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3

Since $a \text{ mod } p = -1$, you already know that $p| a+1$. Since $p|-p$, you can infer $p| ( a+1)+(-p)$ and $p| a-(p-1)$, that is to say $a \text{ mod } p = p-1$.


3

For $A$-modules and homomorphisms $0\to M'\stackrel{u}{\to}M\stackrel{v}{\to}M''\to 0$ is exact, if $M'$ and $M''$ are fintely generated then $M$ is finitely generated. There is exact sequence: $0 \to M_1 \cap M_2 \to M_1 \oplus M_2 \to M_1 + M_2 \to 0 $ .


3

Let $R$ be a ring such that every left $R$-module is free, and let $I \subset R$ be a maximal left ideal. Then $R/I$ is a simple nonzero $R$-module, and is free by hypothesis, so $R/I$ has a basis. Take any basis element $x$, and let $\varphi \colon R \to R/I$ be the $R$-module homomorphism given by $\varphi(r) = rx$. Since $x$ is nonzero and $R/I$ is ...


2

Since $x^p$ and $y^p$ are in the center, they act as scalars on your simple module $V$. This means that in fact $V$ is a module over the algebra $k\langle x,y\mid yx-xy-1,x^p-\alpha,y^q-\beta\rangle$ forsome scalars $\alpha$, $\beta$ in the field. Show that this algebra is central and simple (imitating the proofs for the Weyl algbra in characteristc zero, ...


2

Note that since $A$ is not only semi-simple but also simple, there is a unique simple $A$-module. Call it $I$. Then any finitely generated $A$-module is isomorphic to $I^n$ for some $n$ (in particular they are all projective). Then $\operatorname{End}_A(P) \simeq \operatorname{End}_A(I^n) = M_n(\operatorname{End}_A(I))$ (the last equality is valid for any ...


2

No. To show it's injective, you have to show that, if $\;\sum r_i\otimes m_i\mapsto\sum r_im_i=0$, then $\;\sum r_i\otimes m_i=0$. But that is because $\;\sum r_i\otimes m_i=\sum 1\otimes r_im_i=1\otimes0=0$.


2

The main point is to realise that linear combinations can by definition only have finitely many nonzero coefficients. They must be defined this way, because in pure linear algebra there is no way to take the sum of infinitely many nonzero vectors (this cannot be defined by repeated addition: one never reaches the goal). In analysis some (convergent) infinite ...


2

If you look at the dimensions : $\dim(C) = |S|$, $\dim(D) = |T|$, and $\dim(A\otimes B) = |S|+|T|$. But then $\dim(C\otimes D) = |S|\cdot |T|\neq \dim(A\otimes B)$ (if the dimesnions are finite).


1

You want to find two vectors which, together with $(m,n,k)$ make a matrix with determinant $\pm 1$. Then the three vectors will form a free basis of ${\mathbb Z}^3$. One way of doing that is to perform some elementary unimodular column operations on $(m,n,k)$ to transform it to $(1,0,0)$. This you can easily extend to a free basis of ${\mathbb Z}^3$. Then, ...


1

For one thing, there is a linear map on your vector space which takes the value 1 on each vector in your basis. Is this map a linear combination of the elements of the dual basis?


1

if $ a \equiv -1\ (\textrm{mod}\ p)$, then $a \equiv (- 1) + n * p, \ \forall n \in Z$. Let $n = 1$, then $a \equiv -1 + 1 * p$, Hence $a \equiv p - 1$.


1

$a \bmod p$ is the only integer $r$ such that $a \equiv r \bmod p$ and $0 \le r \le p-1$. Therefore, if $a \equiv -1 \bmod p$, then $a \bmod p = p - 1$, because $p-1 \equiv -1 \equiv a \bmod p$ and $0 \le p-1 \le p-1$.


1

$\ker(B)/\operatorname{im}(A)$ is just the kernel of the map $$\Phi:\operatorname{cok}(A)=R^n/\operatorname{im}(A)\to R^o$$ induced by $B$, from the cokernel of $A$ to $R^o$. But $\operatorname{cok}(A)$ is isomorphic to $$R/a_1R\oplus\dots\oplus R/a_r R \oplus R^{n-r},$$ and $\ker(\Phi)$ must contain the torsion submodule $R/a_1 R\oplus\dots\oplus R/a_r ...


1

If $\;r\otimes m\to rm=0\;$ then $$r\times m=r(1\otimes m)=1\otimes rm=1\otimes0=0$$ and we've finished. I am assuming, of course, the tensor product is over $\;R\;$ without any further conditions.



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