Tag Info

Hot answers tagged

6

Let $\alpha$ be an algebraic integer with minimal polynomial $X^n + a_1 X^{n-1} + \dots + a_{n-1}X+ a_n$. We have that $\alpha^{-1}$ is a root of $a_nX^{n} + a_{n-1}X^{n-1}+ \dots a_1 X+ 1 $ and thus of (of cours $a_n$ is nonzero by the irreducibility of the above polynomial) $$X^{n} + \frac{a_{n-1}}{a_n}X^{n-1}+ \dots \frac{a_1}{a_n} X+ \frac{1}{a_n}.$$ ...


5

If $\alpha \in \overline{\mathbb{Z}}$ $($this is a pretty bad abuse of notation$)$, then $1/\alpha$ is a root of a monic polynomial if and only if the minimum polynomial of $\alpha$ has constant term $\pm 1$. This is because $1/\alpha$'s minimum polynomial must be $\alpha$'s minimum polynomial with its coefficients reversed $($obviously $1/\alpha$ is a root ...


4

$I$ is a prime ideal since clearly $k[x,y,z]/(z-1,x^2-y) = k[x]$.


3

Let $A$ be a commutative ring. Let $M$ be a finitely generated $A$-module and $N$ be an $A$-submodule of $M$. Let $f\colon N \rightarrow M$ be a surjective homomorphism of $A$-modules Then $f$ is injective. Proof. Let $0 \neq x'_0 \in N$. It suffices to prove $f(x'_0) \neq 0$. Set $f(x'_0) = x_0$. Let $x_1, \dots, x_n$ be generators for $M$. Then ...


3

user26857's answer shows how to repair the reduction to the Noetherian case. Here is how to repair the proof of the Noetherian case: Let $M$ be a noetherian $A$-module and let $N \subseteq M$ be a submodule. Let $f : N \to M$ be a surjective linear map. Then $f$ is injective. Proof: Let $n \geq 0$. Although $f^n$ is not a well-defined homomorphism this ...


3

Ideals are not generalized numbers It is true that ideals were introduced by Dedekind as "ideal numbers" in order to restore unique factorization in certain rings where it fails. These rings are now called Dedekind rings and indeed it is a fruitful point of view to treat ideals in these rings as generalized numbers. However these rings have Krull dimension ...


3

The answer is negative since $A\subset B$ flat and $B$ regular implies $A$ regular; see Bruns and Herzog, Theorem 2.2.12. But in this case $A\simeq k[a,b,c]/(ac-b^2)$, so $A$ is not regular. Edit. A simpler approach: let $I=(x^2,xy)$ and $A/I\to A/I$ be the multiplication by $y^2$. Since $A/I\simeq k[y^2]$ this is injective, but on $A/I\otimes_AB\to ...


3

You can proceed as usual by starting with $F\stackrel{f}\to A\to 0$ and $H\stackrel{h}\to C\to 0$, where $F$ and $H$ are free of finite rank. Then show that there is an exact sequence $G=F\oplus H\stackrel{g}\to B\to 0$. Now consider $F'=\ker f$ and so on. You have a short exact sequence $0\to F'\to G'\to H'\to 0$. Now use the result for finitely generated ...


3

The localization of a ring $A$ at a multiplicative submonoid $S$ is the initial ring under $A$ in which $S$ is sent to units. (i.e. there's a localization map $A\to S^{-1}A$ and if $S$ goes to units in $f:A\to B$ then $f$ factors uniquely through $A\to S^{-1}A$.) Diagrammatically, a unit is just an element whose action by multiplication is an isomorphism. A ...


2

Consider the ideals $(x)$ and $(xy)$ in the ring $k[x,y]$ and its localization $k[x,y]_{(x)}$.


2

There is an isomorphism $\mathbf{C} \cong \mathbf{C}_p$ -- i.e. you can extend the p-adic absolute value to the complexes. $\mathbf{C}_p$ is the field of fractions of its ring of integers: the subring of numbers whose $p$-adic absolute value is less than or equal to $1$.


2

The sum of ideals, at least, has a categorical interpretation: It corresponds to the pushout of the epimorphisms that the two ideals being added corespond to. You have a ring $R$ with ideals $\mathfrak I$ and $\mathfrak J$ and canonical homomorphisms $\pi_1:R\to R/\mathfrak I$ and $\pi_2:R\to R/\mathfrak J$. Then whenever you have a homomorphism $f:R\to S$ ...


2

Units are always non zero-divisors. If $x$ is a unit, then $1=xy$ for some $y\in R$. Now, if $rx=0$, then $r=r(xy)=(rx)y=0$. So $x$ is not a zero-divisor.


2

For Dedekind domains we have $\operatorname{Pic}(R)\simeq\operatorname{Cl}(R)$, where $\operatorname{Cl}(R)$ is the ideal class group of $R$. Your case is treated here in detail.


2

I'd use Bruns and Herzog, Cohen-Macaulay Rings, Theorem 4.4.3(b) noticing that the only non-zero local cohomology module occurs for $d=\dim M$. Maybe a simpler approach (at least without local cohomology). One knows that $\max\{i\in\mathbb{Z}\colon P_{M}(i)\neq H(M,i)\}=\deg H_M(t)$ (see Bruns and Herzog, Cohen-Macaulay Rings, Exercise 4.4.10). We can ...


2

This is essentially Mumford's treasure map, or whatever it's called. Let $X=\text{Spec}(\mathcal{O}_K[x])$, $Y=\text{Spec}(\mathcal{O}_K)$, and $X\to Y$ be the obvious map. We're trying to determine $X_y$, for each $y\in Y$. Now, as you've observed, $X_{(0)}=\text{Spec}(K[x])$, whose prime ideals are those of the form $(f(x))$, for $f(x)\in K[x]$ ...


2

If $R=I+J$ then for $a_i\in R$ we get $x_i\in I$ and $y_i\in J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1\bmod I,a_2\bmod J)=\phi(y_1+x_2)$. Conversely, if $\phi$ is surjective and $a\in R$ then $(a\bmod I,0\bmod J)=\phi(x)$, and thus $a=(a-x)+x\in I+J$.


2

Since $R/J\otimes_R A=0$, the second exact sequence implies that the inclusion $$J\otimes_R A\to R\otimes_R A\cong A$$ is surjective. $J\otimes_R A$ is generated by elements of the form $j\otimes a$ for some $a\in A$ and $j\in J$. The homomorphism $J\otimes_R A\to A\otimes_R R\cong A$ sends $j\otimes a\mapsto j\otimes a=ja$. The image is thus $JA$ (as $JA$ ...


2

Incomplete thoughts: Thinking of the elements as written in an $n \times m$ grid, send it to the element $\alpha = \bigotimes_{j=1}^m (v_{1j} \wedge \cdots \wedge v_{nj}) \otimes \bigotimes_{i=1}^n (w_{i1} \wedge \cdots \wedge w_{im})$. (so, putting the $v$'s together in columns and the $w$'s in rows.) To see that this is well-defined, it suffices to ...


2

a) $4=\frac{2}{1}\cdot\frac21$ and similarly for $10$. On the other side, $5$ is irreducible, $6=\frac21\cdot\frac31$ and $\frac31$ is invertible in $P$, so $6$ is also irreducible, and similarly for $15$. It remains $9=3^2$ which is invertible in $P$. b) $P$ is a UFD as a ring of fractions of a UFD; see here. We also have that $P[x]$ is a UFD; see here.


1

Basically you have a ring $R$, a maximal ideal $\mathfrak p$ and $\mathfrak i\subset\mathfrak p$ an ideal. Then in $R/\mathfrak i$ you consider the maximal ideal $\mathfrak p/\mathfrak i$. The localization $(R/\mathfrak i)_{\mathfrak p/\mathfrak i}$ is isomorphic to $R_{\mathfrak p}/\mathfrak iR_{\mathfrak p}$ and $$\displaystyle\frac{\mathfrak pR_{\mathfrak ...


1


1

Suppose that $E$ is a finitely generated $D$-module. Say $E=De_1+\cdots+De_n$. Let's show that the field extension $k\subset K$ is finite. It is clearly algebraic (why?) and of finite type: $K=k(e_1,\dots,e_n)$ (why?).


1

Hints (depending on what you mean, and see for example here) (a) $\mathbb{Z}[i]/(i-5) \cong \mathbb{Z}[x]/(x^2+1, x-5) \cong \mathbb{Z}/26\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/13\mathbb{Z}$ (b) $\mathbb{Z}[i]/(3,i+5) \cong \mathbb{Z}[x]/(x^2+1, x+5,3) \cong \mathbb{Z}/(26,3)\cong 0$ (a) $\mathbb{Z}[i][x]/(x-5) \cong \mathbb{Z}[i] $ (b) ...


1

This can be seen quickly if one uses certain results on the associated primes. For example: answer 1: It is a fact that $\operatorname{Ass}(M) \subset \operatorname{Supp}(M)$ and the minimal elements of $\operatorname{Ass}(M)$ coincide with the minimal elements of $\operatorname{Supp}(M)$. So if $P \in \operatorname{Supp}(M)$ and if $P \not\in ...


1

Here is a quick and easy proof which I have found at the stacks project: If $M$ is cyclic, then $M \cong A/I$ for some ideal $I$. Then replace $A$ by $A/I$ so that wlog $M = A$. But then it's easy. In general, we do induction, but first we do the trick mentioned in Georges Elencwajg's answer: We endow $M$ with an $A[X]$-module structure such that ...


1

Depends what you mean by "interesting". The smallest examples of rings with $\mathbb{C}$ as their field of fractions would be $$R[\{t_\alpha\}_{\alpha\in A}]$$ where $R$ denotes the ring of algebraic integers, and $\{t_\alpha\}_{\alpha\in A}$ is a (edit: pure) transcendence basis for the extension $\mathbb{C}/F$, where $F$ is the field of algberaic numbers ...


1

$\mathbb Z\subset\overline{\mathbb Z}$ is an integral ring extension. By Cohen-Seidenberg Theorem we have $\dim\mathbb Z=\dim\overline{\mathbb Z}$. Since $\dim\mathbb Z=1$ you get exactly what you want.


1

Because of Hartshorne's (not so great ...) book, the word "coherent" is often not used in its proper meaning. Coherent modules are defined as finitely generated modules whose finitely generated submodules are finitely presented. Thus, coherent modules are finitely presented and finitely presented modules are finitely generated, but over a non-noetherian base ...


1

Any ring can be viewed as an additive category with one object and the endomorphisms of that object given by the elements of the ring. Then Gabriel-Zisman localization applied to this special case recovers the usual localization of a ring at a multiplicative subset of elements. This makes sense even when the ring is not commutative, and then one recovers ...



Only top voted, non community-wiki answers of a minimum length are eligible