# Math Help - Ring homomorphism

1. ## Ring homomorphism

Let $\mathbb{Z}[x]$ be the set of all polynomials p(x) in the variable x with with integer coefficients. Show that the function $g: \mathbb{Z}[x] \to \mathbb{Z}$ defined by $g(p(x))=p(0)$ is a ring homomorphism
I don't know I am getting confused with the set up on this one. Can some please help me?

2. ## Re: Ring homomorphism

you need to show 2 things:

for any two polynomials p(x),q(x) in Z[x]:

g(p(x) + q(x)) = g(p(x)) + g(q(x))

g(p(x)q(x)) = (g(p(x))(g(q(x))

here is something that may help:

if p(x) = a0 + a1x + .... + anxn,

then g(p(x)) = p(0) = a0 + a1*0 + .... + an*0 = a0 + 0 + .... + 0 = a0.

in other words p(0) returns the constant term of p(x).

a concrete example:

let p(x) = x - 2, and let q(x) = x + 4

then p(x) + q(x) = 2x + 2, and p(x)q(x) = x2 + 2x - 8

g(p(x)) = -2
g(q(x)) = 4

g(p(x) + q(x)) = 2
g(p(x)q(x)) = -8.

is it not true that -2 + 4 = 2, and (-2)(4) = -8?

3. ## Re: Ring homomorphism

hmmmm so would i show this as follows:
$g(p(x)+q(x))=(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2+... +(a_n+b_n)x^n=(a_0+a_1x+a_2x^2+...+a_nx^n)+(b_0+b_ 1x+b_2x^2+...+b_nx^n)=g(p(x))+g(q(x))$
$g(p(x)*q(x))=(a_0*b_0)+(a_1*b_1)x+(a_2*b_2)x^2+... +(a*n*b_n)x^n=(a_0+a_1x+a_2x^2+...+a_nx^n)*(b_0+b_ 1x+b_2x^2+...+b_nx^n)=g(p(x))*g(q(x))$
I am not sure how g(p(x))=p(0) comes into play though

4. ## Re: Ring homomorphism

no that is *not* the expression for g(p(x)). your very first equation is wrong. it should be:

$g(p(x) + q(x)) = g((a_o + b_0) + (a_1 + b_1)x + \dots + (a_n + b_n)x^n) = a_0 + b_0$

the constant term of p(x) (with a coefficients) and q(x) (with b coefficients) is the term not involving any x's.

what you have is g(p(x) + q(x)) = p(x) + q(x), which is not true.

p(0) is always an integer, it's NOT a polynomial.

5. ## Re: Ring homomorphism

hmm so it wld just be as simple as follows:
$g(p(x)+q(x))=(a_0+b_0)=a_0+b_0=g(p(x))+g(q(x))$
$g(p(x)*q(x))=(a_0*b_0)=a_0*b_0=g(p(x))*g(q(x))$

6. ## Re: Ring homomorphism

yes, it's that simple. in other words:

"the constant term of the sum of two polynomials (over Z) is the sum of the constant terms of the two polynomials"

"the constant term of the product of two polynomials (over Z) is the product of the two product terms"

in fact, this statement remains true if we replace the ring Z by any commutative ring R.

even more amazingly, it remains true if we replace p(0) by p(a), for any any element a in our ring R.

the map p(x) → p(a) is called "the evaluation map at a" and is an important ring homomorphism.

people often get confused and think that p(a) is "a polynomial in a". it is not. it is just a ring element obtained by substituting a for x in the expression p(x).

for example, let's pick a = 2 in the examples i gave earlier:

p(x) = x - 2 and q(x) = x + 4.

note that p(a) = 2 - 2 = 0, q(a) = 2 + 4 = 6. these are just numbers, not "polynomials in 2".

now (p+q)(x) = p(x) + q(x) = 2x + 2, so: (p+q)(2) = 2(2) + 2 = 4 + 2 = 6

and (pq)(x) = p(x)q(x) = x2 + 2x - 8, so (pq)(2) = 4 + 4 - 8 = 0

note that 0 + 6 = 6, and 0*6 = 0.

we use this homomorphism every time we factor a polynomial:

"high-school style":

solve x2 + 2x - 8 = 0 for x.

x2 + 2x - 8 = (x - 2)(x + 4) = 0

thus either x - 2 = 0, or x + 4 = 0

x - 2 = 0 means x = 2.

x + 4 = 0 means x = -4. thus the solutions are {2,-4}.

"abstract algebra style"

find the roots of x2 + 2x - 8 in Z (a in Z such that a2 + 2a - 8 = 0).

since x2 + 2x - 8 = (x - 2)(x + 4) applying the homomorphism p(x) → p(a) to both sides, we get:

(a - 2)(a + 4) = a2 + 2a - 8 = 0.

since Z is an integral domain, one of a - 2 or a + 4 is 0. thus:

a = a - 2 + 2 = 0 + 2 = 2, or:

a = a + 4 - 4 = 0 - 4 = 4.

do you see the (slight) difference? in ordinary high-school algebra, the shift from a function (a polynomial) to a number (substituting in "something for x") is done without even realizing we do it. in ring theory, this is accomplished explicitly (using a homomorphism), before passing to whatever properties the ring R might possess. we know this is OK, *because* we have a homomorphism.