Question

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

(c) What is the kernel of$\mathbf{f}$ ?

Short answer

The kernel of$\mathrm{f}$ is$\mathrm{I}\cap \mathrm{J}$ .

Step-by-step solution

Kernel of a homomorphism.f :

The kernel of a homomorphism $\mathrm{f}:\mathrm{R}\to {\mathrm{R}}^{\text{'}}$ is a set of all elements of$\mathrm{R}$ which is mapped to identity element of${\mathrm{R}}^{\text{'}}$ that is if${\mathrm{e}}^{\text{'}}$ be the identity element of${\mathrm{R}}^{\text{'}}$

$\ker \left(\mathrm{f}\right)=\left\{\mathrm{x}\in \mathrm{R}:\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\text{'}}\right\}$

Conclusion 

By the definition of kernel of$\mathrm{f}$ , if$\mathrm{a}\in \ker \mathrm{f}$ then we have,

$\left(\mathrm{a}+\mathrm{I},\mathrm{a}+\mathrm{J}\right)=\left(\mathrm{I},\mathrm{J}\right)$, which implies that $\mathrm{a}\in \mathrm{I}\cap \mathrm{J}$

Therefore, it is clear that $\ker \left(\mathrm{f}\right)=\mathrm{I}\cap \mathrm{J}$