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Generalizations of the Notion of Primes J
Exercise 0.8
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Thayaden
1353 posts
Thayaden
#1 h  Oct 17, 2024, 10:26 PM
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Consider $f(x)\in R[x]$ and such that $f^{-1}(x)\in R[x]$ we clearly see that,
$$f(x)\cdot f^{-1}(x)=1$$Recall $\text{deg}(1)=0$ thus $\text{deg}(f(x)\cdot f^{-1}(x))=1$ we can clearly see that as they are each other inverse that $\deg(f)=\deg(f^{-1})=1$ thus let,
\begin{align*} f(x)&= a,\\ f^{-1}&=b. \end{align*}Thus we see,
$$a\cdot b=1$$In other words $a$ and $b$ are elements in the group of units of $R[x]$ although since $f(x),f^{-1}(x)\in R$ it is also the group of units of $R$ therefor $R[x]^{\times}=R^{\times}$
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felixgotti
197 posts
felixgotti
#2 h  Oct 31, 2024, 11:43 PM
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Why is $\deg(f) = 1$? And if this is the case, why is $f$ a constant?
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Exercise 0.8
Thayaden   1
N Oct 31, 2024 by felixgotti
Consider $f(x)\in R[x]$ and such that $f^{-1}(x)\in R[x]$ we clearly see that,
$$f(x)\cdot f^{-1}(x)=1$$Recall $\text{deg}(1)=0$ thus $\text{deg}(f(x)\cdot f^{-1}(x))=1$ we can clearly see that as they are each other inverse that $\deg(f)=\deg(f^{-1})=1$ thus let,
\begin{align*} f(x)&= a,\\ f^{-1}&=b. \end{align*}Thus we see,
$$a\cdot b=1$$In other words $a$ and $b$ are elements in the group of units of $R[x]$ although since $f(x),f^{-1}(x)\in R$ it is also the group of units of $R$ therefor $R[x]^{\times}=R^{\times}$
1 reply
Thayaden
Oct 17, 2024
felixgotti
Oct 31, 2024
J
H
Exercise 0.8
hkoo   0
May 26, 2024
What I've tried so far:

<Describe what you have tried so far here. That way, we can do a better job helping you!>

Since $R$ is subset of $R[x]$, group of units of $R$ is subset of group of units of $R[x]$

for any $u\in$ group of units of $R[x]$, there exists $v$ such that $u*v=v*u=1$. It implies that degree of $1$=degree of$u$+degree of $v$ by using exercise 0.7. So degree of$u$ is 0 and u is also element of group of units of R.
it means that group of units of $R[x]$ is group of units of $R$

Conclusively, group of units of $R[x]$ = group of units of $R$. done.

Where I'm stuck:

<Describe what's confusing you, or what your question is here!>
0 replies
hkoo
May 26, 2024
0 replies
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Exercise 0.8
Thayaden   1
N Oct 31, 2024 by felixgotti
Consider $f(x)\in R[x]$ and such that $f^{-1}(x)\in R[x]$ we clearly see that,
$$f(x)\cdot f^{-1}(x)=1$$Recall $\text{deg}(1)=0$ thus $\text{deg}(f(x)\cdot f^{-1}(x))=1$ we can clearly see that as they are each other inverse that $\deg(f)=\deg(f^{-1})=1$ thus let,
\begin{align*} f(x)&= a,\\ f^{-1}&=b. \end{align*}Thus we see,
$$a\cdot b=1$$In other words $a$ and $b$ are elements in the group of units of $R[x]$ although since $f(x),f^{-1}(x)\in R$ it is also the group of units of $R$ therefor $R[x]^{\times}=R^{\times}$
1 reply
Thayaden
Oct 17, 2024
felixgotti
Oct 31, 2024
J