**11.5.6**

Let \(a\) be an element of a ring \(R\), and let \(R’\) be the ring \(R[x]/<ax-1>\) obtained by adjoining an inverse of \(a\) to \(R\). Let \(\alpha\) denote the residue of \(x\) (the inverse of \(a\) in \(R’\)).

**(a)** Show that every element \(\beta\) of \(R’\) can be written in the form \(\beta=\alpha^kb\) with \(b \in R\).

**(b)** Prove that the kernel of the map \(R \to R’\) is the set of elements \(b \in R\) such that \(a^n b=0\) for some \(n > 0\).

**(c)** Prove that \(R’\) is the zero ring if and only if \(a\) is nilpotent.

**Solution**.

**(a)** Show that every element \(\beta\) of \(R’\) can be written in the form \(\beta=\alpha^kb\) with \(b \in R\).

\(\beta \in R’\), WTS \(\beta = \alpha^kb\) for some \(b \in R\). Let \(\pi: R[x] \to R’=R[\alpha]\) be the canonical homomorphism where \(\pi(ax-1)=0\). Then

\begin{align*}

\beta &= \pi(b_0+b_1x+…+b_nx^n)\\

&=b_0+b_1\alpha+…+b_n\alpha^n

\end{align*}

And we want

\[\beta=b_0+b_1\alpha+…+b_n\alpha^n=\alpha^kb\]

So let’s multiply everything through by \(a^k\) to find out what \(b\) has to be (heh). Recall that \(a\alpha=1\)!

\begin{align*}

b_0+b_1\alpha+…+b_n\alpha^n&=\alpha^kb\\

b_0 a^k + b_1 \alpha a^k +…+ b_n\alpha^n a^k&= b

\end{align*}

The only restriction on \(k\) is that it has to be in \(\mathbb{N}\), so let \(k=n\), then

\begin{align*}

b_0 a^n + b_1 \alpha a^n +…+ b_n\alpha^n a^n&= b\\

b_0 a^n + b_1 a^{n-1} + …+ b_n&= b

\end{align*}

And we have

\begin{align*}

\alpha^n b &= \alpha^n (b_0 a^n + b_1 a^{n-1} + …+ b_n)\\

&= b_0(a\alpha)^n+b_1(a\alpha)^{n-1}\alpha +…+ b_n\alpha^n\\

&= b_0+b_1\alpha+…+b_n\alpha^n\\

&= \beta

\end{align*}

So every element \(\beta\) of \(R’\) can be written in the form \(\beta=\alpha^kb\) for some \(k \in \mathbb{N}\).

**(b)** Prove that the kernel of the map \(R \to R’\) is the set of elements \(b \in R\) such that \(a^n b=0\) for some \(n > 0\).

(NB: I’m real iffy on this one, not sure what the map \(R \to R’\) is… Is it the same as the map \(\pi\) above, but restricted to the constants?)

Let \(\varphi\) be our map, let the kernel of the map be \(K\) and let \(B=\{b \in R | a^nb=0 \text{ for some } n \in \mathbb{N}\}\). Consider \(k \in K\) and \(b \in B\). WTS \(b \in K\) and \(k \in B\).

Let’s show \(b \in K\) first. Let \(\varphi(b)=\beta\), WTS \(\beta = 0\). Since \(b \in B\), we have \(a^n b = 0\) for some \(n\). Then \(\varphi\) it:

\begin{align*}

a^n b &= 0\\

\varphi(a^n b) &=\varphi( 0)\\

\varphi(a^n)\varphi(b) &= 0\\

a^n \beta &= 0\\

\alpha^n a^n \beta &= \alpha^n 0\\

\beta &= 0

\end{align*}

Therefore \(b \in K\).

Now let’s show \(k \in B\). Since \(k \in K\), \(\varphi(k)=0\). For contradiction, suppose \(k \notin B\). Then \(a^n k \neq 0\) for all \(n \in \mathbb{N}\) and we have

\begin{align*}

a^n k &\neq 0\\

\varphi(a^n k) &\neq 0\\

\varphi(a^n) \varphi(k) &\neq 0\\

\varphi(a^n) \cdot 0 &\neq 0\\

0 &\neq 0

\end{align*}

Which is not true. So we have to have \(k \in B\).

We’ve got inclusion in both directions, so \(K=B\) and the kernel of the map \(R \to R’\) is the set of elements \(b \in R\) such that \(a^n b=0\) for some \(n > 0\).

**(c)** Prove that \(R’\) is the zero ring if and only if \(a\) is nilpotent.

Let \(\pi: R[x] \to R’=R[\alpha]\) be the canonical homomorphism where \(\pi(ax-1)=0\).

Let’s go forwards first. Suppose \(R’\) is the zero ring. WTS there exists \( n \in \mathbb{N}\) such that \(a^n=0\). Since \(R’\) is the zero ring, \(1=0\) in there. So for all \(p \in R[x]\) we have \(\pi(p)=0\) because \(p\) just doesn’t have anywhere else to go. Let \(p= b_0+b_1 x+ …+ b_m x^m\) where \(b_i \in R\). All the \(b_i\) are in the kernel because everything is in the kernel. So by (b) above we have \(a^k b_i = 0\) for some \( k \in \mathbb{N}\). Let \(n\) be the biggest \(k\) such that \(a^k b_i=0\) for all \(i\). Then we have \(a^n b_i=0\) for all \(i\). Now watch this fancy move:

\begin{align*}

p &= b_0+b_1 x+ …+ b_m x^m\\

a^n p &= a^nb_0+a^nb_1 x+ …+ a^nb_m x^m\\

\end{align*}

Each term on the right hand side is zero, so \(a^n p =0\) for all \(p \in R[x]\). Set \(p=1\), then \(a^n =0\) for some \(n\) and \(a\) is nilpotent. (I now am almost positive there was something much easier to do… forget about \(p\) altogether and just look at \(1 \in R[x]\). blarg.)

Now backwards. Suppose \(a\) is nilpotent. Then there exists an \(n\) such that \(a^n=0\). We want to show that \(R’\) is the zero ring. It suffices to show that 1 is in the kernel. By (b), 1 is in the kernel if there exists \(a^i \cdot 1 = a^i \in R\). And this thing does exist, by assumption. So in \(R’\) 1=0, and \(R’\) is the zero ring. Fine.

We went forwards, we went backwards, so the iff is proved. And we’re done.

P.S. Does this mean if we throw multiplicative inverses into a field everything gets squished? That doesn’t make sense. We should just get the field back. Fields don’t have nilpotent elements. An element can be either nilpotent or a unit, and everything in a field is a unit. Suppose \(\beta\) is nilpotent and a unit. Then \(\beta^n=0\) and there exists a \(\beta^{-1}\) such that \(\beta^{-1}\beta=1\). Then

\begin{align*}

\beta^n&=0\\

\beta^{-n}\beta^n&=\beta^{-n}\cdot 0\\

1&=0

\end{align*}

and that’s not a field. cool, can i haz extra credit plz.