Flaw of proof: polynomial ring is integrally closed if the coefficient ring is integrally closed
$begingroup$
Let $R$ be an integrally closed domain. Then I want to show $R[x]$ is integrally closed.
Let $K$ be the field of fractions of $R$ and if I choose a function $a(x)in K(x)$ which is integral over $R[x]$, then it is also integral over $K[x]$. Since a UFD is integrally closed, we have $a(x)=a_nx^n+cdots+a_1x+a_0in K[x]$. So it is enough to show each $a_iin K$ is contained in $R$.
Since $a(x)$ is integral over $R[x]$, we have an equation $a(x)^m+c_1(x)a(x)^{m-1}+cdots+c_m(x)=0$. Looking at the zeroth degree term, and since $R$ is integrally closed, we have $a_0in R$.
Thus if we can show that $a_1in R$ as well, the remaing $a_iin R$ by the same method.
Since both $a(x)$ and $a_0$ are integrally closed, their difference $a'(x):=a(x)-a_0=a_nx^n+cdots+a_1x$ is integrally closed. Then we have a new equation
begin{equation} label{new_eqn}
a'(x)^k+d_1(x)a'(x)^{k-1}+cdots+d_k(x)=0 , . tag{1}
end{equation}
Here, since $a'(x)$ has no constant term, it is deduced that $d_k(x)$ has no constant term, so a multiple of $x$. So, the equation (ref{new_eqn}) can be divided by $x$ and we get $a_1in R$ once more by looking at the constant terms and using the condition that $R$ is integrally closed.
There are somewhat complicated proofs (e.g. Trouble with proving $A$ is an integrally closed domain $Rightarrow$ $A[t]$ is integrally closed domain) for the same statement, but I don't know why the above proof is not given in any reference. Does this proof has some flaw?
abstract-algebra ring-theory commutative-algebra
edited Dec 15 '18 at 7:54
André 3000
12.7k22243
asked Dec 15 '18 at 6:05
user190964user190964
755
$endgroup$
add a comment |
$begingroup$
Let $R$ be an integrally closed domain. Then I want to show $R[x]$ is integrally closed.
Let $K$ be the field of fractions of $R$ and if I choose a function $a(x)in K(x)$ which is integral over $R[x]$, then it is also integral over $K[x]$. Since a UFD is integrally closed, we have $a(x)=a_nx^n+cdots+a_1x+a_0in K[x]$. So it is enough to show each $a_iin K$ is contained in $R$.
Since $a(x)$ is integral over $R[x]$, we have an equation $a(x)^m+c_1(x)a(x)^{m-1}+cdots+c_m(x)=0$. Looking at the zeroth degree term, and since $R$ is integrally closed, we have $a_0in R$.
Thus if we can show that $a_1in R$ as well, the remaing $a_iin R$ by the same method.
Since both $a(x)$ and $a_0$ are integrally closed, their difference $a'(x):=a(x)-a_0=a_nx^n+cdots+a_1x$ is integrally closed. Then we have a new equation
begin{equation} label{new_eqn}
a'(x)^k+d_1(x)a'(x)^{k-1}+cdots+d_k(x)=0 , . tag{1}
end{equation}
Here, since $a'(x)$ has no constant term, it is deduced that $d_k(x)$ has no constant term, so a multiple of $x$. So, the equation (ref{new_eqn}) can be divided by $x$ and we get $a_1in R$ once more by looking at the constant terms and using the condition that $R$ is integrally closed.
There are somewhat complicated proofs (e.g. Trouble with proving $A$ is an integrally closed domain $Rightarrow$ $A[t]$ is integrally closed domain) for the same statement, but I don't know why the above proof is not given in any reference. Does this proof has some flaw?
abstract-algebra ring-theory commutative-algebra
edited Dec 15 '18 at 7:54
André 3000
12.7k22243
asked Dec 15 '18 at 6:05
user190964user190964
755
$endgroup$
2
$begingroup$
You say "(1) can be divided by $x$". What happens then?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:23
$begingroup$
I just realized that after writing this article. The eq for $a_1$ is not monic, which involves a power of $x$ in the highest degree coefficient. Thanks.
$endgroup$
– user190964
Dec 15 '18 at 10:57
add a comment |
$begingroup$
Let $R$ be an integrally closed domain. Then I want to show $R[x]$ is integrally closed.
Let $K$ be the field of fractions of $R$ and if I choose a function $a(x)in K(x)$ which is integral over $R[x]$, then it is also integral over $K[x]$. Since a UFD is integrally closed, we have $a(x)=a_nx^n+cdots+a_1x+a_0in K[x]$. So it is enough to show each $a_iin K$ is contained in $R$.
Since $a(x)$ is integral over $R[x]$, we have an equation $a(x)^m+c_1(x)a(x)^{m-1}+cdots+c_m(x)=0$. Looking at the zeroth degree term, and since $R$ is integrally closed, we have $a_0in R$.
Thus if we can show that $a_1in R$ as well, the remaing $a_iin R$ by the same method.
Since both $a(x)$ and $a_0$ are integrally closed, their difference $a'(x):=a(x)-a_0=a_nx^n+cdots+a_1x$ is integrally closed. Then we have a new equation
begin{equation} label{new_eqn}
a'(x)^k+d_1(x)a'(x)^{k-1}+cdots+d_k(x)=0 , . tag{1}
end{equation}
Here, since $a'(x)$ has no constant term, it is deduced that $d_k(x)$ has no constant term, so a multiple of $x$. So, the equation (ref{new_eqn}) can be divided by $x$ and we get $a_1in R$ once more by looking at the constant terms and using the condition that $R$ is integrally closed.
There are somewhat complicated proofs (e.g. Trouble with proving $A$ is an integrally closed domain $Rightarrow$ $A[t]$ is integrally closed domain) for the same statement, but I don't know why the above proof is not given in any reference. Does this proof has some flaw?
abstract-algebra ring-theory commutative-algebra
edited Dec 15 '18 at 7:54
André 3000
12.7k22243
asked Dec 15 '18 at 6:05
user190964user190964
755
$endgroup$
Let $R$ be an integrally closed domain. Then I want to show $R[x]$ is integrally closed.
Let $K$ be the field of fractions of $R$ and if I choose a function $a(x)in K(x)$ which is integral over $R[x]$, then it is also integral over $K[x]$. Since a UFD is integrally closed, we have $a(x)=a_nx^n+cdots+a_1x+a_0in K[x]$. So it is enough to show each $a_iin K$ is contained in $R$.
Since $a(x)$ is integral over $R[x]$, we have an equation $a(x)^m+c_1(x)a(x)^{m-1}+cdots+c_m(x)=0$. Looking at the zeroth degree term, and since $R$ is integrally closed, we have $a_0in R$.
Thus if we can show that $a_1in R$ as well, the remaing $a_iin R$ by the same method.
Since both $a(x)$ and $a_0$ are integrally closed, their difference $a'(x):=a(x)-a_0=a_nx^n+cdots+a_1x$ is integrally closed. Then we have a new equation
begin{equation} label{new_eqn}
a'(x)^k+d_1(x)a'(x)^{k-1}+cdots+d_k(x)=0 , . tag{1}
end{equation}
Here, since $a'(x)$ has no constant term, it is deduced that $d_k(x)$ has no constant term, so a multiple of $x$. So, the equation (ref{new_eqn}) can be divided by $x$ and we get $a_1in R$ once more by looking at the constant terms and using the condition that $R$ is integrally closed.
There are somewhat complicated proofs (e.g. Trouble with proving $A$ is an integrally closed domain $Rightarrow$ $A[t]$ is integrally closed domain) for the same statement, but I don't know why the above proof is not given in any reference. Does this proof has some flaw?
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
edited Dec 15 '18 at 7:54
André 3000
12.7k22243
asked Dec 15 '18 at 6:05
user190964user190964
755
edited Dec 15 '18 at 7:54
André 3000
12.7k22243
asked Dec 15 '18 at 6:05
user190964user190964
755
edited Dec 15 '18 at 7:54
André 3000
12.7k22243
edited Dec 15 '18 at 7:54
André 3000
12.7k22243
edited Dec 15 '18 at 7:54
André 3000
12.7k22243
12.7k22243
asked Dec 15 '18 at 6:05
user190964user190964
755
asked Dec 15 '18 at 6:05
user190964user190964
755
asked Dec 15 '18 at 6:05
user190964user190964
755
755
2
$begingroup$
You say "(1) can be divided by $x$". What happens then?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:23
$begingroup$
I just realized that after writing this article. The eq for $a_1$ is not monic, which involves a power of $x$ in the highest degree coefficient. Thanks.
$endgroup$
– user190964
Dec 15 '18 at 10:57
add a comment |
2
$begingroup$
You say "(1) can be divided by $x$". What happens then?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:23
$begingroup$
I just realized that after writing this article. The eq for $a_1$ is not monic, which involves a power of $x$ in the highest degree coefficient. Thanks.
$endgroup$
– user190964
Dec 15 '18 at 10:57
2
2
$begingroup$
You say "(1) can be divided by $x$". What happens then?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:23
$begingroup$
You say "(1) can be divided by $x$". What happens then?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:23
$begingroup$
I just realized that after writing this article. The eq for $a_1$ is not monic, which involves a power of $x$ in the highest degree coefficient. Thanks.
$endgroup$
– user190964
Dec 15 '18 at 10:57
$begingroup$
I just realized that after writing this article. The eq for $a_1$ is not monic, which involves a power of $x$ in the highest degree coefficient. Thanks.
$endgroup$
– user190964
Dec 15 '18 at 10:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I like your approach. You just need to patch the whole in your proof by showing
If $f in K[x]$ is integral over $R[x]$ with lowest nonzero coefficient $f_0$, then $f_0$ is integral over $R$.
This is especially easy if you are familiar with the fact that for domains $R subseteq S$, $a in S$ is integral over $R$ iff there exists a finitely generated ideal $I subseteq R$ such that $aI subseteq I$.
From that characterization, we have $fI subseteq I$ for some f.g. $I subseteq R[x]$. Let $J subseteq R$ be the ideal generated by the lowest coefficients of the elements of $I$. It is clear that $J$ is finitely generated and that $f_0J subseteq J$, thus $f_0$ is integral over $R$.
answered Dec 16 '18 at 17:12
Badam BaplanBadam Baplan
4,611722
$endgroup$
add a comment |
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$begingroup$
I like your approach. You just need to patch the whole in your proof by showing
If $f in K[x]$ is integral over $R[x]$ with lowest nonzero coefficient $f_0$, then $f_0$ is integral over $R$.
This is especially easy if you are familiar with the fact that for domains $R subseteq S$, $a in S$ is integral over $R$ iff there exists a finitely generated ideal $I subseteq R$ such that $aI subseteq I$.
From that characterization, we have $fI subseteq I$ for some f.g. $I subseteq R[x]$. Let $J subseteq R$ be the ideal generated by the lowest coefficients of the elements of $I$. It is clear that $J$ is finitely generated and that $f_0J subseteq J$, thus $f_0$ is integral over $R$.
answered Dec 16 '18 at 17:12
Badam BaplanBadam Baplan
4,611722
$endgroup$
add a comment |
$begingroup$
I like your approach. You just need to patch the whole in your proof by showing
If $f in K[x]$ is integral over $R[x]$ with lowest nonzero coefficient $f_0$, then $f_0$ is integral over $R$.
This is especially easy if you are familiar with the fact that for domains $R subseteq S$, $a in S$ is integral over $R$ iff there exists a finitely generated ideal $I subseteq R$ such that $aI subseteq I$.
From that characterization, we have $fI subseteq I$ for some f.g. $I subseteq R[x]$. Let $J subseteq R$ be the ideal generated by the lowest coefficients of the elements of $I$. It is clear that $J$ is finitely generated and that $f_0J subseteq J$, thus $f_0$ is integral over $R$.
answered Dec 16 '18 at 17:12
Badam BaplanBadam Baplan
4,611722
$endgroup$
add a comment |
$begingroup$
I like your approach. You just need to patch the whole in your proof by showing
If $f in K[x]$ is integral over $R[x]$ with lowest nonzero coefficient $f_0$, then $f_0$ is integral over $R$.
This is especially easy if you are familiar with the fact that for domains $R subseteq S$, $a in S$ is integral over $R$ iff there exists a finitely generated ideal $I subseteq R$ such that $aI subseteq I$.
From that characterization, we have $fI subseteq I$ for some f.g. $I subseteq R[x]$. Let $J subseteq R$ be the ideal generated by the lowest coefficients of the elements of $I$. It is clear that $J$ is finitely generated and that $f_0J subseteq J$, thus $f_0$ is integral over $R$.
answered Dec 16 '18 at 17:12
Badam BaplanBadam Baplan
4,611722
$endgroup$
I like your approach. You just need to patch the whole in your proof by showing
If $f in K[x]$ is integral over $R[x]$ with lowest nonzero coefficient $f_0$, then $f_0$ is integral over $R$.
This is especially easy if you are familiar with the fact that for domains $R subseteq S$, $a in S$ is integral over $R$ iff there exists a finitely generated ideal $I subseteq R$ such that $aI subseteq I$.
From that characterization, we have $fI subseteq I$ for some f.g. $I subseteq R[x]$. Let $J subseteq R$ be the ideal generated by the lowest coefficients of the elements of $I$. It is clear that $J$ is finitely generated and that $f_0J subseteq J$, thus $f_0$ is integral over $R$.
answered Dec 16 '18 at 17:12
Badam BaplanBadam Baplan
4,611722
answered Dec 16 '18 at 17:12
Badam BaplanBadam Baplan
4,611722
answered Dec 16 '18 at 17:12
Badam BaplanBadam Baplan
4,611722
answered Dec 16 '18 at 17:12
Badam BaplanBadam Baplan
4,611722
4,611722
add a comment |
add a comment |
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2
$begingroup$
You say "(1) can be divided by $x$". What happens then?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:23
$begingroup$
I just realized that after writing this article. The eq for $a_1$ is not monic, which involves a power of $x$ in the highest degree coefficient. Thanks.
$endgroup$
– user190964
Dec 15 '18 at 10:57