# Tag Info

6

If you want an honest functor, you can simply choose any representative of the isomorphism class in question; it doesn't have to be "canonical". That is, for every pair $(X,Y)$ of $Z$-schemes, you can choose some triple $(P_{X,Y},p_{X,Y},q_{X,Y})$ where $P_{X,Y}$ is a $Z$-scheme and $p_{X,Y}:P_{X,Y}\to X$ and $q_{X,Y}:P_{X,Y}\to Y$ are maps of $Z$-schemes ...

4

The tensor product is defined up to canonical isomorphism, which is different from being defined up to isomorphism: there is a universal map $M\times N \to M\otimes N$ through which any bilinear map $M \times N \to R$ factors, and if you give one construction of the tensor product (e.g. generators and relations), and I give another one (e.g.perhaps I ...

4

$\mathbb{Q}(t) \otimes_\mathbb{Q} \mathbb{C}\simeq S^{-1}\mathbb C[t]$, where $S=\mathbb Q[t]-\{0\}$. Thus the maximal ideals of your ring are of the form $S^{-1}(t-a)$ with $(t-a)\cap S=\emptyset$. The condition $(t-a)\cap S=\emptyset$ is equivalent to $a$ transcendental over $\mathbb Q$.

3

I am assuming $k$ is algebraically closed. Suppose the statement is false and write $Y = V(f_1, \dots, f_m)$. Consider$$I(Y) = \{g \in k[x_1, \dots, x_n] : g(y) = 0 \text{ for all }y \in Y\}.$$Since $X \subset Y$, we must have$$\{f_1, \dots, f_m\} \subseteq I(Y) \subseteq I(V(f)) = \sqrt{f} = (f),$$where the first equality follows from the Nullstellensatz, ...

3

Let $R$ be a domain possessing two different non-zero prime ideals $p$ and $q$. Then either one of them is contained in the other one, $p\subset q$ say, or they are incomparable with respect to inclusion. In both cases there exists an element $x\in q\setminus p$. The element $x$ is invertible in $R_p$ but not in $R_q$, hence $R_p\neq R_q$. By assumption both ...

3

The natural surjection $A\to A/\text{rad}A$ induces a surjection $A^\times\to (A/\text{rad}A)^\times$, and so $(A/\text{rad}A)^\times$ is path connected. $A/\text{rad}A$ is a finite-dimensional commutative semisimple $\mathbb{R}$-algebra, and so is a finite product of copies of $\mathbb{R}$ and $\mathbb{C}$, and since its group of units is path connected ...

3

The problem with your argument is quite subtle: you can't say $\pi f(m/1)\in N$, because the canonical homomorphism $i:N\to S^{-1}N$ may not be injective (because elements of $S$ might annihilate elements of $N$). That is, there is always some element $n\in N$ such that $i(n)=\pi f(m/1)$, but that $n$ might not be unique, and it is not clear that you can ...

2

Here is my answer. Assume that $A$ is not artinian. Then there exists a maximal ideal not contained in the union of all minimal primes. Hence there exists an element $a$ not invertible and not contained in any minimal prime. Consequently if $ab=0$ then $b$ is contained in all minimal primes and is therefore nilpotent, contradicting the hypothesis.

2

"Qualitatively different" is not a mathematical notion. Either way, if $V$ is defined over reals, it contains some real points since we can always fix one real variable so that we get a cubic equation over reals in the other variable which always has a real solution. Any irreducible plane cubic having a rational point (i.e. a real point in this case) can ...

2

This is not true. For instance, consider the localization $R=\mathbb{Z}_{(p)}$, or more generally the localization of any domain at a height $1$ prime. The only primes in $R$ are $0$ and $(p)$, the nilradical is $0$, but the Jacobson radical is $(p)$.

2

The intersection of ideals is an ideal. Since yours is divisible by infinitely many primes, it must be that it has no presentation as a product of primes. But in a Dedekind domain, every non-zero ideal has a unique expression as the finite product of prime ideals (to some power). Hence it must be that your ideal is the only ideal not covered by this theorem: ...

2

$(x,y)$ is a regular sequence in $R=\mathbf Q[x,y]$ , hence the Koszul complex: \begin{alignat*}{3}0\longrightarrow R&\xrightarrow{\begin{bmatrix}x\!&\!\!y\end{bmatrix}} R^2&\xrightarrow{\smash[t]{\begin{bmatrix}-y\\x\end{bmatrix}}}&R\longrightarrow R/(x,y)\longrightarrow 0 \\ t&\longmapsto \rlap{(xt, yt)}\\ &(u,v)&\longmapsto ...

2

As you were told, $f_i$ can generate a radical ideal without being irreducible, so I try to answer the question if $f_i$ should generate a radical ideal if $f_1,\dots,f_t$, $t\ge2$ is a minimal system of generators of a radical ideal $I$. First note that $(f_i)$ is a radical ideal iff $f_i$ has no multiple irreducible factors. Suppose the contrary and write ...

1

Since $M=IM$, we can write $m_1=a_1m_1+a_2m_2+\dots+a_nm_n$, for $a_i\in I$. Rearranging terms, we find $(1-a_1)m_1\in IM'$. Now use this to show that $(1-a_1)m_i\in IM'$ for $i=2,\dots,n$ as well. Since $a_1\in I$, conclude that in fact $m_i\in IM'$ for $i=2,\dots,n$.

1

Even if you know the prime ideals of a commutative ring $R$ very well, there is nothing substantial you can say about the prime ideals of $R[x]$: if there were algebraic geometers would be out of business! For example it is very easy to determine the prime ideals of $R=\mathbb C[y,z]$ but essentially impossible to classify the height-$2$ prime ideals of ...

1

Upon request, I repost my MathOverflow answer (c.f. MO/218737). There are two questions here, namely, whether the argument given in the question is okay, and whether the criterion really is more general than the one by Raynaud-Gruson. The answer to both questions is yes. The following holds: Let $A$ be a commutative with unity and $M$ a flat finitely ...

1

Since $\dim M=0$ we have $\dim A/\operatorname{Ann}M=0$, so $A/\operatorname{Ann}M$ is a local artinian ring and you are done.

1

(b) $\Rightarrow$ (a) $R$ must have a non-zero prime ideal (why?), say $\mathfrak p$. Then $R_{\mathfrak p}$ is not a field (why?), so $R_{\mathfrak p}=R$. Thus $R$ is local and the only non-zero prime ideal of $R$ is $\mathfrak p$ (why?). Remark. Condition (b) can be interpreted in two ways: If $S\subset R$ is a multiplicative set, then $S^{-1}R=R$ or ...

1

Let $\mathfrak m$ be the maximal ideal of $A$. Since $\operatorname{height}\mathfrak m=2$ we have $\mathfrak m\notin\operatorname{Ass}_A(A/(a))$, so $\mathfrak m$ contains a non-zero divisor on $A/(a)$, in other words, $\operatorname{grade}\mathfrak m\ge2$. (In particular, this shows that $A$ is Cohen-Macaulay.) Now let $\mathfrak ... 1 Let$B$be the set of monomials in$x_1,\ldots,x_n$.$R$is merely a free$\Bbb Z$-module with basis$B$. The action of$G$corresponds to permutations on$B$, let$B/G$be the set of orbits of$B$under$G$. There is an obvious map$B/G \to R$(obtained by summing the monomials of the orbits together). If you remove from$B$one representant of each orbit ... 1 Or you can prove that$K[X,Y]/(XY)$is a finitely generated$K[x+y]$-module, where$x,y$are the residue classes of$X,Y$modulo the ideal$(XY)$. You know that$xy=0$. Then$x,y$are the roots of the polynomial$(Z-x)(Z-y)=Z^2-(x+y)Z\in K[x+y][Z]$, so$K[x+y]\subset K[x,y]$is an integral ring extension, and it is finitely generated for ... 1 Yes, it's true. The ring extension$R = F[X_1, ... , X_n]\subset S = K[X_1, ... , X_n]$has the property that$S$is a free$R$-module (since$K$is a free$F$-module). This shows that it is faithfully flat, and we are done. 1 Take the map$\mathbb{C}[x,y,z]\to\mathbb{C}[t]$given by$x\mapsto t^3,y\mapsto t^4, z\mapsto t^5$. It satisfies your hypothesis, but it is well known that the kernel is a 3-generated ideal and not a two generated ideal. 1$A$is$A_0$-flat (since$k$is$F$-flat), and then the intersection commutes with the extension of ideals (see Matsumura, CRT, Theorem 7.4(ii)). 1 If$S$is a multiplicative subset of$A, the canonical morphism; \begin{align*} \varphi\colon A&\longrightarrow S^{-1}A\\ x&\longmapsto \frac x1 \end{align*} is a flat epimorphism. (This is an example which shows that, in the category of commutative rings, ‘epimorphism’ does not mean ‘surjective’.) 1 By applying the Krull-Kaplansky-Jaffard-Ohm theorem to a totally ordered group of suitable size, you can create valuation domains with whatever Krull dimension you wish. The special case (first proved by Krull himself) of the theorem says that every totally ordered abelian group can be realized by the divisibility group of a valuation domain, and later this ... 1x(1+a)=y(1+b)$with$a,b\in I\Rightarrowx\equiv y\bmod I$. Then from$(x)+I=R$we get$x$invertible in$R/I$. But in$R/I$we have$x=y$, so$y$is also invertible in$R/I$, and thus$(y)+I=R$. Yes, that one is the right map. It's obviously well defined and surjective. For injectivity let$x\bmod I=y\bmod I$and want to show that$x\sim y$. Since ... 1 You can use this: Why does$(A/I)\otimes_R (B/J)\cong(A\otimes_R B)/(I\otimes_R 1+1\otimes_R J)$?. In your case$R=F$,$B=F[X_1,\dots,X_n]$,$A=k$,$I=0$, and$J=I_0$. The only thing to prove is that the extension of$I_0$to$k[X_1,\dots,X_n]$is$I$, but this follows immediately from this answer. (In fact, if$f_1,\dots,f_t$generate$I$, then they ... 1 The cone over a projective variety$A$of dimension$n$will have dimension$n+1$. One way to convince yourself of this is as follows: the natural map$\pi \colon C(A) \setminus \{0\} \to A$is surjective, and for every$p \in A$,$\pi^{-1}(\{p\})$is a line (missing a point) in$\mathbb A^{n+1}$which is a variety of dimension$1$. Intuitively, you have$1$... 1 (a)$\mathbb C[X,Y]/(X^{n}-Y^{m})\simeq\mathbb C[T^m,T^n]$, where$n>m$are coprime positive integers. The field of fractions of$\mathbb C[T^m,T^n]$is$\mathbb C(T)$(why?),$T$is integral over$\mathbb C[T^m,T^n]$, and$T\notin\mathbb C[T^m,T^n]$. (b)$\mathbb C[X,Y]/(XY-1)\simeq S^{-1}\mathbb C[X]$, where$S=\{1,X,X^2,\dots\}\$.

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