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9

Such an isomorphism is not true. What really happens is that, under the bilinear map from $\Omega^*(M)\times\Omega^*(N)$ into $\Omega^*(M\times N)$ you wrote (which, by universality of the tensor product, lifts uniquely to a linear map from $\Omega^*(M)\otimes\Omega^*(N)$ into $\Omega^*(M\times N)$ as user 123456 suggested), $\Omega^*(M)\otimes\Omega^*(N)$ ...


6

No -- you can't say "$k\otimes_{\mathbb Z} m\bar a$" unless $k$ is itself a multiple of $m$. Otherwise it is not a member of the left factor of the tensor product. Instead, we can note that $\mathbb Z$ and $m\mathbb Z$ are isomorphic as groups, so the tensor product ought to be the same (up to isomorphism) as $\mathbb Z\otimes_{\mathbb Z}\mathbb Z/m\mathbb ...


4

This follows at once if you show that $$R/I \otimes R/J \simeq R/(I+J)$$


4

It is surjective because $\phi(1\otimes a) =a$ so every $a\in A$ is in the image. To show that it's a $B$-algebra morphism, first you must show that it's a well-defined $B$-module morphism : this follows from the universal property of tensor products, since $A\times A\to A$ defined by $(a,a')\mapsto aa'$ is $B$-bilinear. Then to check that $\phi(xy)= \phi(...


3

Your argument is correct. If $\{a_1,\dots,a_n\}$ generates $A$ as a $\mathbb{C}$-algebra, then $\{a_1,\dots,a_n,i\}$ generates $A$ as an $\mathbb{R}$-algebra. Indeed, any $\mathbb{R}$-subalgebra of $A$ containing $i$ is a $\mathbb{C}$-subalgebra, and then if such a subalgebra contains $a_1,\dots,a_n$ it must be all of $A$. More generally, a similar ...


3

First show that if $A$ is any ring, $I\subseteq A$ a left ideal and $M$ a right $A$-module, there is a (natural) isomorphism $$ \eta_M : M\otimes_A A/I\longrightarrow M/IM$$ that sends $m\otimes a$ to the class of $ma$. Conclude that in particular $A/J\otimes_A A/I=A/(I+J)$ when $A$ is a commutative ring and finally consider the case when $I,J$ are ...


2

Since $L/K$ is a finite separable extension, there is a $\theta\in L$ such that $L=K(\theta)$ by the primitive element theorem. Let $P$ be the minimal polynomial of $\theta$ over $K$. Let $E/K$ be any field extension, algebraic or not. Since $\theta$ is separable over $K$, the irreducible polynomial $P$ is a separable polynomial. This means that all the ...


2

As noted in the comments, this is false. Intuitively, on the left-hand side in 1) you add one point to both spaces and then take the cartesian product whereas on the right-hand side you first take the product and then add one point. For example let $A=B=C_0((0,1))$. Then $\widetilde{A\otimes B}=\widetilde{C_0((0,1)\times (0,1))}=C(S^2)$ and $\tilde A\otimes ...


2

$K$ is a flat $R$-module, hence the exact sequence $$0 \to I \to R \to R/I \to 0$$ gives rise to an exact sequence $$0 \to I \otimes K \to R \otimes K \to R/I \otimes K \to 0.$$ The middle term is $K$. The right term is $K/IK=0$, hence the left term is isomorphic to the middle term. The fact that tensoring with $K$ is the same as localizing at $\mathfrak ...


1

Alternatively to Eric's answer (+1), you might argue as follows: Writing $$A\cong A_{\mathbb R}:=\{a\otimes 1\ |\ a\in A\}\subset A_{\mathbb C}$$ you have $A_{\mathbb C} = A_{\mathbb R}\oplus i A_{\mathbb R}$ as ${\mathbb R}$ vector spaces. Denote $\text{Re}_A, \text{Im}_A: A_{\mathbb C}\to A_{\mathbb R}$ the respective projections. Then, if $A_{\mathbb C}$ ...


1

This is a question that touches on many issues. On the one hand, things are indeed easier to deal with in the language of the complexification. For $SL(2,\mathbb C)$, there is a specific element $H$ in the Lie algebra $\mathfrak{sl}(2,\mathbb C)$, which acts diagonalizably on any finite dimensional representation. The eigenvalues are called weights, and in ...


1

Let for instance $A$ be a field, and $P$ a prime ideal of $B$ (then $\mathfrak{p}=0$) such that $B\neq B_P$.


1

A different strategy, for the particular case, is considering $r\in M_1$ such that $r\notin M_2$. Then every element of $R/M_2$ is of the form $rs+M_2$ and, for any $t\in R$, we have $$ (t+M_1)\otimes(rs+M_2)=(rt+M_1)\otimes(s+M_2)=0 $$


1

$\nabla u$ is a $(1,0)$ tensor $\nabla u= u^i\frac{\partial}{\partial x^i}$, where $u^i = g^{ij} \frac{\partial u}{\partial x^j}$ , so $\nabla u \otimes \nabla u$, just like $\text{Hess } u$, is a $(2,0)$ tensor given by $$(\nabla u\otimes \nabla u)^{ij} = u^i u^j = g^{ik}g^{jl} \frac{\partial u}{\partial x^k}\frac{\partial u}{\partial x^l}$$


1

It is there so that this is an $F$-algebra; you need a ring homomorphism from $F$ to $T(V)$ in order to have this structure, and being an algebra over a field is a useful structural condition. You also just get better theorems if you don't mutilate $T(V)$ or $Sym(V)$ by removing the field! But I guess you could clarify your question - necessary for what? ...


1

Suppose $U$ is a vector space over a field of characteristic different from $2$. If you have any linear map $\alpha$ on $U$ such that $\alpha^{2} = I$, then either $\alpha = \pm I$, or $x^{2} - 1$ is the minimal polynomial of $\alpha$. In the latter case the minimal polynomial has two distinct roots $1$ and $-1$, thus $U$ decomposes as the direct sum of ...



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