# Prove that if $a\in\operatorname{cent} R$ then $\left<S,a\right>=\{r_0+r_1a+\cdots+r_na^n:n\in\mathbb{Z^+};r_i\in S\}$.

Let $S$ be a subring of $R$, a ring with identity. For any $a\notin S$ the subring generated by $S\cup \{a\}$ is denoted by $\langle S,a\rangle$. It is required to prove that if $a\in\operatorname{cent} R$ then $$\langle S,a\rangle=\{r_0+r_1a+\cdots+r_na^n:n\in\mathbb{Z^+};r_i\in S\}.$$

In my attempt I could show that $$\{r_0+r_1a+...+r_na^n:n\in\mathbb{Z^+};r_i\in S\}\subseteq\langle S,a\rangle$$ as $\langle S,a\rangle$ is the smallest subring containing $S\cup\{a\}$ and consequently $$\forall n\in\mathbb{Z^+}:a^n\in\langle S,a\rangle$$ and for any finite subset $\{r_0,...,r_n\}$ of $S$, $$r_0+...+r_na^n\in\langle S,a\rangle$$ and hence $$\{r_0+r_1a+...+r_na^n:n\in\mathbb{Z^+};r_i\in S\}\subseteq \langle S,a\rangle.$$

But I cannot show the other inclusion and cannot comprehend why the two conditions

1. $R$ is a ring with identity,
2. $a\in\text{cent}R$

• If $a$ is not in the center, an element like $asa$ where $s\in S$ (which is certainly in $\langle S,a\rangle$) might not be equal to an element of the form $\sum_{i = 0}^n r_i a^i$. Jun 30 '17 at 4:47
• what would be a hint to prove the other inclusion? Jun 30 '17 at 5:44
• A hint might be that $\langle S, a \rangle$ is the smallest subring containing $S$ and $a$! Jun 30 '17 at 5:56

Let $S[a] = \{\sum_{i = 0}^n s_i a^i\in R\mid n\in\Bbb N, s_i\in S\}$, and let $s\in S$. Then $s = s + 0a + \cdots$ is clearly of the desired form, so $S\subseteq S[a]$. Since $1\in R$ and $S$ is a subring, $a = 1\cdot a\in S[a]$. Now you simply need to show that $S[a]$ is a ring. The sum of two elements of $S[a]$ is clearly another element of $S[a]$, and because $S$ is a ring, $S[a]$ has additive inverses and $0$. That the product of two elements of $S[a]$ is again in $S[a]$ follows because $a$ is in the center of $R$, and thus when computing the product of two elements, you can collect all the $a$'s in any given term on the right as $a^m$ for some $m$.
• How did you get $a=1.a\in S[a]$? Jun 30 '17 at 6:14
• @Janitha357 Presumably a subring of a ring with unity must contain that unity (this is a common convention). $a$ must be in $\langle S,a\rangle$, but if $1\not\in S$, there is no reason for $a$ to be of the form $s_0 + s_1 a + \dots + s_n a^n$, with $s_i\in S$. Jun 30 '17 at 6:16
• In that case, the claim you want to prove is not true. Suppose $R = \Bbb Z$, $S = 2\Bbb Z$, and $a = 3$. Then $\langle S,a\rangle = R$ because $2,3\in\langle S,a\rangle$, and $3 - 2 = 1$. Adding and subtracting $1$ gives all of $R$. However, your set (which I called $S[a]$) contains only even numbers (in fact, it is the set of even numbers), and so cannot be all of $R$. This demonstrates that you must require $1\in S$. Jun 30 '17 at 6:21
• Yes I agree. But you are saying "since $1\in R$ and $S$ is a subring, $a=1.a\in S[a]$". For $1.a$ to be a member of $S[a]$ we need $1\in S$. But it is not necessary that $1\in S$. Am I wrong? Jun 30 '17 at 6:27