# Thread: Prove k[x]/(f(x)g(x)) \cong k[x]/(f(x)) \bigoplus k[x]/(g(x)).

1. ## Prove k[x]/(f(x)g(x)) \cong k[x]/(f(x)) \bigoplus k[x]/(g(x)).

Suppose k is a field and f(x), g(x) \in k[x] such that (f(x),g(x)) = k[x].
Prove k[x]/(f(x)g(x)) \cong k[x]/(f(x)) \bigoplus k[x]/(g(x)).

2. Originally Posted by joestevens
...................... (f(x),g(x)) = k[x].
what this means?

3. Originally Posted by zoek
what this means?
f(x) and g(x) generate an ideal that is equal to the entire ring.

4. Originally Posted by joestevens
f(x) and g(x) generate an ideal that is equal to the entire ring.
so, it means that every element of k[x] can be written as follows: s(x)f(x)+t(x)g(x) where s(x) and t(x) \in k[x]? or not?

5. Originally Posted by zoek
so, it means that every element of k[x] can be written as follows: s(x)f(x)+t(x)g(x) where s(x) and t(x) \in k[x]? or not?
Correct.

6. Originally Posted by joestevens
Suppose k is a field and f(x), g(x) \in k[x] such that (f(x),g(x)) = k[x].
Prove k[x]/(f(x)g(x)) \cong k[x]/(f(x)) \bigoplus k[x]/(g(x)).
if you define the following map:

p(x)--> (p(x)+I, p(x)+J) where I=(f(x)), J=(g(x)) and p(x) \in k[x]

you can prove that it is a ring epimorphism with kernel = I-intersection-J which is equal to IJ =(f(x)g(x)) since I+J = (f(x), g(x)) = k[x] and then you can apply the 1st isomorphism theorem for rings.

7. Originally Posted by zoek
if you define the following map:

p(x)--> (p(x)+I, p(x)+J) where I=(f(x)), J=(g(x)) and p(x) \in k[x]

you can prove that it is a ring epimorphism with kernel = I-intersection-J which is equal to IJ =(f(x)g(x)) since I+J = (f(x), g(x)) = k[x] and then you can apply the 1st isomorphism theorem for rings.
Thanks. We never covered ring epimorphisms in my class. Is a ring epimorphism just a surjective ring homomorphism?

8. Originally Posted by joestevens
Thanks. We never covered ring epimorphisms in my class. Is a ring epimorphism just a surjective ring homomorphism?
exactly!

I think that this is the most difficult part of the proof.

you can work as follows:

there exist s(x) and t(x) such that 1 = s(x)f(x)+t(x)g(x), since 1 \in k --> 1\in k[x] --> 1\in (f(x), g(x)).

if p(x) = a(x)t(x)g(x)+b(x)s(x)f(x) then you can verify that p(x)+I = a(x) +I and p(x)+J = b(x) +J

so (p(x)+I, p(x)+J) =(a(x)+I, b(x)+J)

with (a(x)+I, b(x)+J) an arbitrary element \in k[x]/(f(x)) \bigoplus k[x]/(g(x)) .