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The ring $\mathbb{C}[x,y]/(xy)$ cannot be a product of two rings since $\text{Spec }\mathbb{C}[x,y]/(xy) = \text{Spec}\mathbb{C}[x,y]/(x) \cup \text{Spec }\mathbb{C}[x,y]/(y)$ (from $(xy) = (x) \cap (y)$) and this is a union of two connected spaces whose intersection is nontrivial (the prime ideal $(x,y)$), hence connected. It's a general fact (exercise ...

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Suppose $x\in D(R)$. Let $Ann_R(x)=\{r\in R\,|\,rx=0\}$. $Ann_R(x)$ is known as the left annihilator of $R$. It is easily seen that $Ann_R(x)$ is a left ideal of $R$. If $a,b\in Ann_R(x)$, then $(a+b)x=ax+bx=0$. Also for any $r \in R,\: (ra)x = r(ax) = r0 = 0$. Moreover there is clearly $Ann_R(x)\subset D(R)$ for any element in annihilator is also a ...

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I think the most explicit description you can get is via partial fraction decompositions. Let $P$ denote the set of monic irreducible polynomials over $k$. Then every $f/g \in k(t)$ has a unique representation of the form $$\frac{f}{g}=\sum_{n \ge 0} a_{0,0,n} t^n +\sum_{p \in P}\sum_{r > 0} \frac{a_{p,r,0} + a_{p,r,1}t + \cdots ... -2 Consider R=\left(\mathbb{Z},+,\cdot\right). Since R has no zero divisors, the only element of D\left(R\right) is 0. R is infinite, D\left(R\right) is finite (contradiction to your claim). 0 It's not true in general, but I think all counterexamples are going to be quite complicated. Suppose we have a counterexample and a_1\otimes b_1+\dots+a_n\otimes b_n\in J(R)\otimes_{\mathbb{Z}}S is not nilpotent. Let A be the non-unital subring of R generated by a_1,\dots,a_n. Then A is a nil ring but not nilpotent. Conversely, if A is any ... 0 Easy: let A be an integral domain (hence it is commutative), and I a non-zero ideal. Then$$\operatorname{Hom}(A/I,A)=\operatorname{Ann}_A(I)=0.$$2 Let A = \mathbb Z and M = \mathbb Z/2\mathbb Z. 7 \mathbb{Z}-modules are precisely abelian groups. As every ring is an abelian group, it is a \mathbb{Z}-module. It is entirely possible to be a module over more than one ring. For example, if M is an R-module then it is also an S-module for any subring S of R (you seem to be interested in the case where M = R). Another example is given by ... 1 There may be simpler examples, but in Lemma 2.2 of "Centres and fixed-point rings of artinian rings" by Christian U. Jensen and Søren Jøndrup (Mathematische Zeitschrift 130, 189-197 (1973)) it's shown that if k is a field and V a k-vector space of dimension at least the cardinality of k, then the ring R=k\oplus V, where V is a square zero ideal, ... 6 xg(x) has no constant term, so the constant term of 2f(x)+xg(x) is twice the constant term of f. But it must be equal to 1, which is a contradiction because 1 is not divisible by 2. 1 The Weyl algebra is a counterexample. It is a noncommutative ring with no proper nontrivial two-sided ideals that has no zero divisors and is not a division ring. 0 A prime p is not prime in \mathbb Z[\sqrt p] because (\sqrt p)^2 \in p\mathbb Z[\sqrt p] but \sqrt p \notin p\mathbb Z[\sqrt p]. Thus p\mathbb Z[\sqrt p] is not a prime ideal of \mathbb Z[\sqrt p]. Therefore, no prime remains prime in the ring of all algebraic integers. 1 There are also plenty of counterexamples apart from fields. For example p could be a unit even if R is not a field, e.g. in \mathbf Z [\frac{1}{p}], or p could fail to be irreducible, e.g in \mathbf Z[i] we have (1+i)(1−i)=2. 5 A silly example : Consider the field of real numbers, \mathbb{R}. Clearly p\mathbb{R}=\mathbb{R}, so \mathbb{R}/p\mathbb{R} \cong \lbrace 0\rbrace. More interesting: Let R=\mathbb{Z}[x]. \mathbb{Z}[x]/p\mathbb{Z}[x] \cong (\mathbb{Z}/p\mathbb{Z})[x] which is not a field! 1 Hint If f_n is not irreducible, then there exists two polynomials q(x) and p(x) in \mathbb Z[X] such that f_n(x)=p(x)q(x) with \deg(p)<n and \deg(q)<n hence for every 1\leq i \leq n we have:$$-1=p(i)q(i)\quad p(i),q(i)\in \mathbb Z$$from this we have for every 1\leq i \leq n$$p(i)=-q(i) $$and this implies that p=-q (two ... 1 No, this argument fails at the point where Ra_0b_0 is replaced by Ra_0. In general Ra_0 is larger than Ra_0b_0. In particular, your claim Rt = R is wrong, since Rt is the ideal generated by t in R. 1 Well, if we assume that [A, A] = \{ [x, y] \mid x, y \in A \}, \tag{1} [\lambda, A ] =\{ [\lambda, x ] \mid x \in A \}, \tag{2} and in general, for C, D \subset A, CD = \{ cd \mid c \in C, d \in D \}, \tag{3} then I think we can proceed as follows: first of all, we describe (C) for C \subset A; we have: (C) =  S = \{ \sum_1^n r_i c_i ... 2 In fact,$$\phi(R)/\phi(I)\simeq R/(I+\ker\phi)$$and this follows easily by the fundamental theorem of isomorphism for rings. (Find the kernel of the surjective homomorphism R\to\phi(R)\to\phi(R)/\phi(I).) 1 As noted above, this is a confusion between the polynomial and the function defined by the polynomial. In particular, if i:R_1\to R_2 is an inclusion of rings (or any homomorphism, really), under you definition, there is no homomorphism between R_1[x]\to R_2[x]. That's because a polynomial that evaluates to zero on all of R_1 does not necessarily ... 9 What is wrong is this: there is a difference between evaluating to zero at all points (as does X^7-X), and being zero. It is only over infinite fields that evaluating to zero at all points implies that the polynomial itself is zero. Your polynomial is not equal to zero, because X and X^7 are different in \mathbb Z_7[X]. In fact, one of the ... 0 Often, when trying to show that there exists an isomorphism between a quotient ring R/I and another ring S, it is easiest to find a surjective homomorphism R\to S which has kernel I, and then to use the isomorphism theorem. In this case, consider the map$$\begin{align}R &\to \mathbb Z/4\mathbb Z\\a+2ib&\mapsto a\pmod 4\end{align}$$Can you ... 1 Given a coalgebra, there is naturally a dual algebra. Namely, the coalgebra map \Delta: A \rightarrow A \otimes A can be dualized to get a map \Delta^*: (A \otimes A)^* \rightarrow A^*. Composing with the natural map \eta: A^* \otimes A^* \rightarrow (A \otimes A)^* gives the multiplication map. Explicitly, this natural map is given by$$\eta(\sum ...

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If $I/I^2$ is generated by $r$ elements, you can pick $r$ elements in $I$ which generates $I/I^2$. Let $J$ be the ideal generated by these $r$ elements and then consider $S=R/J$ and $I'=I/J$ an ideal in $S$. Then $I'/I'^2=0$ and you have your argument for this case, showing $I'$ is generated by one element and thus $I$ is generated by $r+1$ elements. My ...

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Here are some steps you can follow: Learn what a module is. Note that it's similar to a vector space. Note that every abelian group is a $\mathbb{Z}$-module, where the action of an integer $n\in\mathbb{Z}$ on a group element $g$ is just $n\cdot g := \underbrace{(g+g+g+\cdots+g)}_{n \text{ times}}$. Similarly, homomorphisms between abelian groups are the ...

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I can sketch a proof, then you can make all the details: Suppose $I$ is a prime ideal $\Rightarrow$ $\frac {R}{I}$ is a domain. Notice that every finite domain is a field then $\frac{R}{I}$ is a field so ${I}$ is maximal.

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Binary operations It would be more standard to call what you are describing a "binary operator on (a set of) permutations." And likewise, you'd say a "ring of permutations" or "a ring structure on permutations." As has already been mentioned, composition is the most natural binary operation on permutations. The full set of permutations is a group under ...

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Rough outline of how you might prove this: Given any ideal $I$ of $R$, consider the polynomials in the ideal, $I_1=I\cap k[t]$. Show that $I_1$ is an ideal of $k[t]$, hence principal in $k[t]$. Then show that its generator is also a generator for all of $I$ in $R$. The basic property you'll use is that for every element $f\in R$, there is an $m$ so that ...

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$\varphi (n)+\varphi(-n)=\varphi(n+(-n))=\varphi(0)=0$, that's why. And this is valid for any ring homomorphism in any ring.

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You are ordering the equalities wrong, so it seems to assume the result when it isn't: $$\varphi(1)\cdot \varphi(1) = \varphi(1 \cdot 1) = \varphi(1) = 1 \cdot \varphi(1)$$ The last equality is just the fact that for every $a \in D$ , $1 \cdot a = a$. Using $\varphi(1)\cdot \varphi(1) = 1 \cdot \varphi(1)$ we can arrange this as $$\varphi(1)\cdot ... 2 (Revised) Step 1:It is easy to prove that E_{ij}\in [A,A] for all i\neq j. Also E_{ii}-E_{jj}\in [A,A]. Step 2: Using step 1, let's prove that E_{ii}\in ([A,A]). Step 3: Using step 2, it is easy to prove that I\in ([A,A]). Just please use different E_{ij} to obtain all of these..Also we have two sided ideals.:) 1 First of all, the unique ring of characteristic 1 is the ring where 0_R=1_R. 1) You should know that any integral domain has prime characteristic. Hence the unique possibility is that car(D)=2. 2) (A,+) is a group with 3 elements. This means that the additive order of 1 is 3. 3) From$$(1_R+1_R)^6=1_R+1_R$$you get$$62 \cdot 1_R = 0_R$$so ... 4 Multiplication is not defined coordinatewise, rather (n,r)(m,s)=(nm,rs+rm+ns). In particular (1,0)(n,s)=(n,s). Writing (n,r) as n+r should help in understanding why we have defined multiplication in such a way. 0 Hint. Assume the contrary and then reduce all data modulo a maximal ideal. 1 I assume a \in \mathbb{E}\setminus \mathbb{F} You have to ask yourself a fondamental question: WHERE I'm valuating the polynomial f? By your definition you are saying that I is the ideal of the polynomials with coefficients in \mathbb{F} such that have a as rooth as polynomials of \mathbb{E}[x]. Now I is an ideal and you can take its ... 0 Wolfram Alpha can find the inverse in terms of c. The denominators of each term are easy to express in closed form. The polynomials in the numerators seem to follow Euler's triangle. 1 Well in the integers you don't mix positive and negative factors. Why not pick the/a sum of factors with the maximum modulus which would be compatible with the usual definition for integers. 1 Hint: {0, 1} comprises a subring (with unit) of R with two elements. A ring with unit with just 2 elements is a field. Observe that R is a finite-dimensional vector space over this field. 2 Hints: 1. R is abelian. 2. R is a unitary \mathbb{Z}/2\mathbb{Z}-module. 3. R is a unitary free \mathbb{Z}/2\mathbb{Z}-module. 3 The Jacobson radical of R is, by definition, the intersection of maximal ideals of R. In our case R is local with the maximal ideal (x^2+t)/(x^2+t)^2. If we want to see this as an R-module, then we can conclude that it is isomorphic to R/(x^2+t), but I can't see how is this isomorphic to \mathbb F_2(t)[x]/(x^2+t) (which is actually isomorphic ... 3 I assume that you actually mean to say "quotient of R by its Jacobson radical" rather than "Jacobson radical of R", as \mathbb{F}_2(t)[x]/(x^2-t) is naturally a quotient of R, not an ideal of R. The Jacobson radical always contains the nilradical, and x^2-t is nilpotent in R. So it suffices to show S=\mathbb{F}_2(t)[x]/(x^2-t) has trivial ... 2 I had wanted to give some hints and then put the solution into the "hidden" mode. However, the command >! didn't seem to be working with my text (this might have something to do with LaTeX commands), so I will just put my solution then. I would greatly appreciate if anybody can suggest a way to hide my text below. First, suppose ... 7 Let (x_i)_{i\in I} be a transcendence basis of \mathbb R over \mathbb Q, and S=\mathbb Z[x_i:i\in I]. Then Q(S), the field of fractions of S, is \mathbb Q(x_i:i\in I), and Q(S)\subset\mathbb R is an algebraic field extension. Now let R be the integral closure of S in \mathbb R. We have R\subsetneq\mathbb R (why?) and Q(R)=\mathbb R ... 1 Sorry for being so confusing in the comments. Guess that's what happens when you try to answer a question so late at night. I'll first answer using Hartog's lemma. Hartog's Lemma: Let X be a normal algebraic variety. Let U be an open subset of X such that Z, the complement of U, has codimension \ge 2 in X. Then, any function that is regular ... 1 Take G=C_2\times C_2, generated by elements x,y, and let H be the subgroup generated by x. Then you can take u=1+y-xy and a=1+x. 2 Too long for a comment. It looks like there are two conflicting definitions of an ordered ring. The first one was given by @martin-brandenburg in his comment and it is also used by L. Fuchs in his book. The second one is proposed by J.K. Hodge in his book Abstract Algebra: An Inquiry Based Approach and it goes like this: An ordered ring is a commutative ... 2 I'm not 100\% I'm the right person to speak for "modern algebraists," but here's some thoughts. Dietrich mentions semisimple elements in his comment: these generalize diagonalizeable operators to elements of an algebra in the presence of an algebra representation - either an associative algebra or alternatively a Lie algebra. Indeed, given such an element ... 0 This condition is basically asking if Spec(R) is T_1. My favorite equivalent definition of this is "all points are closed." But it is easy to see that if P, Q are distinct primes such that P\subset Q, then Q is in the closure of \{P\}. So, necessarily, a ring must have Krull dimension 0 to be T_1, but the ring you are talking about is 2 ... 1 Just going on the first sentence - I would first mod out by (x-1), which has the effect of setting x =1 (x becomes just another symbol for 1 so we forget about it). Now the quotient is (isomorphic) to the ring Q[y]/(y^2 + 2). Since (y^2 + 2) is irreducible over Q, the ideal it generates is maximal, hence the quotient is in fact a field. But since ... 0 As user hardmath pointed out in the comments, this easily follows from the evaluation homomorphism. For more details, see this. 2 You can prove it is not maximal by finding a prime p such that x^2-5 is irreducible modulo p. Thus we'll have$$(x^2-5)\varsubsetneq(p,x^2-5) and this last one is indeed a maximal ideal since $\mathbf Z[x](p,x^2-5)\simeq\mathbf F_p[x]/(x^2-5)\simeq\mathbf F_{p^2}$. For instance $x^2-5$ is irreducible modulo $3$, since the only squares mod. $3$ ...

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