Question

Suppose I and J are ideals in a ring R and let $f:R\to \frac{R}{I}\times \frac{R}{J}$be the function defined by$f\left(a\right)=\left(a+I, a+J\right)$

b) Is f surjective? [Hint: Consider the case when$R=\mathrm{\mathbb{Z}}, I=\left(2\right), J=\left(4\right)$

Short answer

It is proved that f is surjective mapping.

Step-by-step solution

Homomorphism of rings

Let R and $R\text{'}$be two rings. A mapping $f:R\to R\text{'}$is called an homomorphism of rings if,

• $f\left(a+b\right)=f\left(a\right)+f\left(b\right)$
  
• $f\left(ab\right)=f\left(a\right)f\left(b\right)$, for all$a,b\in R$

Conclusion

We will prove it by contradiction. Let f be not surjective.

We will take the case,$R=\mathrm{\mathbb{Z}}, I=\left(2\right), J=\left(4\right)$so we get,

$\frac{R}{I}\times \frac{R}{J}\cong {\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_4$and there will be no $a\in \mathrm{\mathbb{Z}}$, such that,

$f\left(a\right)=\left({\left[a\right]}_2,{\left[a\right]}_4\right)=\left({\left[1\right]}_2,{\left[0\right]}_4\right)$, since ${\left[a\right]}_2={\left[1\right]}_2$implies that

${\left[a\right]}_4={\left[0\right]}_2$which in turn implies that a is even, that contradicts the assumption that f is not surjective.