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Is the $0$ matrix in $GL_n(\mathbb R)$?

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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} ... 4 I and -I are elements of GL(n), but their sum isn't. 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}). 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. 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 ... 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.) 3 If A\in GL_n (\mathbb {R}) , then -A\in GL_n (\mathbb {R}) , but A+(-A)=0\notin 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. 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 ... 3 Hint: Expand the expression (1+1)^3. 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 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 ... 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 ... 2 Hint:s^3-s=s(s-1)(s+1)=0, s=2, 2(2-1)(2+1)=6=0. 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 ... 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 ... 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 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. 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 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 ... 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. 1 First, here is a general lemma: Lemma: Let I,J\subseteq R be ideals and suppose I\cap J and I+J are finitely generated. Then I and J are finitely generated. Proof: Let x_1,\dots, x_n generate I\cap J, and let y_1+z_1,\dots,y_m+z_m generate I+J, for y_i\in I, z_i\in J. If w\in I, then w\in I+J, so we can write w=\sum ... 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 ... 1 What you say is correct. If you do it in the right generality, it becomes completely obvious that this holds regardless of the fact tht the codomain is \mathbb{Q}. Let me give to the "like a boss" view point. Let A be a commutative ring, I a set. Then the polynomial ring A[(X_i)_{i\in I}] is defined by the existence of a functorial bijection ... 1 I would like to expand on @Alexander's answer to give you a cleaner and more general version of what he's saying. However, to truly understand this you will need some basics in category theory. What is happening here, is that you are working in a category \mathcal{C}, namely the category of commutative rings \mathcal{C}=\bf{CRing}. There is a forgetful ... 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 ...

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@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 ...

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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 ...

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"$\Rightarrow$" Let $S=A\setminus\mathfrak p$. Then $S^{-1}\tilde A=\tilde A_{\mathfrak p}=A_{\mathfrak p}$. If prime ideal $\tilde{\mathfrak p}\subset\tilde A$ lies over $\mathfrak p$ the it survives in $A_{\mathfrak p}$, so there is only one with this property. Then $\mathfrak p\tilde A=\tilde{\mathfrak p}^e$, so \$S^{-1}(\mathfrak p\tilde ...

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