$R$ is a unitary ring iff $R[[x]]$ is a unitary ring

Question

I just started learning about formal power series ring and I came up to this proposition that came without a proof:

$R$ is a unitary ring $\iff$ $R[[x]]$ is a unitary ring

My definition of formal power series ring is the following:

Let $R$ be a ring. $R[[x]]:=\{ \sum_{n=0}^{\infty}a_nX^n \ | \ a_n \in R \}$

I've attempted to prove it but I only proved the right direction:

• $\Rightarrow$ Let $1_R$ be the unit of $R$. We have that $1_R = 1_R X^0 + \sum_{n=1}^{\infty}0X^n$ so $1_R \in R[[X]]$. Also, if $\sum_{n=0}^{\infty}a_nX^n \in R[[X]]$ then $1_R \sum_{n=0}^{\infty}a_nX^n = \sum_{n=0}^{\infty}1_Ra_nX^n = \sum_{n=0}^{\infty}a_n1_RX^n= \sum_{n=0}^{\infty}a_nX^n$, so $R[[X]]$ is unitary.

• $\Leftarrow$ Let $1_{R[[X]]}$ be the unit in $R[[X]]$. Then $1_{R[[X]]} \sum_{i=0}^{\infty}a_nX^n = \sum_{n=0}^{\infty}a_nX^n 1_{R[[X]]}$ for any $\sum_{n=0}^{\infty}a_nX^n \in R[[X]]$. How can we conclude from here that $R$ is unitary?

Answer 1

Let $R[[x]]$ have an identity.

Then $R$ is the homomorphic image of $R[[x]]$ via the ring homomorphism which has kernel $(x)$, and so it has an identity as well. Done.

Answer 2

In such problems, "test" the condition on special elements.

Given that $1_{R[[x]]}$ be a unit element in $R[[x]]$, multiplication with it must fix $x$. Hence, if $1_{R[[x]]}=\sum\limits_{n=0}^{\infty} a_nx^n$ with all the $a_i\in R$, then $\sum\limits_{n=0}^{\infty} a_nx^n \cdot x = x$. In particular, $a_1=a_2=...=0$, so $1_{R[[x]]}=a_0\in R$.

Now test the condition (that $1_{R[[x]]}$ be a unit element in $R[[x]]$) on constants, and you will see that $a_0$ must be a unit element in $R$.