# Tag Info

5

Let $X$ be a compact Hausdorff infinite topological space and $C(X)$ the ring of continuous real functions $X\to \mathbb R$. Then in that ring the zero ideal does not have a primary decomposition. Indeed, if that were the case $C(X)$ would only have finitely many minimal prime ideals. But actually $C(X)$ has infinitely many minimal prime ideals because: ...

4

I'm not sure exactly how Eisenbud intends to argue, because I don't know what he's assuming known at this point. But here is some kind of explanation, which perhaps you can adapt to what you know: The inclusion $(Q_1,Q_2) \subset I$ shows that $V(Q_1,Q_2) \supset V(I) = C$. Now $V(Q_1,Q_2)$ is a curve of degree $4$ (by Bezout), as is $C$ (by assumption) ...

3

In general the function $\phi_S$ isn't even well-defined, since for example $\frac12=\frac24$. You could try to do something with lowest terms, but that would probably not work in general and would no longer be a homomorphism. And, as Karl points out, even in the case of $\mathbb Z\subset \mathbb Q$ no such bijection exists.

3

Finitely generated and torsion free implies projective, but infinitely generated torsion free modules need not be projective. (E.g. the fraction field of $A$ is not projective as an $A$-module; think about $\mathbb Q$ over $\mathbb Z$.) This is why you are having trouble proving it; you have to use finitely generated in the argument. (More generally, over ...

3

Take $R=2\mathbb Z/8\mathbb Z$. This ring has no prime ideals and the multiplication is clearly not trivial. Since the OP took another way, I'd like to add a generalization of the example above: take $R=d\mathbb Z/n\mathbb Z$ with $d\mid n$, and $m=n/d$. The prime ideals of $R$ are of the form $pd\mathbb Z/n\mathbb Z$ with $p$ prime, $p\mid m$ and ...

3

The archimedean places of $K$ correspond to the usual absolute value on $\Bbb C$, after first applying the embeddings $K\hookrightarrow\Bbb C$. Thus if $|x|=1$ for all archimedean places, we know that all of $x$'s conjugates have (the usual) absolute value $1$. Consider the minimal polynomial of $x$. In my Stewart & Tall, the following lemma is used in ...

3

This is indeed true. As Keenan's answer stated, the geometric version of the notion you're talking about is geometric irreducibility and reducedness. Consider the scheme $X$ cut out by $I$. Then $X$ is irreducible if and only if $\sqrt{I}$ is prime. Then $X$ is geometrically irreducible if it stays irreducible after any base extension. Similarly, $X$ is ...

2

It is not totally clear to me what your question is, but let me show the following statement. Let $S = K[x_0,\dots, x_n]$ and $S_+ = (x_0,\dots, x_n)$. Let $F_1,\dots, F_m$ be forms of degree $d$ and $m \le n$. If $(F_1,\dots, F_n) \neq S$, then $S_+ \nsubseteq \sqrt{(F_1,\dots, F_m)}$. In particular, $S_+^l \nsubseteq (F_1,\dots, F_n)$ for all $l$. ...

2

You are asking about a venerable subject in mathematics: elimination theory and resultants. The problem is: given a parametric curve $t\mapsto (x=f(t), y=g(t))$ in the plane, find a cartesian equation $h(x,y)=0$ for its image. In other words find a polynomial satisfying $h(f(t),g(t))\equiv 0$ . For example the parametric curve $x=t^2, y=t^3$ (known as a ...

2

The coordinate ring $k[{\mathbb A}^n]$ is $k[T_1, \dots, T_n]$, the polynomial ring in $n$ variables over $k$; the function field $k({ \mathbb A}^n)$ is $k(T_1, \dots, T_n)$, the field of rational functions in $n$ variables over $k$, which is the field of fractions of $k[T_1, \dots, T_n]$. As for $\text{trdeg}$: this is the transcendence degree and $k(T_1, ... 1 Since a field extension$K/k$is regular iff it is geometrically integral, the equivalence you are asking about is true and follows from the fact that a domain over$k$is geometrically integral if and only if its fraction field$K=\operatorname {Frac}(A)$is geometrically integral: a) Of course if$K$is geometrically integral then$A$will also be, ... 1 To answer your first question, take radicals:$\sqrt{P^n} \subseteq \sqrt{I}$, but$\sqrt{P^n} = P$is a maximal ideal. Therefore,$P \subseteq \sqrt{I}$, but again by the maximality of$P$, it must be the case that$P = \sqrt{I}$. So$I$is$P$-primary. To answer your second question, you should again think about taking radicals of the inclusion$P^n ...

1

Let $(R,m,k)$ be a local ring, and $\phi:F \rightarrow G$ a homomorphism of finite free $R$-modules. Then $\phi \otimes k$ is injective iff $\phi$ is injective, and $\phi(F)$ is a free direct summand of $G$. "$\Leftarrow$" This is not difficult and I leave it to you. "$\Rightarrow$" Let $\{e_1,\dots,e_m\}\subset F$ be an $R$-basis. (In the following we ...

1

You have too many hypotheses on your rings and modules. All you need is a homomorphism $f: R \rightarrow S$ of commutative rings, an $R$-module $M$, an $S$-module $N$, and an $S$-ideal $J$. No flatness or finiteness assumptions are needed anywhere. Then one has the following isomorphisms: $$\frac{(M \otimes_R N)}{J(M \otimes_R N)} \cong (M \otimes_R N) ... 1 If it was not mandatory to follow the hint, you could also say that x^2+y^2+1 is irreducible in \mathbb{C}(x)[y], and it is because x^2+1 is not a square. Hence you have that \mathbb{C}(x)[y]/(x^2+y^2+1) is a field. Now, you know that \mathbb{C}[x][y]/(x^2+y^2+1) is a free \mathbb{C}[x]-module with basis \{1,y\}, hence in particular it's flat. ... 1 Recasting Matt E's answer in terms I feel more comfortable with: Let L=(Q_1,Q_2). By the unmixedness theorem, the associated primes of L are precisely the minimal primes over L and all such have height 2. Since I is such a prime, the minimal primary decomposition of L looks like q \cap q_1 \cap ... \cap q_l, where q is I-primary. Since each ... 1 Hint for 1: What is the most natural homomorphism, from \Bbb Z_p[T] \rightarrow \Bbb Z_p[[T]]? Your isomorphism is induced from that homomorphism. Hint for 2: A similar argument to 1. shows \mathbb Z_p[[T]]/(T,p)^t \cong \mathbb Z_p[T]/(T,p)^t. Note \mathbb Z_p/(p^t) \cong \mathbb Z/p^t\mathbb Z, and now you can just work out what \mathbb ... 1 I based my answer on the following sources, check it out: [A-M] M. Atiyah; I. G. Macdonald, Introduction to Commutative Algebra. [H] T. Hungerford, Algebra (Springer, 1996). [S] J.-P. Serre, Corps Locaux. Let K be the field of fractions of your integral domain R. Your hypothesis is that for every nonzero prime ideal P of R we have ... 1 This has nothing to do with finite extensions of p-adics ... but rather with the basics of algebra. And O could be any ring.$$\Lambda/(\pi,T) = (O[[T]]/(T))/(\pi) = O/(\pi)=O/\mathfrak{p}$$Here are the two trivial facts which I use: If R is a ring and I,J are two ideals of R, then there is a canonical isomorphism$$R/(I,J) \cong (R/I)/ (J) ...

1

Here's what I finally did (no guarantee for correctness): First note that as the localizations $R_\mathfrak{p}$ are given by $\{y \in K\mid ||y||_\mathfrak{p} \leq 1\}$, $x$ is contained in all the $R_\mathfrak{p}$ hence also in $R$ (for $R$ being the integral closure of $\mathbb{Z}$ in $K$). Let $r_1$ be the number of real embeddings of $K$, let $2r_2$ be ...

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