Questions tagged [integral-extensions]

This tag is for questions regarding to the Integral Extensions or, Integral Ring Extensions.

Let $~A ⊆ B~$ be a ring extension. Generalizing to rings the notion of algebraic elements and extensions from field theory, we say that an element $~b~$ in $~B~$ is integral over $~A~$ if there exists a monic polynomial $~f(x)~$ in the variable $~x~$ with coefficients in $~A~$ such that $~f(b) = 0~$; we say that the extension is integral if every element of $~B~$ is integral over $~A~$.

  • A domain that is an integral extension of a field must be a field

Reference:

https://en.wikipedia.org/wiki/Integral_element#Integral_extensions

"Integral Extensions" by Siegfried Bosch

179 questions
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If a polynomial in $B[x]$ is integral over $A[x]$, then are its coefficients integral over $A$?

Suppose $A \subset B$ are integral domains, and $f = b_0 + b_1 x + \ldots + b_m x^m$, where $b_k \in B$. Is it true that if $f$ is integral over $A[x]$ then all the coefficients $b_k$ are integral over $A$? Note that the other way is obvious, i.e.…
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How can I compute the discriminant of the field $\mathbb{Q}(\sqrt[3]{28})$?

$$\newcommand{\Q}{\mathbb{Q}} \newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\al}{\alpha} \newcommand{\bcal}{\mathcal{B}} \newcommand{\qroot}{\sqrt[3]} \newcommand{\froot}{\sqrt[4]} $$ I have a problem which consist in 3…
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Integral domain over which every non-constant irreducible polynomial has degree 1

Let $R$ be an integral domain such that any polynomial $f(X) \in R[X]$ , which is irreducible in $R[X]$, has degree $1$. Then is it true that $R$ is a field ? If this is not true in general , What if we also assume that $R$ is Noetherian (and…
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2 answers

Non-flatness of $k[t]$ as a $k[t^2,t^3]$-module

Let $k$ be a field. How to show that $k[t]$ is not flat as a module over $k[t^2,t^3]$ ? Since the ring extension $k[t^2,t^3]\subseteq k[t]$ is integral, it is clear that $k[t]$ is a finitely generated $k[t^2,t^3]$-module , and also torsion free. I…
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Why does an integral extension over a ring has the same Krull dimension as the ring?

The Krull dimension of a ring R is defined as the supremum of the lengths of chains of prime ideals contained in R. I heard that an integral extension over a ring R has the same Krull dimension as R, however, I don't really see why this is true.
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$R$ is integrally closed in $S$ iff $R[x]$ is integrally closed in $S[x]$

The title is the question. The parts in bold below are where I'm stuck. This is what I've tried (much of this is just me explaining a hint I received): This comes from Eisenbud 3.17. The hint in the back of the book states: First reduce to the case…
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Integral closure and $\bigcap \mathfrak{a}^n$

Let $R$ be a domain such that $\bigcap_{n=1}^\infty \mathfrak{a}^n=0$ holds for all proper ideals $\mathfrak{a}$ of $R$ (this holds, for example, if $R$ is Noetherian). Let $K$ be the quotient field of $R$, and let $\overline{R}$ be the integral…
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4 answers

For a recursion based on $x+y+z+2\sqrt{xy+yz+zx},$ does what happens in $\mathbb Z\left[\sqrt n\right]$ stay in $\mathbb Z\left[\sqrt n\right]$?

This problem has a geometric origin which I'll outline below, but I believe the concepts and explanation are algebraic. Given a function on triples $$K((x,y,z))=x+y+z+2\sqrt{xy+yz+zx}$$ we build a general recursion as follows. Start with an integer…
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Infinite Noetherian ring of dimension $1$ in which distinct non-zero ideals have distinct and finite index

Let $R$ be an infinite commutative ring with unity such that every non-zero ideal has finite index. Then $R$ is Noetherian, every non-zero prime ideal is maximal , and I can also show that $R$ is an integral domain. Now also assume that distinct…
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Terminology of "algebraically closed rings"

There are various approaches how to generalize the notion of an algebraically closed field to the context of commutative rings. A good survey is R. Raphael, On algebraic closures. I am interested in the following notion of an "algebraically closed"…
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In an integral extension $A\subseteq B$, when $B$ Noetherian implies $A$ Noetherian?

Let $A\subseteq B$ be an extension of commutative rings. If $B$ is a Noetherian ring and finitely generated as $A$-module, then $A$ is Noetherian ring. Is there any other such criteria, under possibly some additional conditions, which says that in…
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Is there an integral extension of $\mathbb{Z}$ keeping its prime elements?

Let $R$ be a ring. An extension $S$ of $R$ is called prime-keeping if every element $p$ prime in $R$ is also prime in $S$. Consider the ring $\mathbb{Z}$ and the following two extensions: $\mathbb{Z}[X]$ and $\mathbb{Z}[i]$. The first one is…
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If a finite group acts on an integral domain which is integrally closed, then the fixed point subring is also integrally closed

Let $G$ be a finite group acting on an integral domain $R$, and let $S$ denote the fixed point subring: $$ S=\{x\in R: gx=x \text{ for all }g\in G\}$$ I am asked to show that $R$ is integral over $S$, and that if $R$ is integrally closed then so is…
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Is $\mathbb{Z}[x,y]/(x^3+px+q - y^2)$ integral over $\mathbb{Z}[x]$?

Let $ B = \mathbb{Z}[x,y]/(x^3+px+q - y^2)$ and $A = \mathbb{Z}[x]$. I want to know whether the ring extension $A \subset B$ is integral or not. There can be two possibilities: $x^3+px+q$ is irreducible or is not irreducible. It's enough to…
5
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1 answer

Contraction of (non-prime) ideals in integral extensions

If $A \subset B$ is an integral extension, then any prime $p \subset A$ is the contraction of some prime of $B$ (by lying-over property). Does this hold for more general ideals? That is, given an ideal $I \subset A$, is there always an ideal $J…
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