Thread: maximal ideal question

Conser Z[x] = {all polynomials with integer coefficients} and I={f(x) in Z[x] such that f(0) is even} Note: I = <x,2>

I have already shown that I = <x,2>

Now I must show whether I is prime or maximal or neither and also how many elements Z[x]/I has

I know that I is maximal and that Z[x]/I but I cannot for the life of me understand why it has only 2 elements (odds and evens I'm guessing) I don't think i fully understand Factor rings

So Z[x]/I = {f(x) + <x,2>} which can be rewritten as {x*g(x) + k + <x,2>} and then <x,2> absorbs x*g(x)

so Z[x]/I = {k + <x,2>}

This is as far as I got. Can anyone help?

Originally Posted by mulaosmanovicben

Conser Z[x] = {all polynomials with integer coefficients} and I={f(x) in Z[x] such that f(0) is even} Note: I = <x,2>

I have already shown that I = <x,2>

Now I must show whether I is prime or maximal or neither and also how many elements Z[x]/I has

I know that I is maximal and that Z[x]/I but I cannot for the life of me understand why it has only 2 elements (odds and evens I'm guessing) I don't think i fully understand Factor rings

So Z[x]/I = {f(x) + <x,2>} which can be rewritten as {x*g(x) + k + <x,2>} and then <x,2> absorbs x*g(x)

so Z[x]/I = {k + <x,2>}

This is as far as I got. Can anyone help?

Any element in has free coefficient either even or odd, so awny element modulo the ideal I is

either 0 or 1...

Tonio

[xCan someone please explain this more? I do not see why Z[x]/I has only 2 elements!