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First, the answer to Question 2 is YES and it is not hard to show. For $q=\frac{a}{b}$, by assumption $(a,b)=R$, therefore by standard argument $(a^k,b)=R$ for any positive integer $k$. For any $f\in\mathrm{ker}(\mathrm{eval}_q)$ where $q=\frac{a}{b}$ we have $f(X)=(X-\frac{a}{b})g(X)$ where $g(X)\in K[X]$. Write $g(X)=\sum_{i=0}^{n}a_i X^i$ where $a_i\in K$. Since $f(X)\in R[X]$, by induction we see that $a^{i+1}a_i\in (b)$ for all $i\in\{0,1,\cdots,n\}$. We also have $a_n\in R$ as the leading coefficient of $f(X)$, which implies that $a_n=a_n(a^{n+1}x+by)=a^{n+1}a_nx+b a_n y\in (b)$ for some $x,y\in R$ since $(a^{n+1},b)=R.$ Now we have $a_{n-1}\in R$ since the coefficient of $X^n$ in $f(X)$ is in $R$ and that $a_n\in (b)$. Together with $a^n a_{n-1}\in(b)$ and $(a^n,b)=R$ we conclude that $a_{n-1}\in (b)$. Repeating this process, we see that $a_i\in(b)$ for all $i\in\{0,1,\cdots,n\}$. Therefore $f(X)\in (bX-a)$.

Second, it is true (as shown by proof of 3.) that if $R$ is GCD, then under the assumption that $\mathrm{gcd}(a,b)=1$ (i.e. weak version of being coprime), we also have $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$. However one cannot simply argue which special case is $``stronger''$: on the one hand, confirmation of Question 2 requires only property of the element $q$ while Point 3. requires GCD property of the whole ring $R$; on the other hand, confirmation of Question 2 requires that $a,b$ are "strongly" coprime, i.e. $(a,b)=R$ instead of "weakly" coprime, i.e. $\mathrm{gcd} (a,b)=1$. While for PID they are equivalent, the former one is much stronger than the latter in general.

Third, a very rough answer to Question 1 is NOT REALLY- and NO to the ``In particular'' part: For example, consider a number ring case where $R=\mathbb{Z}[\sqrt{5}]$ and $q=\frac{\sqrt{5}+1}{2}$. Then one can see that $\mathrm{ker}(\mathrm{eval}_q)=\big(2X-(\sqrt{5}+1),(\sqrt{5}-1)X-2, X^2-X-1\big)$ which is not principal any more. We know that $R$ is not integrally closed, let alone GCD.

Last but not least, on the one hand if $R$ is GCD then $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$ for any $a,b\in R$ such that $\mathrm{gcd}(a,b)=1$ and $\frac{a}{b}=q$; on the other hand if $R$ is not GCD then there is an element $q\in K=\mathrm{Frac}(R)$ such that it fails to have such simple property. To see this point, find two nonzero elements $a,b\in R$ which do not have a $\mathrm{gcd}$. Let $q=\frac{a}{b}$ and we claim that $\mathrm{ker}(\mathrm{eval}_q)$ is not a principal ideal. Since $bX-a$ is in the kernel, if it is principal then it must be generated by a degree one polynomial $\beta X-\alpha$ satisfying $r\alpha=a$ and $r\beta=b$ from some element $r\in R$. Since $r$ is not a $\mathrm{gcd}$ of $a$ and $b$, there is another common divisor $d$ of $a$ and $b$ so that $d$ does not divide $r$. Write $b=db_0$ and $a=da_0$ and one sees that $b_0 X-a_0$ is also in the kernel, hence generated by $\beta X-\alpha$, whence $d$ divides $r$, contradiction. QED.

First, the answer to Question 2 is YES and it is not hard to show. For $q=\frac{a}{b}$, by assumption $(a,b)=R$, therefore by standard argument $(a^k,b)=R$ for any positive integer $k$. For any $f\in\mathrm{ker}(\mathrm{eval}_q)$ where $q=\frac{a}{b}$ we have $f(X)=(X-\frac{a}{b})g(X)$ where $g(X)\in K[X]$. Write $g(X)=\sum_{i=0}^{n}a_i X^i$ where $a_i\in K$. Since $f(X)\in R[X]$, by induction we see that $a^{i+1}a_i\in (b)$ for all $i\in\{0,1,\cdots,n\}$. We also have $a_n\in R$ as the leading coefficient of $f(X)$, which implies that $a_n=a_n(a^{n+1}x+by)=a^{n+1}a_nx+b a_n y\in (b)$ for some $x,y\in R$ since $(a^{n+1},b)=R.$ Now we have $a_{n-1}\in R$ since the coefficient of $X^n$ in $f(X)$ is in $R$ and that $a_n\in (b)$. Together with $a^n a_{n-1}\in(b)$ and $(a^n,b)=R$ we conclude that $a_{n-1}\in (b)$. Repeating this process, we see that $a_i\in(b)$ for all $i\in\{0,1,\cdots,n\}$. Therefore $f(X)\in (bX-a)$.

Second, it is true (as shown by proof of 3.) that if $R$ is GCD, then under the assumption that $\mathrm{gcd}(a,b)=1$ (i.e. weak version of being coprime), we also have $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$. However one cannot simply argue which special case is $``stronger''$: on the one hand, confirmation of Question 2 requires only property of the element $q$ while Point 3. requires GCD property of the whole ring $R$; on the other hand, confirmation of Question 2 requires that $a,b$ are "strongly" coprime, i.e. $(a,b)=R$ instead of "weakly" coprime, i.e. $\mathrm{gcd} (a,b)=1$. While for PID they are equivalent, the former one is much stronger than the latter in general.

Third, a very rough answer to Question 1 is NOT REALLY- and NO to the ``In particular'' part: For example, consider a number ring case where $R=\mathbb{Z}[\sqrt{5}]$ and $q=\frac{\sqrt{5}+1}{2}$. Then one can compute to see explicitly $$\mathrm{ker}(\mathrm{eval}_q)=\Big(2X-(\sqrt{5}+1),(\sqrt{5}-1)X-2, X^2-X-1\Big)$$ which is not principal any more. In this case, we know that $R$ is not integrally closed, let alone GCD.

Last but not least, on the one hand if $R$ is GCD then $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$ for any $a,b\in R$ such that $\mathrm{gcd}(a,b)=1$ and $\frac{a}{b}=q$; on the other hand if $R$ is not GCD then there is an element $q\in K=\mathrm{Frac}(R)$ which fails to have such simple property. To see this point, find two nonzero elements $a,b\in R$ which do not have a $\mathrm{gcd}$. Let $q=\frac{a}{b}$ and we claim that $\mathrm{ker}(\mathrm{eval}_q)$ is not a principal ideal. Since $bX-a$ is in the kernel, if it is principal then it must be generated by a degree one polynomial $\beta X-\alpha$ satisfying $r\alpha=a$ and $r\beta=b$ from some element $r\in R$. Since $r$ is not a $\mathrm{gcd}$ of $a$ and $b$, there is another common divisor $d$ of $a$ and $b$ so that $d$ does not divide $r$. Write $b=db_0$ and $a=da_0$ and one sees that $b_0 X-a_0$ is also in the kernel, hence generated by $\beta X-\alpha$, whence $d$ divides $r$, contradiction. QED.

First, the answer to Question 2 is YES and it is not hard to show. For $q=\frac{a}{b}$, by assumption $(a,b)=R$, therefore by standard argument $(a^k,b)=R$ for any positive integer $k$. For any $f\in\mathrm{ker}(\mathrm{eval}_q)$ where $q=\frac{a}{b}$ we have $f(X)=(X-\frac{a}{b})g(X)$ where $g(X)\in K[X]$. Write $g(X)=\sum_{i=0}^{n}a_i X^i$ where $a_i\in K$. Since $f(X)\in R[X]$, by induction we see that $a^{i+1}a_i\in (b)$ for all $i\in\{0,1,\cdots,n\}$. We also have $a_n\in R$ as the leading coefficient of $f(X)$, which implies that $a_n=a_n(a^{n+1}x+by)=a^{n+1}a_nx+b a_n y\in (b)$ for some $x,y\in R$ since $(a^{n+1},b)=R.$ Now we have $a_{n-1}\in R$ since the coefficient of $X^n$ in $f(X)$ is in $R$ and that $a_n\in (b)$. Together with $a^n a_{n-1}\in(b)$ and $(a^n,b)=R$ we conclude that $a_{n-1}\in (b)$. Repeating this process, we see that $a_i\in(b)$ for all $i\in\{0,1,\cdots,n\}$. Therefore $f(X)\in (bX-a)$.

Second, it is true (as shown by proof of 3.) that if $R$ is GCD, then under the assumption that $\mathrm{gcd}(a,b)=1$ (i.e. weak version of being coprime), we also have $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$. However one cannot simply argue which special case is $``stronger''$: on the one hand, confirmation of Question 2 requires only property of the element $q$ while Point 3. requires GCD property of the whole ring $R$; on the other hand, confirmation of Question 2 requires that $a,b$ are "strongly" coprime ($(a,b)=R$) instead of "weakly" coprime ($\mathrm{gcd} (a,b)=1$), for PID they are equivalent but not in general.

Third, a very rough answer to Question 1 is NOT REALLY- and NO to the ``In particular'' part: For example, consider a number ring case where $R=\mathbb{Z}[\sqrt{5}]$ and $q=\frac{\sqrt{5}+1}{2}$. Then one can see that $\mathrm{ker}(\mathrm{eval}_q)=\big(2X-(\sqrt{5}+1),(\sqrt{5}-1)X-2, X^2-X-1\big)$ which is not principal any more. We know that $R$ is not integrally closed, let alone GCD.

Last but not least, on the one hand if $R$ is GCD then $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$ for any $a,b\in R$ such that $\mathrm{gcd}(a,b)=1$ and $\frac{a}{b}=q$; on the other hand if $R$ is not GCD then there is an element $q\in K=\mathrm{Frac}(R)$ such that it fails to have such simple property. To see this point, find two nonzero elements $a,b\in R$ which do not have a $\mathrm{gcd}$. Let $q=\frac{a}{b}$ and we claim that $\mathrm{ker}(\mathrm{eval}_q)$ is not a principal ideal. Since $bX-a$ is in the kernel, if it is principal then it must be generated by a degree one polynomial $\beta X-\alpha$ satisfying $r\alpha=a$ and $r\beta=b$ from some element $r\in R$. Since $r$ is not a $\mathrm{gcd}$ of $a$ and $b$, there is another common divisor $d$ of $a$ and $b$ so that $d$ does not divide $r$. Write $b=db_0$ and $a=da_0$ and one sees that $b_0 X-a_0$ is also in the kernel, hence generated by $\beta X-\alpha$, whence $d$ divides $r$, contradiction. QED.

First, the answer to Question 2 is YES and it is not hard to show. For $q=\frac{a}{b}$, by assumption $(a,b)=R$, therefore by standard argument $(a^k,b)=R$ for any positive integer $k$. For any $f\in\mathrm{ker}(\mathrm{eval}_q)$ where $q=\frac{a}{b}$ we have $f(X)=(X-\frac{a}{b})g(X)$ where $g(X)\in K[X]$. Write $g(X)=\sum_{i=0}^{n}a_i X^i$ where $a_i\in K$. Since $f(X)\in R[X]$, by induction we see that $a^{i+1}a_i\in (b)$ for all $i\in\{0,1,\cdots,n\}$. We also have $a_n\in R$ as the leading coefficient of $f(X)$, which implies that $a_n=a_n(a^{n+1}x+by)=a^{n+1}a_nx+b a_n y\in (b)$ for some $x,y\in R$ since $(a^{n+1},b)=R.$ Now we have $a_{n-1}\in R$ since the coefficient of $X^n$ in $f(X)$ is in $R$ and that $a_n\in (b)$. Together with $a^n a_{n-1}\in(b)$ and $(a^n,b)=R$ we conclude that $a_{n-1}\in (b)$. Repeating this process, we see that $a_i\in(b)$ for all $i\in\{0,1,\cdots,n\}$. Therefore $f(X)\in (bX-a)$.

Second, it is true (as shown by proof of 3.) that if $R$ is GCD, then under the assumption that $\mathrm{gcd}(a,b)=1$ (i.e. weak version of being coprime), we also have $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$. However one cannot simply argue which special case is $``stronger''$: on the one hand, confirmation of Question 2 requires only property of the element $q$ while Point 3. requires GCD property of the whole ring $R$; on the other hand, confirmation of Question 2 requires that $a,b$ are "strongly" coprime, i.e. $(a,b)=R$ instead of "weakly" coprime, i.e. $\mathrm{gcd} (a,b)=1$. While for PID they are equivalent, the former one is much stronger than the latter in general.

Third, a very rough answer to Question 1 is NOT REALLY- and NO to the ``In particular'' part: For example, consider a number ring case where $R=\mathbb{Z}[\sqrt{5}]$ and $q=\frac{\sqrt{5}+1}{2}$. Then one can see that $\mathrm{ker}(\mathrm{eval}_q)=\big(2X-(\sqrt{5}+1),(\sqrt{5}-1)X-2, X^2-X-1\big)$ which is not principal any more. We know that $R$ is not integrally closed, let alone GCD.

Last but not least, on the one hand if $R$ is GCD then $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$ for any $a,b\in R$ such that $\mathrm{gcd}(a,b)=1$ and $\frac{a}{b}=q$; on the other hand if $R$ is not GCD then there is an element $q\in K=\mathrm{Frac}(R)$ such that it fails to have such simple property. To see this point, find two nonzero elements $a,b\in R$ which do not have a $\mathrm{gcd}$. Let $q=\frac{a}{b}$ and we claim that $\mathrm{ker}(\mathrm{eval}_q)$ is not a principal ideal. Since $bX-a$ is in the kernel, if it is principal then it must be generated by a degree one polynomial $\beta X-\alpha$ satisfying $r\alpha=a$ and $r\beta=b$ from some element $r\in R$. Since $r$ is not a $\mathrm{gcd}$ of $a$ and $b$, there is another common divisor $d$ of $a$ and $b$ so that $d$ does not divide $r$. Write $b=db_0$ and $a=da_0$ and one sees that $b_0 X-a_0$ is also in the kernel, hence generated by $\beta X-\alpha$, whence $d$ divides $r$, contradiction. QED.

First, the answer to Question 2 is YES and it is not hard to show. For $q=\frac{a}{b}$, by assumption $(a,b)=R$. In particular, $1$ is generated by $a$ and $b$ so we have $1=ax+by$ for some $x,y\in R$. Take $n$-th power on both sides we see that $1$ is generated by $a^k$ and $b$. Therefore $(a^k,b)=R$ for any positive integer $k$.

  For any $f\in\mathrm{ker}(\mathrm{eval}_q)$ where $q=\frac{a}{b}$ we have $f(X)=(X-\frac{a}{b})g(X)$ where $g(X)\in K[X]$. Write $g(X)=\sum_{i=0}^{n}a_i X^i$ where $a_i\in K$. Since $f(X)\in R[X]$, by induction we see that $a^{i+1}a_i\in (b)$ for all $i\in\{0,1,\cdots,n\}$. We also have $a_n\in R$ as the leading coefficient of $f(X)$, which implies that $a_n=a_n(a^{n+1}x+by)=a^{n+1}a_nx+b a_n y\in (b)$ for some $x,y\in R$ since $(a^{n+1},b)=R.$ Now we have $a_{n-1}\in R$ since the coefficient of $X^n$ in $f(X)$ is in $R$ and that $a_n\in (b)$. Together with $a^n a_{n-1}\in(b)$ and $(a^n,b)=R$ we conclude that $a_{n-1}\in (b)$. Repeating this process, we see that $a_i\in(b)$ for all $i\in\{0,1,\cdots,n\}$. Therefore $f(X)\in (bX-a)$.

Second, it is true (as shown by proof of 3.) that if $R$ is GCD, then under the assumption that $\mathrm{gcd}(a,b)=1$ (i.e. weak version of being coprime), we also have $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$. However one cannot simply argue which special case is $``stronger''$: on the one hand, confirmation of Question 2 requires only property of the element $q$ while Point 3. requires GCD property of the whole ring $R$; on the other hand, confirmation of Question 2 requires that $a,b$ are "strongly" coprime ($(a,b)=R$) instead of "weakly" coprime ($\mathrm{gcd} (a,b)=1$), for PID they are equivalent but not in general.

Third, a very rough answer to Question 1 is NOT REALLY- and NO to the ``In particular'' part: For example, consider a number ring case where $R=\mathbb{Z}[\sqrt{5}]$ and $q=\frac{\sqrt{5}+1}{2}$. Then one can see that $\mathrm{ker}(\mathrm{eval}_q)=\big(2X-(\sqrt{5}+1),(\sqrt{5}-1)X-2, X^2-X-1\big)$ which is not principal any more. We know that $R$ is not integrally closed, let alone GCD.

Last but not least, on the one hand if $R$ is GCD then $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$ for any $a,b\in R$ such that $\mathrm{gcd}(a,b)=1$ and $\frac{a}{b}=q$; on the other hand if $R$ is not GCD then there is an element $q\in K=\mathrm{Frac}(R)$ such that it fails to have such simple property. To see this point, find two nonzero elements $a,b\in R$ which do not have a $\mathrm{gcd}$. Let $q=\frac{a}{b}$ and we claim that $\mathrm{ker}(\mathrm{eval}_q)$ is not a principal ideal. Since $bX-a$ is in the kernel, if it is principal then it must be generated by a degree one polynomial $\beta X-\alpha$ satisfying $r\alpha=a$ and $r\beta=b$ from some element $r\in R$. Since $r$ is not a $\mathrm{gcd}$ of $a$ and $b$, there is another common divisor $d$ of $a$ and $b$ so that $d$ does not divide $r$. Write $b=db_0$ and $a=da_0$ and one sees that $b_0 X-a_0$ is also in the kernel, hence generated by $\beta X-\alpha$, whence $d$ divides $r$, contradiction. QED.

First, the answer to Question 2 is YES and it is not hard to show. For $q=\frac{a}{b}$, by assumption $(a,b)=R$, therefore by standard argument $(a^k,b)=R$ for any positive integer $k$. For any $f\in\mathrm{ker}(\mathrm{eval}_q)$ where $q=\frac{a}{b}$ we have $f(X)=(X-\frac{a}{b})g(X)$ where $g(X)\in K[X]$. Write $g(X)=\sum_{i=0}^{n}a_i X^i$ where $a_i\in K$. Since $f(X)\in R[X]$, by induction we see that $a^{i+1}a_i\in (b)$ for all $i\in\{0,1,\cdots,n\}$. We also have $a_n\in R$ as the leading coefficient of $f(X)$, which implies that $a_n=a_n(a^{n+1}x+by)=a^{n+1}a_nx+b a_n y\in (b)$ for some $x,y\in R$ since $(a^{n+1},b)=R.$ Now we have $a_{n-1}\in R$ since the coefficient of $X^n$ in $f(X)$ is in $R$ and that $a_n\in (b)$. Together with $a^n a_{n-1}\in(b)$ and $(a^n,b)=R$ we conclude that $a_{n-1}\in (b)$. Repeating this process, we see that $a_i\in(b)$ for all $i\in\{0,1,\cdots,n\}$. Therefore $f(X)\in (bX-a)$.

Second, it is true (as shown by proof of 3.) that if $R$ is GCD, then under the assumption that $\mathrm{gcd}(a,b)=1$ (i.e. weak version of being coprime), we also have $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$. However one cannot simply argue which special case is $``stronger''$: on the one hand, confirmation of Question 2 requires only property of the element $q$ while Point 3. requires GCD property of the whole ring $R$; on the other hand, confirmation of Question 2 requires that $a,b$ are "strongly" coprime ($(a,b)=R$) instead of "weakly" coprime ($\mathrm{gcd} (a,b)=1$), for PID they are equivalent but not in general.

Third, a very rough answer to Question 1 is NOT REALLY- and NO to the ``In particular'' part: For example, consider a number ring case where $R=\mathbb{Z}[\sqrt{5}]$ and $q=\frac{\sqrt{5}+1}{2}$. Then one can see that $\mathrm{ker}(\mathrm{eval}_q)=\big(2X-(\sqrt{5}+1),(\sqrt{5}-1)X-2, X^2-X-1\big)$ which is not principal any more. We know that $R$ is not integrally closed, let alone GCD.

Last but not least, on the one hand if $R$ is GCD then $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$ for any $a,b\in R$ such that $\mathrm{gcd}(a,b)=1$ and $\frac{a}{b}=q$; on the other hand if $R$ is not GCD then there is an element $q\in K=\mathrm{Frac}(R)$ such that it fails to have such simple property. To see this point, find two nonzero elements $a,b\in R$ which do not have a $\mathrm{gcd}$. Let $q=\frac{a}{b}$ and we claim that $\mathrm{ker}(\mathrm{eval}_q)$ is not a principal ideal. Since $bX-a$ is in the kernel, if it is principal then it must be generated by a degree one polynomial $\beta X-\alpha$ satisfying $r\alpha=a$ and $r\beta=b$ from some element $r\in R$. Since $r$ is not a $\mathrm{gcd}$ of $a$ and $b$, there is another common divisor $d$ of $a$ and $b$ so that $d$ does not divide $r$. Write $b=db_0$ and $a=da_0$ and one sees that $b_0 X-a_0$ is also in the kernel, hence generated by $\beta X-\alpha$, whence $d$ divides $r$, contradiction. QED.