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6

Let $A = k\left[x\right]/\left(x^2\right)$, and let us denote the projection of $x \in k\left[x\right]$ onto $A$ by $x$ (by abuse of notation). I assume that your $\left(x\right)$ means the $A$-submodule of $A$ spanned by $x$. And your question is: Is $\left(x\right) \otimes_A \left(x\right) = 0$ ? I claim that the answer is "No". Indeed, there is an ...


6

$x\in F[t,t^{-1}]$ is a power of $t$ iff $x$ is invertible (or, equivalently $x\mid 1$) and $t-1\mid x-1$. If $x=t^n$, then $xt^{-n}=1$, and as you already noticed $t-1\mid t^n-1$. For the converse write $x=f(t)/t^n$ with $n\ge0$ and $f(t)\in F[t]$. Since $x$ is invertible there exists $y\in F[t,t^{-1}]$ such that $xy=1$. Write $y=g(t)/t^m$ with ...


6

Let $R$ be an integral domain that is not a field. Then $R$ contains a nonzero element that is not invertible. For every nonzero element $r$ that is not invertible, $r$ is not a multiple of $r^2$: if it were, then $r=r^2s$ for some $s\in R$. Since $R$ is an integral domain, $1=rs$, contradicting that $r$ is not invertible. Now, for $I$ any nonzero ...


5

If $m = n$, then $0 \to \mathbb{Z}/n\mathbb{Z} \xrightarrow{\operatorname{id}} \mathbb{Z}/n\mathbb{Z} \to 0$ is a free resolution. If $m < n$, it will be useful to write $n = km$, where $k > 1$. The first step of the resolution looks like $F_0 \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0$ for some free $\mathbb{Z}/n\mathbb{Z}$-module $F_0$. ...


5

If it is not a field, it has exactly one prime element (up to associates, of course.) (explain why) It is a UFD, and apparently everything that is nonzero has a factorization which is a power of p times a unit. From this, observe the nontrivial ideals are just $(p^i)$ for integers $i>0$.


4

Let $p\in\ker(f)$, that is, $p(t^2-1,t^3-t)=0$. Write $$p(x_1,x_2)=(x_2^2-x_1^2(x_1+1))q(x_1,x_2)+r(x_1,x_2)$$ with $\deg_{x_2}r\le1$. Then $r(x_1,x_2)=a(x_1)+b(x_1)x_2$ and from $p(t^2-1,t^3-t)=0$ we get $a(t^2-1)+b(t^2-1)(t^3-t)=0$. Now conclude that $a=b=0$. (In order to do this look at the degree of polynomials involved in the last equation.)


4

The ring $$R=\mathbb{C}[x_1,x_2,\ldots]/(x_1,x_2,\ldots)^2\cong\mathbb{C}[\epsilon_1,\epsilon_2,\ldots]$$ is non-noetherian (where $\epsilon_i$ denotes the image of $x_i$ in the quotient), and the radical of the (obviously finite) zero ideal $I=(0)$ in $R$ is equal to the ideal $(\epsilon_1,\epsilon_2,\ldots)$ which is not finitely generated, much less ...


3

Suppose $\{R_i\}$ is a directed system of local rings, let $R:=\varinjlim R_i$ and suppose further $R\neq \{0\}$. Consider $\mathfrak{m}:=R\setminus R^\times$. We will show that $\mathfrak{m}$ is an ideal and therefore $R$ is local with maximal ideal $\mathfrak{m}$. If $x,y\in \mathfrak{m}$ then for some large enough index $i$ there is a homomorphism ...


3

Here's a proof that uses Krull dimension. Observations: $\ker f$ and $\mathfrak a$ are prime. $\ker f$ is prime because $k[x_1,x_2]/\ker f$ is an integral domain, $\mathfrak a$ is prime because is generated by an irreducible element of $k[x_1,x_2]$ which is an UFD, hence it is generated by a prime element. In $\text{Im} f=k[t^2-1,t^3-t]$ the element ...


3

Here's a proof: Call $Q:=\mathbb{F}_p[X,Y]/A$. It is a field as $A$ is maximal, and it's prime field is $\mathbb{F}_p$ since $\mathbb{F}_p\hookrightarrow Q$. Now, $Q$ is finitely generated as an algebra over $\mathbb{F}_p$ so that Noether's normalization lemma applies. So there is a polynomial ring $S=\mathbb{F}_p[x_1,...,x_d]$ in $d$ algebraically ...


2

The following observations are made en passant in the accepted answer. I think they deserve a full proof. Let $R=K[X]/(X^2)$, and $x$ be the residue class of $X$ modulo the ideal $(X^2)$. Then $$\operatorname{Ann}_R(x)=Rx.$$ Since $x^2=0$ we have $Rx\subseteq\operatorname{Ann}_R(x)$. For the converse, let $f(x)\in\operatorname{Ann}_R(x)$. Then ...


2

Any Dedekind domain is dimension one, but not all Dedekind domains are PID's. For instance $\mathbb{Z}[\sqrt{-5}]$ is a standard example. The ideal $(2,1+\sqrt{-5})$ is maximal and therefore prime and is not the zero ideal, so it has rank $1$, but is not principal.


2

$I, J$ are each generated by a single element, say $x, y$ respectively. Every ring ideal is contained in a maximal ideal, in this case the (principal) maximal ideal, say, $(t)$. If $x$ is a unit, then $I=R$, so $J\subseteq I$. Similarly if $y$ is a unit. If not, then $x=ut^j$, $y=vt^k$ for some units $u,v$ and $j,k\in\mathbb{N}$. If $j\leq k$, then ...


2

Right before that he says Let us keep the hypotheses of prop. 9. If $\mathfrak{P}$ is a non-zero prime ideal of $B$... Proposition 9 says Proposition 9. If $A$ is Dedekind then $B$ is Dedekind. and in a Dedekind domain, every non-zero prime ideal is maximal. Thus $\mathfrak{P}$ and $\mathfrak{p}=\mathfrak{P}\cap A$ are indeed maximal, and thus ...


1

Yes. More generally: Let $A = \bigoplus_{i\in\mathbb{Z}} A_i$ be a graded domain und $f\in A\setminus \{0\}$ homogeneous. If $f$ factors in $A$ as $f=gh$ then $g,h$ are homogeneous. Proof: Since we are in a domain and $f\neq 0$, the factors $g,h$ are non-zero as well. Let $g$ have non-zero component of lowest degree $d_{\min}$ and of highest degree ...


1

If $S\subset A$ is a multiplicative set then $$\operatorname{gldim}(S^{-1}A)\le\operatorname{gldim}(A)$$ since $\operatorname{pd}_{S^{-1}A}(S^{-1}M)\le\operatorname{pd}_AM$ and every $S^{-1}A$-module is of the form $S^{-1}M$ with $M$ an $A$-module. Suppose now $$\sup_{m\in\operatorname{SpecMax}A} ...


1

Let $R=k[x]$ and $M=k[x^{\pm1}]$ viewed as an $R$-module in the obvious way. One can easily check that it is not finitely generated, not divisible and it is flat because it is a localization. To check that it is not projective, consider the map $\phi:\bigoplus_{n\in\mathbb Z}Re_n\to M$ from the free $R$-module with basis $\{e_n:n\in\mathbb Z\}$ to $M$ such ...


1

No. If $R$ has dimension greater than one, take any element that is not a zero-divisor. This is a regular sequence and definitely not a SOP for $R$.


1

The question reduces to prove that there are at most $t$ maximal ideals in $S$ containing $mS$. We know that $S/mS$ is an $R/m$-vector space of dimension at most $t$, and therefore $S/mS$ is an artinian ring. One knows that $S/mS$ is isomorphic to a finite direct product of artinian local rings (which are the localizations of $S/mS$ at its maximal ideals), ...


1

It's obvious that $$\dim B=\sup\{\dim B/P:P\text{ minimal prime}\}.$$ Let $P$ be a minimal prime of $B$. Then $P=\mathfrak p/(XY,XZ)$ with $\mathfrak p$ prime in $A[X,Y,Z]$ minimal over $(XY,XZ)$. Then $\mathfrak p$ is minimal over $(X)$ or $\mathfrak p$ is minimal over $(Y,Z)$. Conversely, every minimal prime over $(X)$, respectively $(Y,Z)$ gives rise to ...


1

The dimension is precisely $\dim A+2$. In Mariano Suarez-Alvarez comment above, I think the inequality should be reversed, since $\dim B\geq\dim B/x=\dim A+2$. On the other hand, since any prime ideal of $B$ must either contain $x$ or both $y,z$, one can check that the above inequality is actually an equality.



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