Question

Let $\mathbf{f}\mathbf{:}\mathbf{\mathbb{Z}}\mathbf{\to }{\mathbf{\mathbb{Z}}}_{\mathbf{6}}$be the homomorphism in example 6. Let$\mathit{K}\mathbf{=}\left\{a\in \mathbb{Z}|f\left(a\right)=\left[0\right]\right\}$Prove that $\mathit{K}$is a subring of $\mathbf{\mathbb{Z}}$ .

Short answer

Hence proved that$K$ is a subring of $\mathrm{\mathbb{Z}}$ .

Step-by-step solution

Homomorphism

If any ring$R$has elements such that,$a,b\in R$, then, the condition for homomorphism of the function of its elements is given by:

$$\begin{gathered} f\left(a+b\right)=f\left(a\right)+f\left(b\right) \\ f\left(ab\right)=f\left(a\right)f\left(b\right) \end{gathered}$$

Homomorphism

We have a function of rings as:$f:\mathrm{\mathbb{Z}}\to {\mathrm{\mathbb{Z}}}_6$ given by $f\left(a\right)=\left[a\right]$.

Clearly, the given function is surjective, then, in this case, we have:

$$\begin{gathered} K=\left\{a\in \mathrm{\mathbb{Z}}|f\left(a\right)=\left[0\right]\right\} \\ =\left\{a\in \mathrm{\mathbb{Z}}\left|6\right|a\right\} \\ =6k,\forall k\in \mathrm{\mathbb{Z}} \end{gathered}$$

Now, for some elements $\left\{a,b\in K\right\}$, we get:

$$\begin{gathered} a-b=6\left(a_1-b_1\right) \\ \Rightarrow a_1-b_1\in \mathrm{\mathbb{Z}}\Rightarrow a-b\in K \\ ab=36\left(a_1{b}_1\right)=6\left(6a_1{b}_1\right) \\ \Rightarrow 6a_1{b}_1\in \mathrm{\mathbb{Z}}\Rightarrow ab\in K \end{gathered}$$

Hence proved, $K$ is a subring of $\mathrm{\mathbb{Z}}$.