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integrally

integrally, adv. (ˈɪntɪgrəlɪ) [f. as prec. + -ly2. Cf. med.L. integrāliter entirely, wholly.] a. In an integral manner; as a whole, in its entirety; completely, entirely, wholly.1471 Ripley Comp. Alch. ii. v. in Ashm. (1652) 136 When the Erth ys integrally yncynerat. 1649 Jer. Taylor Gt. Exemp. ii. ... Oxford English Dictionary

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Integrally closed

ring is said to be integrally closed in if is equal to the integral closure of in . An integral domain is said to be integrally closed if it is equal to its integral closure in its field of fractions. wikipedia.org

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integrally

integrally/-grəlɪ; -ɡrəlɪ/ adv. 牛津英汉双解词典

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Integrally convex set

Therefore X is not integrally convex. In contrast, the set Y = { (0,0), (1,0), (2,0), (1,1), (2,1) } is integrally convex. Properties Iimura, Murota and Tamura have shown the following property of integrally convex set. Let be a finite integrally convex set. wikipedia.org

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Integrally closed domain

A completely integrally closed domain is integrally closed. Conversely, a noetherian integrally closed domain is completely integrally closed. A direct limit of integrally closed domains is an integrally closed domain. wikipedia.org

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Integrally Closed implies reduced Suppose $A$ is a commutative ring with unit and that $A$ is integrally closed in $A[x]$. Show that $A$ is reduced?

Thus $p(x)$ is an integral element of $A[x]\setminus A$, which implies $A$ is not integrally closed in $A[x]$, which is what we wanted.

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Geometric meaning of being integrally closed in some overring The geometric counterpart of integrally closed rings (in their fraction fields) are normal varieties, as described in this MathOverflow post. Is their a s...

So the analogue of "$B$ being integrally closed in $A$" would be "the normalization of $Y$ in $X$ is $Y$", in other words, $Y = Y'$.

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An integral domain that is Noetherian, integrally closed, but not one-dimensional A Dededind domain is defined as an integral domain which is integrally closed, one-dimensional, and Noetherian. Also I know an equival...

This is Noetherian and integrally closed but $2$-dimensional.

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Overring of an integrally closed domain that is not integrally closed > Assume that $A$ is an integrally closed integral domain, and $K$ is its fraction field. Is every overring of $A$ between $A$ and $K$ also integra...

To show that $A$ is integrally closed notice that $A\subset K[X,Y]$, and the last ring is integrally closed. Clearly $R$ is not integrally closed since $X$ is integral over $R$ and $X\notin R$.

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$R_1\subsetneq R_2$ integral domains, $R_1$ integrally closed. Is $R_2$ integrally closed? > Let $R_1\subset R_2$ be integral domains (i.e., commutative rings with unit without zero divisors). > > Suppose $R_1, R_2$ ...

Here is a counterexample: Let $S_{1},S_{2} \subset \mathbb{Z}^{\oplus 2}$ be the submonoids $S_{1} := \mathbb{N} \cdot (0,1) + \mathbb{N} \cdot (1,0)$ and $S_{2} := \\{\mathbb{N} \cdot (-1,1) + \mathbb{N} \cdot (1,0)\\} \setminus \\{(-1,1)\\}$ and let $R_{i} := \mathbb{Z}[S_{i}]$ be the monoid algeb...

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When is an integrally closed local domain a valuation ring? Every valuation ring is an integrally closed local domain, and the integral closure of a local ring is the intersection of all valuation rings containing it....

Now the following is true: the local integrally closed domain $R$ is a valuation domain if and only if $R$ is coherent and there exist $r,s\in R$, $s\not

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Ring of entire functions is integrally closed or not? > Is the ring $\mathscr{O}(\mathbf{C})$ of entire functions integrally closed (in its field of fractions, the meromorphic functions)? I know it's not factorial, ...

Yes, $\mathscr{O}(\mathbb{C})$ is a GCD domain, and as such it is integrally closed.

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$2$-dimensional Noetherian integrally closed domains are Cohen-Macaulay Any 1-dimensional Noetherian domain is Cohen-Macaulay (C-M). > For the $2$-dimensional case, a condition of being integrally closed is necessar...

If $R$ is integrally closed, then $R$ is Cohen-Macaulay. From Serre's normality criterion we have that $R$ satisfies $(R_1)$ and $(S_2)$. $k[x^4,x^3y,xy^3,y^4]$ is $2$-dimensional, _not_ Cohen-Macaulay and _not_ integrally closed.

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Prove that GCD domain is integrally closed In the case $R$ is a UFD, the proof of that $R$ is integrally closed essentially uses a key fact: If gcd($a,b)=1$, then gcd$(a^n,b)=1$. The proof of this relies on UFD prop...

$\color{#0a0}{(a,b)=1}\,=\,(c,\,b)\,\Rightarrow\,\color{#c00}1 = (a,b)(c,b) = (ac,ab,bb,bc)= (ac,\color{#0a0}{(a,b},c)b) = \color{#c00}{(ac,b)},\ $ therefore $\color{}{(a,b)=1}=(a^n,b)\,\Rightarrow\,\color{#c00}{1=(a^{n+1},b)}\ $ yields the induction step.

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Finite extension of integrally closed ring again integrally closed Let $S\subset R$ be a finite ring extension, i.e. $R$ is finitely generated as an $S$-module. Assume that $S$ is integrally closed. Does this imply th...

Then $R$ is f.g. over $S$ (in fact free of rank $2$) but now $R$ is not integrally closed: The quotient field of $R$ is $\Bbb{Q}(\sqrt{-3})$ and $ \frac

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integrally

Answers

MindMap

Sources

integrally

Integrally closed

integrally

Integrally convex set

Integrally closed domain

Integrally Closed implies reduced Suppose $A$ is a commutative ring with unit and that $A$ is integrally closed in $A[x]$. Show that $A$ is reduced?

Geometric meaning of being integrally closed in some overring The geometric counterpart of integrally closed rings (in their fraction fields) are normal varieties, as described in this MathOverflow post. Is their a s...

An integral domain that is Noetherian, integrally closed, but not one-dimensional A Dededind domain is defined as an integral domain which is integrally closed, one-dimensional, and Noetherian. Also I know an equival...

Overring of an integrally closed domain that is not integrally closed > Assume that $A$ is an integrally closed integral domain, and $K$ is its fraction field. Is every overring of $A$ between $A$ and $K$ also integra...

$R_1\subsetneq R_2$ integral domains, $R_1$ integrally closed. Is $R_2$ integrally closed? > Let $R_1\subset R_2$ be integral domains (i.e., commutative rings with unit without zero divisors). > > Suppose $R_1, R_2$ ...

When is an integrally closed local domain a valuation ring? Every valuation ring is an integrally closed local domain, and the integral closure of a local ring is the intersection of all valuation rings containing it....

Ring of entire functions is integrally closed or not? > Is the ring $\mathscr{O}(\mathbf{C})$ of entire functions integrally closed (in its field of fractions, the meromorphic functions)? I know it's not factorial, ...

$2$-dimensional Noetherian integrally closed domains are Cohen-Macaulay Any 1-dimensional Noetherian domain is Cohen-Macaulay (C-M). > For the $2$-dimensional case, a condition of being integrally closed is necessar...

Prove that GCD domain is integrally closed In the case $R$ is a UFD, the proof of that $R$ is integrally closed essentially uses a key fact: If gcd($a,b)=1$, then gcd$(a^n,b)=1$. The proof of this relies on UFD prop...

Finite extension of integrally closed ring again integrally closed Let $S\subset R$ be a finite ring extension, i.e. $R$ is finitely generated as an $S$-module. Assume that $S$ is integrally closed. Does this imply th...