Question

Let $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$be a homomorphism of rings and T a subring of S .

Let$\mathit{P}\mathbf{=}\left\{r\in R|f\left(r\right)=T\right\}$ . Prove that P is a subring of R .

Short answer

It is proved that$P$ is subring of $R$.

Step-by-step solution

Determine is a homomorphism and P={r∈R|f(r)∈T}

Consider the given equation,

$P=\left\{\mathrm{r}\in \mathrm{R}|\mathrm{f}\left(\mathrm{r}\right)\in \mathrm{T}\right\}$

Here,$f:R\to S$ is a homomorphism of ring and $T$is a subring of$S$ .

It is easy to show that is closed under multiplication and subtraction under Theorem 3.1.

P={r∈R|f(r)∈T}

Let’s consider that $x$and $y$are arbitrary elements in$P$ .

Then,

$$\begin{gathered} x,y\in P\Rightarrow f\left(x\right),f\left(y\right)\in T \\ \Rightarrow f\left(x-y\right)=f\left(x\right)-f\left(y\right)\in T \end{gathered}$$

According to Theorem 3.10, ring $T$is closed under subtraction.

$f\left(xy\right)=f\left(x\right)f\left(y\right)\in T$

Ring $T$ is closed under multiplication.

Therefore, $P$is subring of R.