# Tag Info

2

When $R$ is commutative, you have $$\mathbf{A}\, \mathrm{adj}(\mathbf{A}) = \mathrm{adj}(\mathbf{A})\, \mathbf{A} = \det(\mathbf{A})\, \mathbf I_n \qquad$$ where $\mathrm{adj}(\mathbf{A})$ is the adjugate or classical adjoint of $\mathbf{A}$, that is, the transpose of its cofactor matrix. This equation proves that $\mathbf{A}$ is invertible in $M_n(R)$ ...

0

Depending on how much is proven prior it might be okey but it's not pretty Ring with unity does not exclude the possibility that $ab=0$ with neither $a$ nor $b$ being $0$. Better way is that we have $x^{-1}x=xx^{-1}=1$ and $xy=yx=1$, then we have $xy=xx^{-1}$ and my left multiplication of $x^{-1}$ we get $x^{-1}xy=y=x^{-1}xx^{-1}=x^{-1}$ It is extremely ...

0

You always have $$I_1 \cdots I_n \subseteq \bigcap\limits_{i=1}^n I_i$$ Try to prove by induction that if $I_1 \cdots I_n \subseteq P$, one of the $I_i$ is contained in $P$ (some people take this as the definition of prime ideal). For example, here's the $n = 2$ case: let $IJ \subseteq P$, and suppose there is a $y \in J$ which is not in $P$. Then for ...

0

Must all finite groups be cyclic? Certainly not. Just look at the Klein 4-group. To see a little more clearly why, just remember that a cyclic group of order $n$ must contain an element of order $n$, but as you see in the case of the Klein 4-group, all nonidentity elements have order $2$, and none have order $4$. All finite rings? (=Does the fact ...

1

In fact, it is an integral domain: the polynomial $f^p-a$ is irreducible in $k[x,y]$. In order to show this we use the Eisenstein's criterion. Note that $f^p-a=\alpha^px^p+\beta^py^p-a$. Multiplying by $\alpha^{-p}$ doesn't change anything, so we may instead consider the polynomial $x^p+b^py^p-c$, $c\notin k^p$. Now note that the polynomial $b^py^p-c$ is ...

0

Hint. Use that any $A$-module $M$ equals a direct sum of $A_i$-modules. See Decomposition of rings gives decomposition of modules.

0

The shortest proof of the first fact is the following: $1_H$ is an element of $F \setminus \{0\}$ with $1_H^2=1_H$. Multiplication with $1_H^{-1}$ (the inverse being with respect to $1_F$) shows $1_H=1_F$.

0

proof 1: What is your definition of a subfield. You assume that the multiplicative inverse is the same in F and H. Is this part of your definition of a subfield? I don't think so. So this should be proven, too, if you need it for your proof. proof 2: which nonzero/zero elements? If $ab=0$ then in a ring this does not mean that either $a$ or $b$ is $0$. ...

4

Yes. $A=2\mathbb{Z}/10\mathbb{Z} = \left\{ 2k+10\mathbb{Z}\;|\; k \in \mathbb{Z} \right\} = \left\{ 10\mathbb{Z}, 2+10\mathbb{Z}, \dots, 8+10\mathbb{Z}\right\}$. Also, by the third isomorphism theorem: $R/A = (\mathbb{Z}/10\mathbb{Z})/(2\mathbb{Z}/10\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$ (the field of order 2).

1

We want to show that if $P$ is the projective cover of $M$, then no proper submodule $P'$ of $P$ surjects onto $M$. The key is to know that $Ker(\theta)\ll P$ means $Ker(\theta)$ is a superfluous submodule of $P$, i.e. for all submodules $L\subseteq P$, $P=Ker(\theta)+L$ implies $L=P$. Suppose $P'\subsetneq P$ is a submodule and $P'\twoheadrightarrow M$. We ...

3

No, you're not right. The ring you are searching has the particular relation that x=0, so it will be all the polynomials of grade zero, and so all the costants, which is actually $\mathbb{Z}$

5

If there are no such $m$ and $n$, then the elements $a,a^2,a^3,a^4,\dots$ are all distinct elements of $R$. But these elements then form an infinite subset of $R$, which contradicts the assumption that $R$ was finite. (Note that this does not require $R$ to be a domain or $a$ to be nonzero.)

2

If $C=B/f(A)$ is not torsion, let $T$ be its torsion subgroup and use the quotient map $B\to C/T$ to show $f$ is not an epimorphism. Conversely, if $g_0,g_1:B\to D$ are such that $g_0f=g_1f$ and $D$ is torsion-free, $(g_0-g_1)f=0$, so $g_0-g_1$ factors through $B/f(A)$. Now use the fact that any map from a torsion group to a torsion-free group is $0$.

1

This is typically not true for $n>2$. For instance, suppose $R=k$ is a field, $M=k^2$, and the $M_i$ are a bunch of distinct $1$-dimensional subspaces. Then the $M_i$ satisfy your hypotheses, but $M/\bigcap M_i=M$ is $2$-dimensional while $\bigoplus M/M_i$ is $n$-dimensional.

2

Get a feel for the pattern: $$bx-xb \\ b^2x-2bxb+xb^2\\ b^3x-3b^2xb+3bxb^2-xb^3$$ By this point, the formula $d^n(x)=\sum_{i=0}^{n} (-1)^i\binom{n}{i}b^{n-i}xb^{i}$ suggests itself. Prove it using induction. You see that in order for all terms to hit zero, a power of b in every term has to hit n. $2n$ certainly works, but you can further show that $2n-1$ ...

2

In point 1. make $k$ concrete as $k = n+m$ (which will work, mostly for the reasons you discussed). You are right that $(1 + a + \ldots + a^{n-1})$ is the inverse of the element $1-a$. Is this element of the form $1-b$ for some nilpotent $b$? (it should be in $H$). Here 1. will help, plus the observation that the powers of nilpotent elements are also ...

3

Your approach is totally valid. However, you missed one possibility: $f([1])$ could also be $[21]$. And in fact, this one works, since $[21][nm]=[21n][21m]$ for all $n,m$ (since $21^2\equiv 21$ mod $28$). So besides the homomorphism sending $[1]$ to $[0]$, there's also one sending $[1]$ to $[21]$. (All of this is assuming you are talking about non-unital ...

2

You have already found the inverse, it is: $X^2 + I$. You know that: $(X^3 + X + 1)(X + 1) + (X^2 + X + 1)X^2 = [X^4 + X^3 + X^2 + (X+X) + 1] + [X^4 + X^3 + X^2]$ Now in $F_2[X]$, for any element, we have $f(x) + f(x) = (1 + 1)(f(x)) = 0(f(x)) = 0$. So in the product above, we continue: $[X^4 + X^3 + X^2 + (X+X) + 1] + [X^4 + X^3 + X^2] = [X^4 + X^3 + ... 2 What you have is correct: Since$+1 = -1$, you can write it as $$(x^3 + x + 1)(x+1) - (x^2 + x + 1)x^2 = x^4 + x^3 + x^2 + 2x + 1 - x^4 - x^3 - x^2 = 1$$ since$2x=0$When you apply the quotient map$\pi : F_2[x] \to F_2[x]/(x^3+x+1)$to this expression, you get $$\pi(x^2+x+1)\pi(x^2) = \pi(1)$$ Hence,$\pi(x^2+x+1)$is invertible in the quotient. 1 Let$R_0$be any ring and let$R=R_0[x_1,x_2,x_3,\dots]$be a polynomial ring over$R_0$in infinitely many variables. Then$R/I\cong R$, where$I=(x_1)$(since$R/I=R_0[x_2,x_3,\dots]$, and you can just shift the variables over by one to get an isomorphism with$R$). There are many other similar examples. For instance, you could take a product ... 1 You're right! A small nitpick in your proof: as stated, you've only shown that$\ker\pi\subseteq(b)(R/(a))$, not the reverse inclusion. However, the reverse inclusion is easy to show by a similar argument. More generally, by a similar argument, you can show that if$I$and$J$are ideals, then there is a canonical isomorphism between$R/(I+J)$and ... 1 @AnalysisStudent0414 's answer is much cleaner and more direct, but you can follow the idea in the question as follows: Suppose that$a+b\sqrt{d}=$(we may safely assume that$a$and$b$are relatively prime). Then, we know that$\sqrt{d}=-\frac{a}{b}$. Then,$d=\frac{a^2}{b^2}$by squaring both sides. Since$d$is an integer and$a$and$b$are ... 1 Give the explicit application $$\phi: \Bbb{Z}[x]\to\Bbb{Z}[\sqrt{d}]$$ $$\; \; x \mapsto \sqrt{d}$$ Now check that, if you quotient left side by$x^2 - d$, the application is well-defined on the quotient (that is,$\phi(x^2 -d) = 0$) This gives you one side. For the other one you have to give an explicit inverse, which is pretty straightforward (let me ... 4 Let$K$be a field and$\bar K$its algebraic closure. Since every element of$\bar K$has an unique monic minimal polynomial over$K$, we can define an equivalence relation on$\bar K$by calling two elements of$\bar K$equivalent iff they have the same minimal polynomial. If you know some Galois theory, you will recognise these equivalence classes as the ... 1 I came across this old post and, even if the following remark isn't of great importance, I just wanted to point out that the statement below is wrong: We can define$J=\{a\in A:\exists f\in I \;\text{ such that } a=\lambda (f)\}$but this is not necessarily an ideal of$A$. It is an ideal of$A$. If$a,b\in J-\{0\}$with$a\neq b$, then there exists ... 1 Most probably, this fact is also true for a real closed field, because its algebraic closure has dimension$2$. Choosing one of the square roots of$-1$will give you an analogue of the upper half plane. 0 Yes, it is. There is a theorem saying that given G is a finite group, G is a p-group iff |G|=p^k for some k. The proof requires Lagrange Theorem and Cauchy's Theorem. 1 Yes, in fact, the structure of the group$\mathbb{Z}/(n)^\times$is known completely (see Wikipedia or the answers to this question; in particular, Hurkyl's answer sketches an elegant proof using the$p$-adic logarithm). In particular, from these descriptions one can easily deduce the following values for$k$. Let$a_p(n)$denote the number of distinct ... 2 For a general polynomial f with integer coefficients, f(i) = a + bi where a is determined by the coefficients of even degree, and b is determined by the coefficients of odd degree. Write f(x) = g(x^2) + x h(x^2). By definition, f is in the kernel of phi if and only if f(i) is an integer multiple of p. This is equivalent to the simultaneous conditions that ... 1 Intuitively I understand the first definition but not the second. In fact the standard notation for a polynomial with integer coefficients and variables taken from the set X suggests first taking products, then integer linear combinations so that is exactly what a free abelian group on a free abelian monoid means. The second definition is less intuitive. ... 3 If$A\in GL_n (\mathbb {R}) $, then$-A\in GL_n (\mathbb {R}) $, but$A+(-A)=0\notin GL_n (\mathbb {R})$. 4 Yes, property (1) fails, because if$+$denotes the usual addition of matrices,$GL_n(\mathbb{R})$isn't even closed under$+$. For instance, if$I$is the identity matrix, then$I\in GL_n(\mathbb{R})$and$-I\in GL_n(\mathbb{R})$, but$I+(-I)=0\not\in GL_n(\mathbb{R})$. 3 In addition, the sum of two elements in$GL_n(\mathbb{R})$is not necessarily invertible, so it is not closed under addition. 4$I$and$-I$are elements of$GL(n)$, but their sum isn't. 5 Is the$0$matrix in$GL_n(\mathbb R)$? 4 You need a zero element in the ring. If$A+O = A$then$O$is necessarily in the ring. But the zero matrix isn't invertible as it has a zero determinant. 0 Observe that$x\in 1+fb^{-1}$if and only if$x-1 \in fb^{-1}$. But this means that, as fractional ideals, $$(x-1)\mathcal O\cdot a\subseteq\ fb^{-1}a = f\cdot(x\mathcal O)$$and hence$$\left(\frac{x-1}x\right)\mathcal O\cdot a \subseteq \ f$$and $$\left(\frac{x-1}x\right)\mathcal O = \left(\frac{1-x}x\right)\mathcal O=(x^{-1}-1)\mathcal O\subseteq \ ... 0 Sure, that's also correct. There are many different ideals that properly contain J; both (J,Y^2) and (J,Y) are examples, as are (J,y+1), (J, y^7+y+2), and many others. I'm not sure why they chose to give you the example (J,Y^2); there's nothing particularly special about it, and (J,Y) is in some sense the "simplest" example. 0 Let f(x) = a_m x^m + \ldots + a_0, a_m \neq 0, m \leq n be the polynomial in D[x]. If we evaluate f(c_i) for each i we get a sequence of equations:$$f(c_0) = a_m c_0^m + a_{m-1} c_0^{m-1} + \ldots a_0f(c_1) = a_m c_1^m + a_{m-1} c_1^{m-1} + \ldots a_0\ldotsf(c_n) = a_m c_n^m + a_{m-1} c_n^{m-1} + \ldots a_0$$These equations ... 2 We want to prove: If p is not a sum of two squares in \mathbb Z, then p is prime in \mathbb{Z}[i]. Let's prove the contrapositive: If p is not prime in \mathbb{Z}[i], then p is a sum of two squares in \mathbb Z. Indeed, if p is not prime in \mathbb{Z}[i], then we can write p=\alpha \beta, with \alpha, \beta \in ... 0 The ring \mathbb{Z} \times \mathbb{Z} \cong \mathbb{Z}[X]/(X^2-X) has the universal property$$\mathrm{Hom}_{\mathsf{Ring}}(\mathbb{Z} \times \mathbb{Z},R) \cong \{a \in R : a^2=a\},$$and the ring \mathbb{Z}[X]/(X^2-1) has the universal property$$\mathrm{Hom}_{\mathsf{Ring}}(\mathbb{Z}[X]/(X^2-1),R) \cong \{a \in R : a^2=1\}.$$Thus, if \mathbb{Z} ... 1 Before trying to understand the endomorphism ring, try to understand the set of endomorphisms. Since \mathbb{Z}_2 has only two elements, this isn't so bad to do by brute force: there are only four different functions f:\mathbb{Z}_2\to\mathbb{Z}_2. You can list them out and check them one-by-one to see which ones are homomorphisms (you should get that ... 3 b(z)=f(z)/g(z) with f,g\in\mathbb C[[z]], g\ne0. Write f(z)=z^kf_1(z) with f_1(0)\ne0, and g(z)=z^lg_1(z) with g_1(0)\ne0. Then b(z)=z^{k-l}f_1(z)(g_1(z))^{-1}. Now set a(z)=f_1(z)(g_1(z))^{-1} and notice that a(0)\ne0. The uniqueness is left as an exercise. (You can ask for help if need it.) 2 Very generally, any model of a finitary algebraic theory is a filtered colimit of finitely presented objects. For any such object X has a presentation \langle G \mid R\rangle, where G is some set of generators and R is some set of relations (and each relation only involves finitely many generators). Now consider the directed poset P consisting of ... 3 If \mathfrak a\mathfrak p^{-1}=\mathfrak a, then \mathfrak a\mathfrak p=\mathfrak a. Moreover (\mathfrak aA_{\mathfrak p})(\mathfrak pA_{\mathfrak p})=\mathfrak aA_{\mathfrak p}. By Nakayama lemma we get \mathfrak aA_{\mathfrak p}=0, so \mathfrak a=0 which I suppose you don't want. 2 (ii) (-a)b=-(ab) \Leftrightarrow (-a)b+ab=0 \Leftrightarrow ((-a)+a)b=0 \Leftrightarrow 0·b=0 which is true by (i) (iii) (-a)(-b)=-a(-b) by (ii). But -a(-b)=ab \Leftrightarrow ab+a(-b)=0 \Leftrightarrow a(b+(-b))=0 \Leftrightarrow a·0=0 which is true by (i) (iv) If R is non trivial \exists a\in R-\{0\}. Then (1·a)-a=0. If 1=0 then 1·a=0 and ... 2 It is quite unclear to me what you are trying to do there, in particular it seems you want to use (-1)(-1)=1 which is a special case of what you should show in (iii). Moreover, there is no need to assume 1 is in R, it appears your definition enforces this. Now, to show (ii) it can be a good idea to get clear what is even written there, it seems so ... 0 Hint for (ii): show that ab is the additive inverse of the RHS and the LHS. By uniqueness of inverses (you probably will have seen the proof of this), RHS = LHS. Hint for (iii): Use (ii). Hint for (iv): Prove the contrapositive. Let R be a ring where 0=1 and pick a x \in R. You should use the fact that 0=1 to show that R is trivial. 5 Not necessarily. If there is a nontrivial ring honomorphism \psi: A\to A, then composing an evaluation homomorphism with \psi gives you something that is not an A-algebra homomorphism. For example, in the case A=\mathbb C, n=1 and \psi being complex conjugation we could have$$ \phi(a_0 + a_1X + a_2X^2 + \cdots + a_kX^k) \mapsto \overline{a_0} ... 1 For a simple example where (2) does not imply (1), let$R$be any nonzero ring and let$S$be the rng with the same underlying abelian group as$R$but with the multiplication$ab=0$for all$a,b\in R$. Consider the$S$-module$M=R\times R$, with$S$acting by$a\cdot(b,c)=(0,ab)$. There is an epimorphism of$S$-modules$p:M\to S\$ given by the projection ...

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