New answers tagged commutative-algebra
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Annihilators of elements in the associated graded ring
Consider the element $u+I^{r+1}\in T$. Since $uas\in I^{r+1}$, we have $(as)(u+I^{r+1})=uas+I^{r+1}=0$, so $u+I^{r+1}\in \text{Ann}_T(as)=\text{Ann}_T(s)$. Thus, $s(u+I^{r+1})=0$, that is, $us\in I^{r+...
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Cardinal of the injective hull
Yes. It is enough to embed a countable abelian group $A$ into a countable divisible (i.e., injective) abelian group $D$, because such a $D$ will contain an isomorphic copy of the injective hull of $A$....
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kernel of subgroup homomorphism
(I include in this the additional assumptions mentioned in the comments.)
Prove both directions of containment: one direction is trivial since if $x\in \ker(g)$, then $g(x) = 0 \Rightarrow f(x) = 0$ ...
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If $B$ is a flat $A$-algebra, $M$ an $A$-module, and $x\neq 0\in M$, why does $Ax \cong A/\mathfrak a$ imply $B(1\otimes x) \cong B/\mathfrak{a}^e$
There is an exact sequence $$0\to\mathfrak{a}\to A\stackrel{x}\to M.$$ Since $B$ is flat, this exact sequence remains exact after tensoring with $B$, giving an exact sequence $$0\to\mathfrak{a}\otimes ...
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projective and injective ideals of $\mathbb Z/n\mathbb Z$ as $\mathbb Z/n\mathbb Z$ module
It's known that $\mathbb Z/n\mathbb Z$ is always a quasi-Frobenius ring, and since the injective modules coincide with the projective modules, solving one problem solves the other.
Considering that $R$...
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Is every ring homomorphism between real algebras also real-linear?
No, this is indeed not correct. For instance, there are two $\mathbb{R}$-algebra automorphisms of $\mathbb{C}$ (the identity and conjugation), but under the axiom of choice there are infinitely many ...
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What does it mean for a prime ideal to lie over another ideal in the case of a morphism of rings?
$\def\frp{\mathfrak{p}}
\def\frq{\mathfrak{q}}
\def\spm{\operatorname{Spm}}$Given an algebraically closed field $k$, by a “classical $k$-(pre)variety,” I will mean what Serre calls in FAC a “(pre)...
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If $A$ is a $K$-algebra and a field, and is contained in an affine $K$-domain, then $A$ is algebraic over $K$
Why are there any irreducible factors that happen to lie in $k[a_1]$?
It doesn't matter. There doesn't need to be any.
Why applying $\phi$ to $p^{−1}$ yields a diagonal matrix with all entries equal ...
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Ideal of definition of local ring
In order to prove that $R/I$ has dimension $0$ you need to prove that there is no prime ideal $\frak p\neq\frak m$ such that
$$
{\frak m}^n\subset I\subset\frak p\subset \frak m.
\qquad (\ast)
$$
That'...
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Rees algebras: why do we have this bijection on minimal primes?
I don't see how the text explicitly shows that $P$ minimal implies $PR[t,t^{-1}] \cap R[It]$ is a minimal prime, either. But maybe Huneke and Swanson expect the reader to do a little bit of work from ...
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If $B$ is an integrally closed domain and $B \to A$ is an integral extension, then $A$ is the integral closure of $B$ in $K(A)$ or not?
The claim is not true. Choose $A=\mathbb{Z}[\sqrt{5}]$ and $B=\mathbb{Z}$. $B$ is integrally closed, and the extension $B \subseteq A$ is finite, hence integral. However, in this case
$$C=\mathbb{Z}\...
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Where does it require $B$ an integrally closed domain (in a lemma used for proving the going down theorem)?
The norm $N_{L/K(B)}(a)$ is an element of $K(B)$ and integral over $B$, so you need $B$ to be integrally closed to have $N_{L/K(B)}(a)\in B$.
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Irredundant primary decomposition of radical from irredundant primary decomposition under certain conditions
Let's suppose that there exists $S\subsetneq \{1,\dots,r\}$ such that $\sqrt{\mathfrak{a}}=\bigcap_{i∈S}\mathfrak{p}_{i}$. Then, there exists $j\in\{1,\dots,r\}\setminus S$ such that $\mathfrak{p}_{j}\...
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Presentation of Witt vectors of polynomial rings
The main trick I know of for getting your hands on Witt vectors (of $\mathbb F_p$-algebras) is to reduce to the case of perfect $\mathbb F_p$-algebras, where we can appeal to the theorem that $W(-)$ ...
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Questions about the Cayley-Hamilton theorem for modules
I think I'm a bit late! Either way:
Yes, we require $\varphi(M) \subseteq IM$ to guarantee that $p_j \in I^j$ (this is particularly useful for commutative algebra when $I$ is prime). Note that $I = R$...
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Geometric interpretation of Minimal Prime Ideals
First, note that when one is looking for minimal primes over an ideal $I$, one can work with the radical $\sqrt{I}$ instead: $\sqrt{I}=\bigcap_{P\supset I, \text{ P prime}} P$ (ref), so if $P$ is a ...
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Commutative algebra: Irreducible components of $\operatorname{Spec}(A)$
The minimal primes of $A$ correspond to the minimal primes over $I$. If $P$ is a minimal prime containing $I$, then $X(Y+1)\in P$, so either $X\in P$ or $Y+1\in P$. If $X\in P$, then $I\subseteq (X)\...
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Freeness of $I/I^2$ implies that of $\dfrac{I}{I^2+xR}$ over $R/I$?
Counterexample: let $k$ be a field, $A=k[[t]], B=k[[t^3, t^7, t^8]], \mathfrak{m}=t^3B+t^7B+t^8B, I= t^6A\subset B, x=t^{11}.$
$(B,\mathfrak{m})$ is a noetherian local ring and $\sqrt{I}=\mathfrak{m}$ ...
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Tensor product over several modules
For each $p \in P$, the map
\begin{align}
M \times N &\to M \otimes N \otimes P \\
(m,n) & \mapsto m \otimes n \otimes p
\end{align}
is bilinear, and so there is a unique linear map
$$
f_p \...
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Fraction field of $\mathbb Z_p[[X]]$
$\newcommand{\Z}{\mathbb{Z}}$
Every element $f\in\mathbb{Z}_p[[x]]$ can be written as
$$
f(x) = p^e u(x) q(x)
$$
with $e \geq 0$, $u(x)\in \mathbb{Z}_p[[x]]^*$ and $q(x)\in \mathbb{Z}_p[x]$ monic.
...
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Tensor of ring of formal power series and a field
For the canonical map:
$$A=K[[t_1,\ldots,t_n]]\otimes_K L$$ is the subring of $B=L[[t_1,\ldots,t_n]]$ of formal power series whose coefficients lie in a finite dimensional $K$-vector subspace of $L$.
...
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Primary decomposition of ideal saturation
Let $I=\bigcap_{i=1}^n\mathfrak q_i$ be a primary decomposition of $I$ and suppose that $\newcommand{\qq}{\mathfrak q} \newcommand{\pp}{\mathfrak p}J\not\subseteq\sqrt{\qq_i}:=\pp_i$ for $1\leq i\leq ...
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Prove that $\mathbb{Z}/ 3\mathbb{Z}$ is a projective $\mathbb{Z} / 6\mathbb{Z}$ module which is not free.
In addition to Diego's answer, another approach is to use the fact that a module $P$ is projective $R$-module if it is a direct summand of a free $R$-module $F$ (i.e. $P$ is an $R$-submodule of $F$ ...
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Prove that $\mathbb{Z}/ 3\mathbb{Z}$ is a projective $\mathbb{Z} / 6\mathbb{Z}$ module which is not free.
The fact that $\mathbb{Z}/3\mathbb{Z}$ is not a free $\mathbb{Z}/6\mathbb{Z}$-module is obvious, as any finite free $\mathbb{Z}/6\mathbb{Z}$-module has cardinality a multiple of $6$.
Why is it ...
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Inverse image of maximal ideals under finite type ring maps.
The property you claimed above is equivalent to "$R$ is a Jacobson ring", i.e. we have
$R$ is Jacobson if and only if for any finite type ring map $R\to A$, inverse image of maximal ideals ...
- 1,958
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Primary decomposition and zero locus
Set $J=\cap I_p$ and fix an element $f\in J$. We consider the ideal quotient $I’=(I:f):=\{g\in R\mid fg\in I\}$ (which is an ideal too). Clearly $I\subset I’$ and $V(I)\supset V(I’)$. But the ...
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when the fractional ideal $S^{-1}I$ of the localization $S^{-1}A$ of a dedekind domain $A$ is principal
If $S$ is a multiplicative subset of $A$ then
some $A$-fractional ideal $I$ becomes principal in $ S^{-1}A$ iff for some $c\in Frac(A)$, $cI\subset A$ and $cI\cap S\ne \emptyset$.
This follows from $...
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Why is the identity condition needed in the definition of a direct system?
This is necessary because a directed set is just a preorder, not a partial order; i.e., it’s not required to be antisymmetric, as a partial order is.
If a directed set were defined as a partial order ...
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UFD implies GCD
As the previous answers have addresses your problem(and clearly the proof you posted is, admittedly, confusing), I just wish to provide a clean and easy-to-read proof to my taste. Our goal is to show ...
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If a polynomial, in $n\geq2$ variables, over an infinite field $k$, vanishes everywhere on $k^n$, is it the zero polynomial?
The answer is yes. We proceed by induction on $n$. For $n=1$, the number of roots of a non-zero polynomial is at most its degree and in particular finite. For general $p\in k[x_1,\dots,x_n],$ we write ...
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Residue field of DVR
Talking about DVRs immediately is the wrong way to solve this problem. The right way is to use the fact that you're dealing with a scheme of finite type over $k$:
Suppose $C$ is our curve and let $U\...
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Prove this is an exact sequence
If one of $a,b$ is 0, then the proof is easy.
So assume $a,b\not=0$.
The last map in the sequence should be
$$f: ao + bo \longrightarrow o/(ao:bo)$$
$$ as + br \mapsto r$$
To see $f$ is well-defined, ...
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Tensor product of integral $k$-algebras is integral domain
In fact, this is not true. The linked post talks about filtered colimits, but a tensor product is a colimit of a discrete diagram (and nontrivial discrete diagrams are not filtered).
Counterexample: ...
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Gröbner Basis for a sum of ideals
The observation in the last part of your question is correct. In general, given two polynomials $f$ and $g$ in some ideal $I$ of $k[x_1,\dots,x_n]$, it may happen that some combination $af+bg$ ...
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On the ideal of entries of a morphism between free modules
Yes, they are equivalent.
Write down the matrices representing the maps $f \otimes_R M : M^{\oplus a} \to M^{\oplus b}$ and $\operatorname{Hom}_R(f,M) : M^{\oplus b} \to M^{\oplus a}$. The former is ...
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Gathmann commutative algebra Exercise 1.13
There isn't a 1-1 correspondence between subvarieties and ideals because the ideal of a subvariety is radical. What you have shown above by looking at the zero loci is that the radicals of the two ...
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How to show $(k[x,y,z]/(xz,yz))_z\cong k[z]_z$?
$$\begin{align}\left(k[x,y,z]/(xz,yz)\right)_z&\cong k[x,y,z,t]/(xz,yz,1-zt)\\&\cong k[x,y,z,t]/(x,y,1-zt)\\&\cong k[z,t]/(x,y,1-zt)\\&\cong \left(k[z]\right)_z.\end{align}$$
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Simple Zorn Lemma application doubt.
Although the union of ideals is in general not an ideal, the union of a chain of ideals is always an ideal.
Let $C$ a chain of ideals and $C^*$ the union of the elements of $C$. We can easily check ...
- 8,138
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Fraction field of $\mathbb Z_p[[X]]$
I wanted to post this as a comment, but I don't have enough reputation. The accepted answer is wrong. It is not true that a nonzero element of $\mathbb{Z}_p[[X]]$ is of the form $X^np^m\sum_k b_kX^k$ ...
- 51
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A zero-dimensional ring is Noetherian?
I would like to point out another example which occurs quite often in nature: $\mathcal{O}_{\mathbb{C}_p}/p^n\mathcal{O}_{\mathbb{C}_p}$ for $n\geqslant 1$. Here $\mathcal{O}_{\mathbb{C}_p}$ is the ...
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Is the ring of regular functions of a simple complex connected linear algebraic group factorial?
The answer is no: a semi-simple algebraic group $G$ over a field $k$ has the property that $A_G:=\mathcal{O}_G(G)$ is a UFD if and only if $G$ is simply connected.
First observe that as $G$ is smooth ...
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UFD implies GCD
The PW argument is either incomplete (or incorrect) and it is impossible to decide which from what is written. Likely the author assumes the reader can complete it because they assume it is "...
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UFD implies GCD
Then the author claims $f$ must contain in its prime factorisation a irreducible element which does not divide $d$.
Why should this be true? For example $\mathbb{Z}$ is a UFD and 4 does not divide 6 ...
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Characterization of closed points $x$ of affine $k$-varieties with $[\kappa(x):k]=1$
Yes to the first question, no to the second. The point is that you want a $k$-algebra isomorphism $k[x_1, \ldots, x_n]/\mathfrak{m} = k$, i.e. your inclusion (1) is the identity. Given this, you get a ...
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Görtz-Wedhorn Lemma 3.20
For your first question, the $m_{ij}$, for a given $i$, generate the localization of $M$ at $f_i$, and $N$ contains all of the $m_{ij}$, so its localization at $f_i$ must contain $M_{f_i}$ since it ...
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Give an example about associated primes where two containments are proper.
We can go up one dimension. Take $R=\mathbb Z[x]$,
$M=R/(2x)$, $L=6M$ and $N=R/(6)=(\mathbb Z/6\mathbb Z)[x]$.
Then Ass$(M)=\{(2),(x)\}$, Ass$(L)=\{(x)\}$ and Ass$(N)=\{(2),(3)\}$.
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Show that the tensor $(x+2)\otimes( x+2) -x\otimes x -2\otimes 2\in \langle x, 2\rangle^{\otimes 2}$ isn't pure
As you are tensoring over $\mathbb Z[x]$, you can start with a projective presentation of your ideal
$$ \mathbb Z[x]\to\mathbb Z[x]^2\to(x,2)\to0.$$
The left hand map sends $r$ to $(2r,-xr)$, and the ...
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the intersection of all curves is not empty.
This is a consequence of the Hilbert Nullstellensatz.
One version of the Hilbert Nullstellensatz says that every maximal ideal of $\mathbb C[x, y]$ is of the form $(x - a, y - b)$, where $a, b \in \...
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Torsion and divisible module
I think I've found an answer to my question.
We construct a sequence $(x_i\in M)$ by: $x_0=0$ and $\pi x_i=x_{-1}$. Now define a map $K\rightarrow M$ by $\pi^{-i}a\mapsto a.x_i$, then the kernel is ...
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How to show the localization of a commutative ring is unique up to an unique isomorphism
Proving this only requires the concepts of category theory and does not depend on the specific structure of the objects. The proof also applies to demonstrating isomorphisms between free groups (or ...
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