Question

Let R be a ring and b a fixed element of R. Let $\mathit{T}\mathbf{=}\left\{n{1}_R|n\in \mathbb{Z}\right\}$. Prove that T is a subring of R.

Short answer

It is proved that T is a subring of R.

Step-by-step solution

Theorem

According to theorem 3.2, if S is a non-empty subset of a ring Rthen, S is a subring of Rif it follows the mentioned properties:

1. The set is closed under addition.
2. The set is closed under multiplication.
3. Zero elements of the ring should belong to the set S.
4. The solution of the equation is in the set, if $a+x=0_R$.

Check for closed under addition

Let $t_1,t_2\in T$, then$t_1=r_{1}b,t_2=r_{2}b$ for any $r_1,r_2\in R$.

Now, add $t_1,t_2$.

role="math" localid="1648187952946" $\begin{array}{rcl}{t}_1+t_2& =& r_{1}b+r_{2}b\\ & =& (r_1+r_2)b\\ & =& rb\in T\end{array}$

Hence, T is closed under addition.

Check for closed under multiplication

Now, find the product of $t_1,t_2$.

$\begin{array}{rcl}{t}_1{t}_2& =& \left(r_{1}b\right)\left(r_{2}b\right)\\ & =& \left(r_1{r}_2\right)b\\ & =& rb\in T\end{array}$

Hence, T is closed under multiplication.

Check for zero element multiple

Now, for $0_R$.

$0_R·b=0_R\in T$

Hence, Thas a zero-element multiple also.

Check for equation

Now, for $t\in S$. Then, find the solution of $\begin{array}{rcl}a+x& =& 0_R\end{array}$for the given set.

$\begin{array}{rcl}a+x& =& 0_R\\ & \Rightarrow & x=0_R-t\\ & =& -rb\left\{\mathrm{Since}t\mathit{=}rb\right\}\\ & =& (-r)b\end{array}$

Which implies that $x\in T$, so$x+t=0_R$ has zero solution in R.

Hence, T is a subring of R.