Exercise 6.1.5 (Properties of induced homomorphism)

Question

Show that if a ring homomorphism $\psi$ is injective then
so is $\psi_*$, and if $\psi$ is an isomorphism then so is $\psi_*$.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
We have that $\psi: R ightarrow S$ and $\psi_*:R[t] ightarrow S[t]$. Since $\psi$ is injective we that
$\forall x,y \in R$ if $\psi(x)=\psi(y) \implies x=y$. Then we choose $f, f' \in R[t]$ and assume that
$\psi_*f = \psi_*f'$. From 	extbf{Definition 3.1.7} we then have that:
\begin{equation*}
    \begin{aligned}
        \psi_*f &= \psi_*f'\
        \psi_* \left(\sum_{i} a_it^i ight) &= \psi_* \left(\sum_{i} b_it^i ight)\
        \sum_{i} \psi(a_i)t^i &= \sum_{i} \psi(b_i)t^i\
        \psi(a_i) &= \psi(b_i)\
        \implies a_i &= b_i 
    \end{aligned}
\end{equation*}
Hence we have that $f=\sum_{i}a_it^i = f'$. Hence $\psi_*$ is injective.
\\
If $\psi$ is an isomorphism then $\psi$ is both surjective and injective. We have just shown that
$\psi_*$ is injective, so we show that it is also surjective to prove it is an isomorphism. We know that
$\phi_*$ is surjective $\iff \forall s \in S[t] \ \exists r \in R[t]:\psi_*r=s$. We choose $s \in S[t]$ and
let $e$ be the identity homomorphism. Then we have:
\begin{equation*}
    \begin{aligned}
        s&=es\
        &=(\psi_*\psi_*^{-1})s\
        &=\psi_*(\psi_*^{-1}s)\
        &=\psi_*\left(\psi_*^{-1}\left(\sum_{i}a_it^iight)ight)\
        &=\psi_*\left(\sum_{i}\psi^{-1}(a_i)t^iight)\
        &=\psi_*r
    \end{aligned}
\end{equation*}
where $\psi^{-1}(a_i) \in R$ exists since $\psi$ is an isomorphism, and $r=\sum_{i}\psi^{-1}(a_i)t^i \in R[t]$.
Hence $\psi_*$ is also surjective, hence it is an isomorphism.

Richard Ganaye · Answer

\begin{proof} Let $\psi : K 	o K'$ be a homomorphism, and
$$
\psi_*
\left\{
\begin{array}{ccl}
K[t] & 	o &K'[t]\
\sum_{i=0}^d a_i t^i & \mapsto &\sum_{i=0}^d \psi(a_i)t^i.
\end{array}
ight.
$$
the induced homomorphism.
\begin{itemize}
\item Assume that $\psi$ is injective. Let $p = \sum_{i=0}^d a_i t^i \in \ker(\psi_*)$. Then $\sum_{i=0}^d \psi(a_i)t^i = 0$, therefore $\psi(a_i) = 0\ (0\leq i \leq d)$. Since $\psi$ is an injective homomorphim, $a_i = 0$, so $p = 0$. Since $\ker(\psi_*) = \{0\}$, $\psi_*$ is injective.

\item Assume that $\psi$ is surjective. Let $q(t) = \sum_{i=0}^d b_i t^i$ be any polynomial in $K'[t]$. Since $\psi$ is surjective, there are elements $a_i \in K$ for each $i = 0,\ldots, d$ such that $b_i = \psi(a_i)$ Therefore 
$$q(t) =  \sum_{i=0}^d b_i t^i = \sum_{i=0}^d \psi(a_i) t^i = \psi_*\left(\sum_{i=0}^d a_i t^iight).$$
Thus, if $p(t) = \sum_{i=0}^d a_i t^i$, then $q = \psi_*(p)$. This proves that $\psi_*$ is surjective.
\end{itemize}

To conclude, if $\psi$ is an isomorphism, so is $\psi_*$.
\end{proof}

Exercise 6.1.5 (Properties of induced homomorphism)

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Exercise 6.1.5 (Properties of induced homomorphism)

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