# Tag Info

4

Find $u_i \in A$ such that $$\alpha^{-n} + u_{n-1} \alpha^{1-n} + \dotsc + u_1 \alpha^{-1} + u_0 = 0.$$ Multiplying with $\alpha^n$ yields $$1+ (u_{n-1} + \dotsc + u_0 \alpha^{n-1})\alpha = 1 + u_{n-1} \alpha + \dotsc + u_1 \alpha^{n-1} + u_0 \alpha^n = 0.$$ Now you see that $$\alpha^{-1} = -(u_{n-1} + \dotsc + u_0 \alpha^{n-1}) \in A.$$

4

If you want $\operatorname{Hom}_R(P,R)\simeq P$ for $P$ a finitely generated projective $R$-module, then forget it. $\bullet$ If $R$ is an integral domain, and $I\subset R$ is a non-zero ideal, then $\operatorname{Hom}_R(I,R)\simeq I^{-1}$, where $I^{-1}=\{x\in Q(R):xI\subseteq R\}$. (Here $Q(R)$ stands for the field of fractions of $R$.) $\bullet$ If ...

3

Let $R\subset\mathbb Z_2^{\mathbb N}$ be the set of sequences with elements in $\mathbb Z_2$ which are constant from a rank on. Then $R$ is a boolean ring. If $R\simeq\prod_{i\in I}R_i$ with $R_i$ indecomposable, then $R_i$ is isomorphic to a subring of $R$, hence $R_i$ is boolean. Moreover, since $R_i$ is indecomposable we must have $R_i\simeq\mathbb Z_2$. ...

3

Consider the inclusions $k[x^2,y^2]\subseteq k[x^2,xy,y^2]\subseteq k[x,y]$. The middle algebra is isomorphic to $k[u,v,w]/(uv-w^2)$, which has infinite global dimension because its localization "at the origin" is not a local regular ring.

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This answers the question before the last edit. In that case $I$ were the edge ideal of a non-complete bipartite graph. The only thing to prove is the following: $\operatorname{depth}R/I\ge\min(2,\dim R/I)$. It's obvious that $\dim R/I\ge 2$, so we have to prove $\operatorname{depth}R/I\ge2$. Since the graph $G$ is supposed non-complete the simplicial ...

2

We use that $k\subset k'$ is separable, a result of Grothendieck which says: If $k'$ and $K$ are extension fields of $k$ such that either $k'$ or $K$ is finitely generated over $k$ and if $k'$ is separable over $k$, then $k'\otimes_kK$ is regular. and the Theorem 33.2(i) from Matsumura, CRT, which asserts: If $A\to B$ is a faithfully flat ...

1

If $P$ is a finitely generated and projective $R$-module, then there is a canonical isomorphism $$\zeta:P^*\otimes_RM\simeq\operatorname{Hom}_R(P,M)$$ given by $\zeta(f\otimes m)(x)=f(x)m$. (For a proof see Proposition 6 from these notes.) Now, coming back to the question we have $\mathfrak a^*\otimes_RM\simeq\operatorname{Hom}_R(\mathfrak a,M)$. But ...

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To compute the radical of a binomial ideal, in some cases it can be advantageous (that is, faster) to use the Binomials package in Macaulay2. Here is the documentation. In Charles Boyd's answer you can just replace the call to radical by binomialRadical.

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First, note that there always exists a left-$add(X)$-approximation $\varphi:C\to X^{\widehat{m}}$ for any module $C$ in mod-$A$. This comes from the fact that $Hom_A(C,X)$ is a finitely generated module over the (implicit) base ring $k$; if $f_1, \ldots, f_r$ are generators, then the map given in matrix form by $(f_1, \ldots, f_r)^t$ from $C$ to $X^r$ is an ...

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The original sequence being exact at $X^n$ is equivalent to the map $\theta:C\to X^m$ induced by $\beta$ being injective. If $\varphi:C\to X'$ is any left $\text{add}(X)$-approximation, then $\theta$ factors through $\varphi$, and so $\varphi$ must also be injective, and the resulting sequence $0\to Y\to X^n\to X'$ is exact.

1

Let $Fn(\omega, 2)$ be the set of all finite partial functions from $\omega$ to $2$. For $s \in Fn(\omega, 2)$, let $[s] = \{x \in 2^{\omega} : s \subseteq x\}$. Then $\mathcal{B} = \{[s] : s \in Fn(\omega, 2)\}$ is a basis for the product topology $\mathcal{T}$ on $2^{\omega}$ where $2$ has discrete topology. Note that each set in $\mathcal{B}$ is also ...

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One can uses a parametrization of $x^2-y^2+y^3=0$: $x=t^3-t$, and $y=1-t^2$. This shows that $k[x,y]/(x^2-y^2+y^3)\simeq k[t^3-t,t^2-1]$. But the integral closure of $k[t^3-t,t^2-1]$ is $k[t]$, so you can conclude that the integral closure of $k[x,y]/(x^2-y^2+y^3)$ is isomorphic to $k[t]$ (or, if you like, equals $k[x/y]$).

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