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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 $W(-)$ ...
• 7,103
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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 $I = R$...
• 2,424
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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 ...
• 56.6k
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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 $I\subseteq (X)\... • 11.6k 1 vote ### Freeness of$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}$... • 514 1 vote ### Tensor product over several modules For each$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 $$f_p \... • 19.6k 0 votes ### Fraction field of \mathbb Z_p[[X]] \newcommand{\Z}{\mathbb{Z}} Every element f\in\mathbb{Z}_p[[x]] can be written as$$ f(x) = p^e u(x) q(x) $$with e \geq 0, u(x)\in \mathbb{Z}_p[[x]]^* and q(x)\in \mathbb{Z}_p[x] monic. ... • 2,475 2 votes ### Tensor of ring of formal power series and a field For the canonical map:$$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. ... • 75.8k 1 vote Accepted ### 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 ... • 11.6k 3 votes Accepted ### 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 ... • 26.9k 3 votes ### 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 ... 1 vote ### 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 1 vote Accepted ### 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 ... • 514 0 votes ### 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 ... • 75.8k 0 votes ### 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 ... • 226k 0 votes ### 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 ... • 690 1 vote Accepted ### 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 ... • 514 2 votes ### 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\... • 56.6k 1 vote Accepted ### 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$$To see f is well-defined, ... • 310 0 votes ### Tensor product of integral k-algebras is integral domain In fact, this is not true. The linked post talks about filtered colimits, but a tensor product is a colimit of a discrete diagram (and nontrivial discrete diagrams are not filtered). Counterexample: ... • 12.1k 0 votes ### Gröbner Basis for a sum of ideals The observation in the last part of your question is correct. In general, given two polynomials f and g in some ideal I of k[x_1,\dots,x_n], it may happen that some combination af+bg ... 2 votes Accepted ### On the ideal of entries of a morphism between free modules Yes, they are equivalent. Write down the matrices representing the maps f \otimes_R M : M^{\oplus a} \to M^{\oplus b} and \operatorname{Hom}_R(f,M) : M^{\oplus b} \to M^{\oplus a}. The former is ... • 12.1k 1 vote Accepted ### Gathmann commutative algebra Exercise 1.13 There isn't a 1-1 correspondence between subvarieties and ideals because the ideal of a subvariety is radical. What you have shown above by looking at the zero loci is that the radicals of the two ... • 992 1 vote Accepted ### How to show (k[x,y,z]/(xz,yz))_z\cong k[z]_z?$$\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}$$• 18.4k 4 votes Accepted ### 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 3 votes ### 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 1 vote ### 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 ... • 51.3k 2 votes Accepted ### 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 ... • 51.3k 0 votes ### 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 "... • 263k 4 votes ### 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 ... • 146k 1 vote Accepted ### 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 ... • 338 0 votes Accepted ### 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 ... • 121 1 vote Accepted ### 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)\}. • 4,491 0 votes ### 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)$, and the ... • 4,491 1 vote Accepted ### the intersection of all curves is not empty. This is a consequence of the Hilbert Nullstellensatz. One version of the Hilbert Nullstellensatz says that every maximal ideal of$\mathbb C[x, y]$is of the form$(x - a, y - b)$, where$a, b \in \...
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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 ...