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9

Is the product of prime numbers prime?


6

The answer is no, consider $\mathbb{Z}$ for example.


4

The original question suggests that you want to ask the following: Is the product of two prime ideals radical? Well, $\sqrt{\mathfrak p\mathfrak q}$ equals the intersection of all (minimal) primes containing $\mathfrak p\mathfrak q$, and therefore $\sqrt{\mathfrak p\mathfrak q}=\mathfrak p\cap \mathfrak q$. So the answer is: $\mathfrak p\mathfrak ...


4

It is a common misconception that every element of $S \otimes T$ has the form $s \otimes t$ and that you can define multiplication on $S \otimes T$ via such a formula $(s \otimes t) \cdot (s' \otimes t') = ss' \otimes tt'$ and verify the ring axioms with elements. Rather, one has to use the universal property of the tensor product in order to construct a ...


3

Smooth morphisms are locally of finite presentation by definition. Since $k[x_1,x_2,\dotsc]$ is not finitely generated (hence, not finitely presented) as a $k$-algebra, the morphism is not smooth. But it is formally smooth: If $B$ is a commutative $k$-algebra and $I \subseteq B$ is a nilpotent ideal (in fact, every ideal works here), then every $k$-algebra ...


3

Notice that the proof of Noether Normalization is constructive. Define $t = x - x^{-1}$. Then $xt = x^2 - 1$, hence $x$ is integral over $k[t]$. Also $t$ is algebraically independent and $k[x,x^{-1}]=k[t][x]$. You can also interpret this geometrically: $k[x,x^{-1}]$ is the coordinate ring of the hyperbola $V(xy-1) \subseteq \mathbb{A}^2$. We project it ...


3

$R=k[x]/(x^n)$ is an artinian local ring. Let $M$ be an $R$-module having a finite free resolution, or equivalently $\operatorname{pd}_RM<\infty$. Now we can apply the Auslander-Buchsbaum formula and get $\operatorname{pd}_RM=0$, that is, $M$ is projective, hence free.


3

We have $(A\cap B)(A+B)\subseteq AB$, but the other containment isn't always true. In the ring $K[X,Y]$, where $K$ is a field and $X$ and $Y$ are indeterminates, we have $$\Big((X)\cap(Y)\Big)\Big((X)+(Y)\Big)\subsetneq(X)(Y).$$ (Note that $(X)\cap(Y)=(XY)$, and therefore $(XY)(X,Y)\subsetneq(XY)$; otherwise $1\in(X,Y)$, a contradiction.)


3

In fact, under the given hypothesis there is a nonzero element $x$ of $R$ with $x \mathfrak{m} = \{0\}$. This is proved in Lemma 5.85 of T.Y. Lam's GTM Lectures on Modules and Rings. Lam's proof is reproduced in Corollary 10.8 of these notes.


2

I think it's useful to promote my comments on Pete Clark's answer to a new answer. We know that the set of zero divisors of $R$ is equal to the union of the associated primes of $R$. Moreover, the set $\operatorname{Ass}(R)$ of associated primes is finite. We have $$\mathfrak m=\mathfrak m^2\cup(\mathfrak m-\mathfrak m^2)\subseteq\mathfrak ...


2

$x+ 1/x = x+ 1/x$ and $x \cdot 1/x = 1$ so $x$, $1/x$ are roots of $$\lambda^2 - (x+1/x) \lambda + 1$$ and so $k[x,1/x]$ is integral (and finite ) over $k[x+1/x]$. $t = x + 1/x$ ( or $t = ax + b/x$ except when $ab=0$).


2

The product $\mathfrak p\cdot \mathfrak q$ of two prime ideals $\mathfrak p\cdot \mathfrak q\subset A$ in a commutative ring $A$ can only be prime if $\mathfrak p= \mathfrak q$ and $\mathfrak p= \mathfrak p^2$. If $\mathfrak p$ is finitely generated (for example if $A$ is noetherian) Nakayama then implies that $\mathfrak p=(e)$ is principal, generated by an ...


2

Proposition: Let $R$ be a local ring, and $M, N$ $R$-modules with $M \otimes_R N \cong R$. Then $M, N \cong R$. Proof: By the answers to this question, $M, N$ are finitely generated and projective. Since $R$ is local, $M, N$ are in fact free, say $M \cong R^m, N \cong R^n$. Then $R \cong M \otimes N \cong R^m \otimes R^n \cong R^{mn}$, so $mn = 1 \implies m ...


2

Yes, the structure theorem for f.g. modules over a PID gives a quick resolution: such a module is a direct sum of a free module and a torsion module. Another method: $M$ is flat iff $\text{Tor}_1^R(M, R/I) = 0$ for every $R$-ideal $I$. If $R$ is a PID, every (nonzero) ideal is principal and generated by a nonzerodivisor, say $I = (a)$. Then using the long ...


1

The answer to your question is yes: let $\widehat{K}$ be the completion of $K$ with respect to the absolute value $v$. Then by the extension theory for absolute values we know that there exists a field compositum $L.\widehat{K}$ within the algebraic closure of $\widehat{K}$, such that $v(x)=(v(N(x))^{1/n}$, where $N$ is the norm function of the finite (!) ...


1

Actually you want to prove the following: If $R$ is a $G$-domain, then $R[X]$ can't be a $G$-domain. This is a well known result (see e.g. Kaplansky, Commutative Rings, Theorem 21). Just in case, I add a sketch of the proof: If $R[X]$ is a $G$-domain, then $K[X]$ is also a $G$-domain, where $K$ is the field of fractions of $R$. But $K[X]$ is in the ...


1

Any valuation domain $R$ satisfies the condition "P". If $I$ is a proper ideal of a valuation domain $R$, then $\sqrt I$ is a prime ideal. Let $a,b\in R$ such that $ab\in\sqrt I$. Then there is $n\ge 1$ such that $(ab)^n\in I$. If $a\mid b$ write $b=ax$ and note that $b^{2n}\in I$, so $b\in\sqrt I$. The conclusion: it is not possible to bound the ...


1

The Krull dimension of $A$ is $1$, so almost every choice of $y\in A-k$ is good.


1

In EGA II, Proposition 2.3.6, there is a nice proof of the fact that the map $\psi_f: D_+(f)\rightarrow Spec(S_{(f)})$ given by $\mathfrak{p}\rightarrow \mathfrak{p}_f\cap S_{(f)}$ is a homeomorphism. Let $f\in S_d$. The map $\psi$ is continuous because for $g\in S_e$ you have $g^d/f^e\notin\psi_f(\mathfrak{p})=\{x/f^m:x\in \mathfrak{p}_{md}\}$ if and only ...


1

The answer is yes, even in the much more general case where $R=\mathcal O(X)$ is the space of global holomorphic functions on a Stein space $X$ (allowed to have singularities). Indeed, given $n$ global holomorphic functions $s_1,\cdots, s_n\in \mathcal O(X)$ we get an exact sequence of coherent sheaves on $X$: $$0\to \mathcal K\to \mathcal O_X^n\ \stackrel ...


1

You should think of $\mu(\mathfrak p,M)$ as it is: $\dim_{k(\mathfrak p)}M_{\mathfrak p}/\mathfrak pM_{\mathfrak p}$. If $V$ is a $k$-vector space, and $x\in V$, then $\dim V/\langle x\rangle=\dim V-1$ iff $x\ne 0$. If $\mathfrak p'\subset \mathfrak p$, then there is a homomorphism $R_{\mathfrak p}\to R_{\mathfrak p'}$, $a/s\mapsto a/s$. If $M_{\mathfrak ...


1

No. I don't think that is correct. Let $S = k[x,y]$, where $k$ is a field. Let $I = (x)$ and $J = (y)$. Then both $I,J$ are saturated. But $I+J = (x,y)$ is not.


1

$A$ is integral over $\mathbb{Z}$, thus has the same dimension as $\mathbb{Z}$, which is $1$.


1

$P \cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}$. It cannot be zero: Choose $p \in P \setminus \{0\}$. There is some polynomial equation $p^n+z_{n-1} p^{n-1} + \dotsc + z_0=0$ with $z_i \in \mathbb{Z}$. Choose $n$ minimal. Then we cannot have $z_0 = 0$. Thus, $z_0 \in P \cap \mathbb{Z}$ and $z_0 \neq 0$. Hence, $P \cap \mathbb{Z} = (p)$ for some prime ...


1

Hint: Show that the constant term of the minimal polynomial of some nonzero element $z \in P$ also belongs to $P$.


1

$f=y^2-x^2(x+1)$ is an irreducible polynomial in $\mathbb C[x,y]$. This can be proved elementary by supposing that $f$ is a product of two polynomials necessarily of degree one in $y$ and getting a contradiction, or one can use the Eisenstein's criterion for the prime element $x+1$. To conclude, $S$ is indeed an integral domain. $T$ is not an integral ...


1

$k[X_{ij}]_{0\le i\le m, 0\le j\le n}/(X_{ij}X_{kl}-X_{il}X_{kj})_{0\le i,k\le m, 0\le j,l\le n}$ is isomorphic to the Segre product of two polynomial rings in $m+1$, respectively $n+1$ indeterminates over $k$; see here. This is at its turn a subring of their tensor product, and therefore an integral domain.


1

I think the answer is yes and no. In finite games (finite players, finite strategies), you are mostly working with mixed strategies and your utility functions are polynomial functions in your strategies. But the equilibrium condition is really a stability condition which you can express with a bunch of polynomial inequalities. This and the fact that ...


1

You should look at a much more general statement: For any ideal $I\subset R$, and any homomorphism $f: N\to N'$ of $R$-modules, $f(IN) \subset IN'$. This statement is immediate from the definition of a module homomorphism. If $f$ is an isomorphism, then we can apply the same claim to $f^{-1}$, and see that $f$ restricts to an isomorphism $IN\to IN'$. ...


1

The sentence you are citing, "For each $i$, there exists $x_i\in I$ such that $x_i\notin p_k$ for all $k\neq i$", is not the logical negation of the statement, but the application of induction. That is, they assume that the statement is true for $n-1$ prime ideals and are applying to all subsets of size $n-1$ of $\{p_1,\ldots, p_n\}$, i.e., the set of the ...



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