Question

Integral extension of polynomial rings.

score 927 · Answer 1 · 2026-02-23 09:46:50

927

Views

Integral extension of polynomial rings.

Published on 23 Feb 2026 - 9:46

score 322 · Answer 2 · 2026-02-23 09:49:32

322

Views

Finite extension of domains : preimage of non-zero ideal is non-zero?

Published on 23 Feb 2026 - 9:49

score 791 · Answer 3 · 2026-02-23 09:51:32

791

Views

Contraction of extension of prime ideals in an integral ring extension

Published on 23 Feb 2026 - 9:51

score 251 · Answer 4 · 2026-02-23 09:46:33

251

Views

About localization of integral closure at maximal ideals.

Published on 23 Feb 2026 - 9:46

score 346 · Answer 5 · 2026-02-23 09:43:52

346

Views

Integral closure and algebraic closure

Published on 23 Feb 2026 - 9:43

score 257 · Answer 6 · 2026-02-23 09:47:42

257

Views

Normality of $\mathbb{C}[x, y]/(y^2-x^3+x)$

Published on 23 Feb 2026 - 9:47

score 50 · Answer 7 · 2026-02-23 09:50:34

50

Views

$S$ is a subring of an integral domain $R$ which is finitely generated algebra over $S$, Jac(S)=0 implies Jac(R)=0

Published on 23 Feb 2026 - 9:50

score 2.1k · Answer 8 · 2026-02-23 09:51:32

2.1k

Views

$R$ is integrally closed in $S$ iff $R[x]$ is integrally closed in $S[x]$

Published on 23 Feb 2026 - 9:51

score 622 · Answer 9 · 2026-02-23 09:43:23

622

Views

Checking whether a given element is integral over $\mathbb{Z}$.

Published on 23 Feb 2026 - 9:43

score 406 · Answer 10 · 2026-02-23 09:49:25

406

Views

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

Published on 23 Feb 2026 - 9:49

score 141 · Answer 11 · 2026-02-23 09:49:14

141

Views

Show $\mathbb{Z}\left[i,\frac{1+\sqrt{5}}{2}\right]$ is integrally closed

Published on 23 Feb 2026 - 9:49

score 59 · Answer 12 · 2026-02-23 09:49:25

59

Views

Understanding the proof of the existence of absolute integral closure

Published on 23 Feb 2026 - 9:49

score 69 · Answer 13 · 2026-02-23 09:50:50

69

Views

$1$-dimensional reduced local ring $R$ such that $(R:_{Q(R)} \overline R)$ is non-zero and consists of zero-divisors

Published on 23 Feb 2026 - 9:50

score 117 · Answer 14 · 2026-02-23 09:51:32

117

Views

$C$ is integral closure of $A$ then $C[x]$ is of $A[x]$

Published on 23 Feb 2026 - 9:51

score 49 · Answer 15 · 2026-02-23 09:51:10

49

Views

Show $\left(F\left[X_{1}, \ldots, X_{n}\right]\right)^{A_{n}}$ is an integral extension of $\left(F\left[X_{1}, \ldots, X_{n}\right]\right)^{S_{n}}$

Published on 23 Feb 2026 - 9:51

List Question

Integral extension of polynomial rings.

Finite extension of domains : preimage of non-zero ideal is non-zero?

Contraction of extension of prime ideals in an integral ring extension

About localization of integral closure at maximal ideals.

Integral closure and algebraic closure

Normality of $\mathbb{C}[x, y]/(y^2-x^3+x)$

$S$ is a subring of an integral domain $R$ which is finitely generated algebra over $S$, Jac(S)=0 implies Jac(R)=0

$R$ is integrally closed in $S$ iff $R[x]$ is integrally closed in $S[x]$

Checking whether a given element is integral over $\mathbb{Z}$.

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

Show $\mathbb{Z}\left[i,\frac{1+\sqrt{5}}{2}\right]$ is integrally closed

Understanding the proof of the existence of absolute integral closure

$1$-dimensional reduced local ring $R$ such that $(R:_{Q(R)} \overline R)$ is non-zero and consists of zero-divisors

$C$ is integral closure of $A$ then $C[x]$ is of $A[x]$

Show $\left(F\left[X_{1}, \ldots, X_{n}\right]\right)^{A_{n}}$ is an integral extension of $\left(F\left[X_{1}, \ldots, X_{n}\right]\right)^{S_{n}}$

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