# Thread: Ideal Test - Abstract Algebra

1. ## Ideal Test - Abstract Algebra

Let Z[x] be the ring of all polynomials with integer coefficients. Let I = {f(x) in Z[x]| f(2) = 0}. Prove that I is an ideal.

I know the definition: A subring A of a ring R is called a (two-sided) ideal of R if for every r in R and every a in A ra and ar are in A.

There is also an Ideal Test: A nonempty subset A of a ring R is an ideal of R if 1. (a-b) belongs to A whenever a, b belong to A. 2. ra and ar are in A whenever a is in A and r is in R.

I don't think this should be a hard problem at all, but we haven't really had any examples in class and I am not sure how to apply the test.

2. This shoudn't be too hard of a problem..
1.) If a and b are in A,then a-b is in A.
Proof: Since a and b are in A, then a(2) =b(2)=0. Because we are in Z[x] this implies that 2 is a root and they are divisible by (x-2). So we can write them in the form a = (x-2)a', b = (x-2)b', where a' and b' are in Z[x]. Then factoring the (x-2) out we see a-b = (x-2)(a'-b'). This means a-b is divisible by x-2 and has a root of 2, so a-b is in A.

2.) ar is in A for a in A and r in Z[x]
Similar to above, we see that a = (x-2)a', and ar = (x-2)(a')(r) which is divisible by x-2 and therefore has a root at 2, so ar is in A for a in A and r in Z[x].

,

,

,

,

# test on rings and ideals

Click on a term to search for related topics.