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**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 4x^{3}+1 + (2Z)[x] and 2x^{2}+4x+3 + (2Z)[x]. note that these are actually the SAME coset, since:

4x^{3}+1 - (2x^{2}+4x+3) = 4x^{3}-2x^{2}-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, x^{2}+1, and x^{3}+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(a_{0}+a_{1}x+...+a_{n}x^{n}) = φ(a_{0}) + φ(a_{1})x +....φ(a_{n})x^{n}.

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

ψ(4x^{3}+1) = φ(4)x^{3} + φ(1) = 0x^{3} + 1 = 1,

ψ(2x^{2}+4x+3) = φ(2)x^{2} + φ(4)x + φ(3) = 0x^{2} + 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(a_{j}) = 0, for every coefficient a_{j} 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 +....+0x^{n} = 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 x^{2} 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 2^{n} distinct polynomials of degree n. for example, we have, of degree 2:

x^{2}

x^{2}+1

x^{2}+x

x^{2}+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 ax^{j}, 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].