# Thread: Polynomial Rings -- Z/4Z

1. ## Polynomial 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

2. ## Re: Polynomial Rings -- Z/4Z

for (1):

(1 + 2x)(1 - 2x) = 1 - 4x2 = 1 (since 4 = 0 (mod 4)).

for (2):

consider the homomorphism:

φ:(Z/4Z)[x]-->(Z/2Z)[x] given by:

a0 + a1x +...+ anxn --> [a0]+[a1]x+...+[an]xn, where [ak] = ak (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)).

3. ## Re: Polynomial Rings -- Z/4Z

"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

4. ## Re: 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

5. ## Re: Polynomial Rings -- Z/4Z

Originally Posted by Bernhard

"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
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?

6. ## Re: 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?

7. ## Re: Polynomial Rings -- Z/4Z

exactly. that is a theorem stated in most algebra texts.

8. ## Re: 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?

9. ## Re: 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 Zn, since Z/nZ is a chore to write.

let's look at zero divisors in Z12:

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 Z12.

let's look at the ideals of Z12 (this is easy since Z12 is a cyclic ring).

(0) = {0}
(1) = (5) = (7) = (11) = Z12 (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 Z12 are precisely the maximal ideals of Z12, 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) = Z12 (an easy way to see this is that 5 is in (2) + (3), and (5) = Z12).

the two possible ways we might create an integral domain from Z12, are thus these:

Z12-->Z12/(2) -this is ring-isomorphic to Z2.
Z12-->Z12/(3) -this is ring-isomorphic to Z3 (convince yourself of this...it's educational).

we can't get rid of all the zero-divisors of Z12 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 Z4-->Z4/(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:

Zpq ≅ Zp x Zq, 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 Zpq" (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"?

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# the polynomials in z/4z[x]

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