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Let $f(x)$, $g(x)$ be two univariate, coprime, integer polynomials and let $I=\big(f(x),g(x)\big)$ the ideal of $\mathbb{Z}[x]$ generated by $f, g$. Let $I \cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which can be expressed as linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. $I \cap \mathbb{Z}$ is clearly an ideal in $\mathbb{Z}$. The following argument shows that, given two coprime integer polynomials $f,g$, the ideal $I \cap \mathbb{Z}$ of $\mathbb{Z}$ has always non zero elements:

If $f,g$ are considered as elements of $\mathbb{Q}[x]$, Bezout's identity tells us that there exists a pair of unique rational polynomials $U(x)$, $V(x)$ with $\deg U<\deg g$, $\deg V<\deg f$ such that $$ U(x)f(x)+V(x)g(x)=1 $$ Thus, clearing denominators in the above identity, we get that there exists a pair of unique integer polynomials $u(x)$, $v(x)$ with $\deg u<\deg g$, $\deg v<\deg f$ such that $$ u(x)f(x)+v(x)g(x)=c $$ where $c$ is the $lcm$ of the denominators of $U,V$ and: $u=cU$, $v=cV$. Thus: $0 \neq c\in I \cap \mathbb{Z}$.

Question $1$: Can we determine a sufficient and necessary condition such that: $\mathbb{Z}\subset I \cap \mathbb{Z}$?

Question $2$: Is there a general method for determining the least positive generator of $I \cap \mathbb{Z}$ ?

References:

P.S.: The second question above, is actually part of the question posted in the first of the references above. However, it is posted here rather as a problem of commutative algebra than a problem of number theory.

Let $f(x)$, $g(x)$ be two univariate, coprime, integer polynomials and let $I=\big(f(x),g(x)\big)$ the ideal of $\mathbb{Z}[x]$ generated by $f, g$. Let $I \cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which can be expressed as linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. $I \cap \mathbb{Z}$ is clearly an ideal in $\mathbb{Z}$. The following argument shows that, given two coprime integer polynomials $f,g$, the ideal $I \cap \mathbb{Z}$ of $\mathbb{Z}$ has always non zero elements:

If $f,g$ are considered as elements of $\mathbb{Q}[x]$, Bezout's identity tells us that there exists a pair of unique rational polynomials $U(x)$, $V(x)$ with $\deg U<\deg g$, $\deg V<\deg f$ such that $$ U(x)f(x)+V(x)g(x)=1 $$ Thus, clearing denominators in the above identity, we get that there exists a pair of unique integer polynomials $u(x)$, $v(x)$ with $\deg u<\deg g$, $\deg v<\deg f$ such that $$ u(x)f(x)+v(x)g(x)=c $$ where $c$ is the $lcm$ of the denominators of $U,V$ and: $u=cU$, $v=cV$. Thus: $0 \neq c\in I \cap \mathbb{Z}$.

Question $1$: Can we determine a sufficient and necessary condition such that: $\mathbb{Z}\subset I \cap \mathbb{Z}$?

Question $2$: Is there a general method for determining the least positive generator of $I \cap \mathbb{Z}$ ?

References:

P.S.: The second question above, is actually part of the question posted in the first of the references above. However, it is posted here rather as a problem of commutative algebra than a problem of number theory.

Let $f(x)$, $g(x)$ be two univariate, coprime, integer polynomials and let $I=\big(f(x),g(x)\big)$ the ideal of $\mathbb{Z}[x]$ generated by $f, g$. Let $I \cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which can be expressed as linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. $I \cap \mathbb{Z}$ is clearly an ideal in $\mathbb{Z}$. The following argument shows that, given two coprime integer polynomials $f,g$, the ideal $I \cap \mathbb{Z}$ of $\mathbb{Z}$ has always non zero elements:

If $f,g$ are considered as elements of $\mathbb{Q}[x]$, Bezout's identity tells us that there exists a pair of unique rational polynomials $U(x)$, $V(x)$ with $\deg U<\deg g$, $\deg V<\deg f$ such that $$ U(x)f(x)+V(x)g(x)=1 $$ Thus, clearing denominators in the above identity, we get that there exists a pair of unique integer polynomials $u(x)$, $v(x)$ with $\deg u<\deg g$, $\deg v<\deg f$ such that $$ u(x)f(x)+v(x)g(x)=c $$ where $c$ is the $lcm$ of the denominators of $U,V$ and: $u=cU$, $v=cV$. Thus: $0 \neq c\in I \cap \mathbb{Z}$.

Question $1$: Can we determine a sufficient and necessary condition such that: $\mathbb{Z}\subset I \cap \mathbb{Z}$?

Question $2$: Is there a general method for determining the least positive generator of $I \cap \mathbb{Z}$ ?

References:

P.S.: The second question above, is actually part of the question posted in the first of the references above. However, it is posted here rather as a problem of commutative algebra than a problem of number theory.

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Let $f(x)$, $g(x)$ be two univariate, coprime, integer polynomials and let $I=\big(f(x),g(x)\big)$ the ideal of $\mathbb{Z}[x]$ generated by $f, g$. Let $I \cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which can be expressed as linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. $I \cap \mathbb{Z}$ is clearly an ideal in $\mathbb{Z}$. The following argument shows that, given two coprime integer polynomials $f,g$, the ideal $I \cap \mathbb{Z}$ of $\mathbb{Z}$ has always non zero elements:

If $f,g$ are considered as elements of $\mathbb{Q}[x]$, Bezout's identity tells us that there exists a pair of unique rational polynomials $U(x)$, $V(x)$ with $\deg U<\deg g$, $\deg V<\deg f$ such that $$ U(x)f(x)+V(x)g(x)=1 $$ Thus, clearing denominators in the above identity, we get that there exists a pair of unique integer polynomials $u(x)$, $v(x)$ with $\deg u<\deg g$, $\deg v<\deg f$ such that $$ u(x)f(x)+v(x)g(x)=c $$ where $c$ is the $lcm$ of the denominators of $U,V$ and: $u=cU$, $v=cV$. Thus: $0 \neq c\in I \cap \mathbb{Z}$.

Question $1$: Can we determine a sufficient and necessary condition such that: $\mathbb{Z}\subset I \cap \mathbb{Z}$?

Question $2$: Is there a general method for determining the least positive generator of $I \cap \mathbb{Z}$ ?

References:

P.S.: The second question above, is essentially the same withactually part of the question posted in the first of the references above. However, it is posted here rather as a problem of commutative algebra than a problem of number theory.

Let $f(x)$, $g(x)$ be two univariate, coprime, integer polynomials and let $I=\big(f(x),g(x)\big)$ the ideal of $\mathbb{Z}[x]$ generated by $f, g$. Let $I \cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which can be expressed as linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. $I \cap \mathbb{Z}$ is clearly an ideal in $\mathbb{Z}$. The following argument shows that, given two coprime integer polynomials $f,g$, the ideal $I \cap \mathbb{Z}$ of $\mathbb{Z}$ has always non zero elements:

If $f,g$ are considered as elements of $\mathbb{Q}[x]$, Bezout's identity tells us that there exists a pair of unique rational polynomials $U(x)$, $V(x)$ with $\deg U<\deg g$, $\deg V<\deg f$ such that $$ U(x)f(x)+V(x)g(x)=1 $$ Thus, clearing denominators in the above identity, we get that there exists a pair of unique integer polynomials $u(x)$, $v(x)$ with $\deg u<\deg g$, $\deg v<\deg f$ such that $$ u(x)f(x)+v(x)g(x)=c $$ where $c$ is the $lcm$ of the denominators of $U,V$ and: $u=cU$, $v=cV$. Thus: $0 \neq c\in I \cap \mathbb{Z}$.

Question $1$: Can we determine a sufficient and necessary condition such that: $\mathbb{Z}\subset I \cap \mathbb{Z}$?

Question $2$: Is there a general method for determining the least positive generator of $I \cap \mathbb{Z}$ ?

References:

P.S.: The second question above, is essentially the same with the question posted in the first of the references above. However, it is posted here rather as a problem of commutative algebra than a problem of number theory.

Let $f(x)$, $g(x)$ be two univariate, coprime, integer polynomials and let $I=\big(f(x),g(x)\big)$ the ideal of $\mathbb{Z}[x]$ generated by $f, g$. Let $I \cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which can be expressed as linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. $I \cap \mathbb{Z}$ is clearly an ideal in $\mathbb{Z}$. The following argument shows that, given two coprime integer polynomials $f,g$, the ideal $I \cap \mathbb{Z}$ of $\mathbb{Z}$ has always non zero elements:

If $f,g$ are considered as elements of $\mathbb{Q}[x]$, Bezout's identity tells us that there exists a pair of unique rational polynomials $U(x)$, $V(x)$ with $\deg U<\deg g$, $\deg V<\deg f$ such that $$ U(x)f(x)+V(x)g(x)=1 $$ Thus, clearing denominators in the above identity, we get that there exists a pair of unique integer polynomials $u(x)$, $v(x)$ with $\deg u<\deg g$, $\deg v<\deg f$ such that $$ u(x)f(x)+v(x)g(x)=c $$ where $c$ is the $lcm$ of the denominators of $U,V$ and: $u=cU$, $v=cV$. Thus: $0 \neq c\in I \cap \mathbb{Z}$.

Question $1$: Can we determine a sufficient and necessary condition such that: $\mathbb{Z}\subset I \cap \mathbb{Z}$?

Question $2$: Is there a general method for determining the least positive generator of $I \cap \mathbb{Z}$ ?

References:

P.S.: The second question above, is actually part of the question posted in the first of the references above. However, it is posted here rather as a problem of commutative algebra than a problem of number theory.

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