, and that R/P is an integral domain. so to "get rid of unwanted zero divisors", we just need to find the right prime ideal that takes them out.

i'm going to use the notation Z_{n}, since Z/nZ is a chore to write.

let's look at zero divisors in Z_{12}:

0 <---well, it's 0, what can you say?

1 <---has an inverse (itself), so not a zero divisor.

2 <---zero divisor (2*6 = 0)

3 <---zero divisor (3*4 = 0)

4 <---zero divisor (see 3)

5 <---has an inverse (itself)

6 <---zero divisor (see 2)

7 <---has an inverse (itself)

8 <---zero divisor (3*8 = 0)

9 <---zero divisor (4*9 = 0)

10 <---zero divisor (6*10 = 0)

11 <---has an inverse (itself)

so the set of zero divisors (including 0) is {0,2,3,4,6,8,9,10} <--most of Z_{12}.

let's look at the ideals of Z_{12}(this is easy since Z_{12}is a cyclic ring).

(0) = {0}

(1) = (5) = (7) = (11) = Z_{12}(this is because all of these are co-prime to 12)

(2) = (10) = {0,2,4,6,8,10}

(3) = (9) = {0,3,6,9}

(4) = (8) = {0,4,8}

(6) = {0,6}

the prime ideals of Z_{12}are precisely the maximal ideals of Z_{12}, in this case, since a maximal ideal is proper, we can straight off the bat eliminate (1),(5),(7) and (11).

that leaves (2),(3),(4) and (6).

but (6) is contained in (3) and (2) (in fact (6) = (2)∩(3)), so (6) is not maximal.

similarly, (4) is contained in (2), so (4) is not maximal. this leaves (2) and (3), which are, in fact, prime ideals (note that these are the prime factors of 12).

in fact, (2) and (3) are co-prime ideals, that is: (2) + (3) = Z_{12}(an easy way to see this is that 5 is in (2) + (3), and (5) = Z_{12}).

the two possible ways we might create an integral domain from Z_{12}, are thus these:

Z_{12}-->Z_{12}/(2) -this is ring-isomorphic to Z_{2}.

Z_{12}-->Z_{12}/(3) -this is ring-isomorphic to Z_{3}(convince yourself of this...it's educational).

we can't get rid of all the zero-divisors of Z_{12}with one ideal (because 12 has more than one prime factor), unless we make our homomorphism the 0-map. however, 4 has no such problem (it has only one prime factor, 2).

that is, the ring-homomorphism Z_{4}-->Z_{4}/(2) gets rid of all those pesky zero-divisors.

there is a deep connection between the notion of prime number (of the integers) and prime ideals (in an arbitrary ring). prime ideals are the natural "generalizations" of the properties of prime numbers for general rings.

in practice, this often boils down to an application of the chinese remainder theorem:

as rings:

Z_{pq}≅ Z_{p}x Z_{q}, for primes p and q (there are more general versions of the CRT, but this will serve for purposes of this discussion).

this moves "factoring pq" (as a number) to "factoring the ring Z_{pq}" (as a direct product of rings).

a bit more simply, from high-school algebra, we know that pq/p = q. we'd like to know under what conditions we have (for ideals): I/(IJ) ≅ J (this looks "upside down" because with ideals (a) and (b), (a) is contained in (b) if b divides a:

"the smaller the generator, the larger the ideal" (where "smaller" is with respect to divisibility)). when is it kosher to "cancel the I"?