Question

Suppose I and J are ideals in a ring R and let f: R -> R/I X R/J be the function defined by f(a)=(a+I, a+J).

a) Prove that f is a homomorphism of rings
b) Is f surjective? (Look at the care when R=Z, I=(2) and J=(4))
c) What is the kernel of f?

Answer 1

write down exactly that you have to prove, that
\[f(a+b)=f(a)+f(b)\] which means that
\[(a+b)+I, (a+b)+J=((a+I)+(b+I),(a+J) +(b+J))\] which it does, by definition of \((a+I)+(b+I)\)

Answer 2

left hand side of the above equal sign should be an ordered pair, that is
\[\left((a+b)+I,(a+b)+J\right)\]

Answer 3

then you have to show that
\[f(ab)=f(a)f(b)\] which means that
\[(ab+I, ab+J)=(a+I, a+J)\times (b+I,b+J)\] which again is true by the definition of multiplication of factor rings

Answer 4

kernel is everything that gets sent to zero
in \(R/I\) the zero element is \(I\) so kernel should be \(K=\{a\in R|a\in I, a\in J\}\) which is another way of saying \(K=I\cap J\)

Answer 5

\(\mathbb{Z}/(2)=\mathbb{Z}_2=\{0,1\}\) the other one is \(\{0,1,2,3\}\)

Answer 6

Thank you! I think I have it figured out now!

Answer 7

yw