# Tag Info

1

You need to remember the proof of the proposition you quote. Basically suppose $p(x)$ is reducible in $R[x]$, then $p(x) = q(x) r(x)$ where neither $q$ nor $r$ is a unit (of $R$). When you reduce $\bar{p}(x) = p(x) \pmod{I}$, you get $\bar p = \bar q \bar r$. And the key fact is that if a polynomial in $R[x]$ is not a unit, then its reduction mod I is not a ...

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I think the other factor is what you are being asked to consider. Yes, mod $y$ the factor $x+1$ remains irreducible but the factor $y+1$ becomes a unit. Hence the product $(x+1)(y+1)$ winds up mod $y$ being an irreducible polynomial.

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Z⊂Q is such an example. For an example in a noncommutative setting we can take the ring of quaternions (which is simple since it is a division ring) and it's subring only the elements with integer coefficients - Z[i,j,k]

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Note that by assumption $a_1b_1, a_2b_2 \in \langle a \rangle$, since $\langle a \rangle \supseteq I$. Multiplying by $-1$, we find that $(-1)a_2b_2 \in \langle a \rangle$, using the property that ideals absorb multiplication. Since $\langle a \rangle$ is an ideal, then it is closed under addition, so $a_1b_1 + (-1)a_2b_2 = a_1b_1- a_2b_2 \in \langle a ... 1 Proof: Let$I$be a nontrivial prime ideal for$F[x]$. Since$F$is a field, that means$F$is a Euclidean Domain which also implies$F$is a PID. So$I$is a principal ideal which is generated by$f$for some$f \in F[x].I$is maximal if and only if$f$is irreducible. Think you can go from there? 1 Let$R$be a noetherian integral domain and$I=(0)$. If$\dim R=2$and$R$is not Cohen-Macaulay, then there is$x\in R$,$x\ne 0$, such that$xR$is not unmixed. (For more details look here.) 1 I don't think we need to go far as to find prime ideals. Given a maximal element$(b)$of$M$, observe that since$(b) \in M$, the element$b$is not irreducible. So we can write it as a product of two non-units$b = b_1 b_2$. But then$(b_1) \supsetneq (b)$and$(b_2) \supsetneq (b)$, so by the maximality of$(b)$in$M$, we see that$b_1$and$b_2$can ... 0 Hint If$\alpha_1, \alpha_2, \alpha_3$are the roots of$f(t)$, then$f(t) = (t-\alpha_1)(t-\alpha_2)(t-\alpha_3)$. Multiply this out and compare to the definition of$f(t)$with which you started. 1 This looks pretty good, assuming your ring is commutative (as noted in a comment above). Usually we discuss closure first, because without it, there's not really anything to talk about, but that's not essential. What's required to establish is just: 0) The product of two units is a unit - you discovered that proof, and it will be fully general if you use ... 0 The title statement "localization of an integer quotient is a field" is incorrect in general. For example when$n$is not squarefree (as is the case for$24=2^3\cdot 3$) then the localization at the prime which isn't squarefree contains nonzero nilpotent elements, so it is not a field. In the$n=24$case, localizing at$2$demonstrates this: if you localize ... 0 Mumble! Gripe! Once again I seem to have answered the pre-edited version of the question! Ah well, at least I can take consolation in the fact that I do not appear to be alone! I won't attempt to prove the title assertion, because it is false. I will however give a simple counterexample: Let$N_1 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} ...

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What is to be proved is the following: $$e^{A \otimes I_b +I_a \otimes B} = e^A \otimes e^B~$$ where $I_a,A \in M_n$ , $I_b, B \in M_m$ This is true because $$A \otimes I_b~~~~\text{and}~~~~ I_a \otimes B$$ commute, which can be shown by using the so called mixed-product property of the Kronecker product. i.e. $$(A \otimes B)\cdot (C \otimes D) = ... 0 First and foremost, the result is not true as stated. It is only true of A and B commute, which is a very restrictive condition for matrices. To handle the commutative case, one can first consider the formal power series case. In the ring \Bbb Q[[X,Y]] of formal power series with rational coefficients in commuting indeterminates X,Y, one defines ... 1 If p and q are coprime and greater than 1, \mathbb Z_{pq} does contain a copy of \mathbb Z_q. Proof: Bezout's identity says that$$ap-bq=1$$for some positive integers a,b and a<q. Say that \mathbb Z_{pq}=\{0,1,\ldots,pq-1\}. We shall see that the subset A=\{0,ap,2ap,(q-1)ap\} is a subring. The products are computed in \mathbb Z_{pq}, ... 2 Denote \bar r the class of r\in R modulo I. Then$$ \bar r=\overline{r\cdot1}=r\cdot\bar1 $$i.e. R/I is generated as R-module by \bar1. More generally, if M is a R-module generated by m_1,...,m_s, then the classes \bar m_1,...,\bar m_s generate M/IM over R. 1 If A and B are n\times n, then by Taylor expansion we have:$$e^A=\sum_{k=0}^{\infty}\frac{A^k}{k!}$$Therefore:$$e^Ae^B=\sum_{k_1=0}^{\infty}\frac{A^{k_1}}{k_1!}\sum_{k_2=0}^{\infty}\frac{B^{k_2}}{k_2!}\Rightarrow e^Ae^B=\sum_{k_1=0}^{\infty}\sum_{k_2=0}^{\infty}\frac{A^{k_1}}{k_1!}\frac{B^{k_2}}{k_2!}\Rightarrow ...

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A way to proceed. If $A$ and $B$ commute they are simultaneously diagonalizable (if they are diagonalizable, otherwise one must fall back to Jordan decomposition). For diagonal matrices the formula is easy, because you reduce to the property of exponential for real numbers.

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If your subrings are required to contained the unit element of the former ring, then the answer is NO. Let $X$ be any ring, and let $A, B$ are its subrings, assume that $A \cong \mathbb{Z}_n$, and $B \cong \mathbb{Z}_m$, where $n \neq m$. From $A \cong \mathbb{Z}_n$, we can deduce that the unit element $e$ of $X$ has additive order $n$ (note that $e \in A$ ...

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user49685 got there first and should get credit for his work. (I've upvoted his answer...) Final answer: That a subring must contain the additive identity and the multiplicative identity requires that all subrings contain the characteristic subring. (I.e., they must contain the subring generated by $0$ and $1$, and since $0$ isn't doing any heavy lifting ...

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The notation $R_M$ means the following. Since you are viewing $R$ as an $R'$-module, $R_M = R'_M \otimes_{R'} R_M$. In other words, $R_M = W^{-1} R$ where $W = R' \setminus M$ not $R \setminus M$. In your example the element $1+x \notin R'$.

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Hint 1: What are the maximal ideals in $\mathbb{Z}$? What about the prime ideals in $\mathbb{Z}$? Use this to help you find your answer. Hint 2: Maximal ideals are always prime ideals (as you seem to already know), so if you have an idea of what the maximal ideals are, the prime ideals that are not maximal should be slightly smaller in some sense...

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For $F[x]$ to be a field, you need to show there is an inverse for each element that isn't 0. Now $x \in F[x]$, and clearly $x \ne 0$ (considered as a polynomial). But if you multiply $x$ by any non-zero polynomial, the result will always contain $x$ or higher powers, so it has no inverse.

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A classical example is the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$ ($p$ a prime); for each proper subgroup $H$ of it, it holds $\mathbb{Z}(p^{\infty})\cong \mathbb{Z}(p^{\infty})/H$ (and there's plenty of subgroups). If $R$ is the endomorphism ring of an infinite dimensional vector space, then, as (left) modules, $R\oplus R\cong R$, so $$\frac{R\oplus ... 1 Hint   The smallest subring of \,\Bbb R\, containing \,\Bbb Q,\ r,\ s is \,\Bbb Q[r,s],\, i.e. the set of all polynomials in \,r,s\, with rational coefficients: clearly they form a ring, and must be contained in any subring containing \,\Bbb Q,\ r,\ s.\, But since \,r,s = \sqrt{2},\,\sqrt{3}\, satisfy \,r^2 =2,\, s^2=3,\, all \,r^k,s^k ... 3 Well, suppose you take any element of the form a+b\sqrt{2} + c\sqrt{3} + d\sqrt{6} = a + b\sqrt{2} + c\sqrt{3} + (d\sqrt{2})\sqrt{3}. Isn't this clearly in \mathbb{Q}(\sqrt{2},\sqrt{3})? 0 I think you want to translate the condition of being cyclic into the condition that there exists v\in V for which$$\varphi:k[t]\rightarrow Vp(t)\mapsto p(T)v$$is a surjective map of k[t]-modules. In this case, the kernel has a monic generator. 0 The integers are a principal ideal domain, which means that every ideal I \subseteq \mathbb{Z} is generated by a single element. The notation \mathbb{Z} / n is more commonly written \mathbb{Z} / (n) or \mathbb{Z} / n\mathbb{Z} or \mathbb{Z}_n. This ring \mathbb{Z}/n\mathbb{Z} should be understood as the quotient by the ideal (n), which is the ... 1 Well, the idea is to assume that for a ring R and this special subset H, we want H to be an ideal of R, that is,$$ \forall x,y \in H, \forall r \in R, \quad x+y, rx \in H. $$Then, for each element r \in R, we group together all the elements \{ r+h \, | \, h \in H \}. Note that H doesn't have to be finite as in the notation you suggested by ... 1 To show this is a genuine subring: Firstly, if s,t \in R' then for any a \in \mathfrak a we have sa, ta \in \mathfrak a, and since \mathfrak a is an ideal it follows that sa + ta = (s+t)a \in \mathfrak a. Clearly zero and 1 (if your ring has an identity) satisfy 0\mathfrak a \subseteq \mathfrak a and 1\mathfrak a \subseteq \mathfrak a. ... 1 Did you mean to write$$\frac{\mathbb{Q}[x]}{<16x^4-30x^3+15x^2+6>}$$?? The "as a vector space over \mathbb Q" thing means that there is a natural action of \mathbb Q on this ring and that makes the ring a vector space over the field \mathbb Q. Every vector space has a basis so you should be able to list elements of the ring which span and are ... 1 Collecting the observations above by user121097 into an answer: Let I = (x_1, \ldots, x_n). Then m(M/IM) \ne M/IM iff (M/IM) \otimes_A A/m \ne 0, but M/IM \otimes_A A/m \cong M \otimes_A A/I \otimes_A A/m \cong M \otimes_A A/(I+m) \cong M \otimes_A A/m, so asking when m(M/IM) \ne M/IM and M \otimes_A A/m \ne 0 are equivalent. Thus replacing M ... 0 As should be abundantly clear from the comments yesterday, they are far from similar. Jorst pointed out that \frac23\cdot 3\in (2)\subseteq\Bbb Z, and yet neither one of \frac23 or 3 lies in (2). I'll answer talking about a domain R with field of quotients Q. As for your question "what is the difference?" the first thing to point out is that the ... 0 McConnell/Robson Lemma 9.6.9 says the following: Let U be a simple ring with centre k and V any k-algebra. Then (i) the map A \mapsto A \otimes U provides a 1-1 correspondence between ideals of V and ideals of V \otimes_k U; (ii) if V is a prime ring then V \otimes U is a prime ring so your 1:1 correspondence appears even if B is not ... 0 A slightly more general statement: If M is a minimal right ideal of a ring R, then either M^2=\{0\} or else M^2=M and M is generated by an idempotent. If you ask for R to be simple, the annihilator of M is zero, so you are automatically in the second case. Anyhow, the argument is simple (har har). Since M^2=M\neq \{0\}, aM=M for ... 0 Here is an example that is perhaps simpler than the common quadratic integer examples. \: Let \rm R = \mathbb R + x\:\mathbb C[x],\: i.e. the ring of all polynomials with complex coefficients and real constant coefficient. Here \rm\:x^2\: has infinitely many distinct factorizations into irreducibles$$\rm x^2\ =\ (c\: x)\: (c^{-1}\: x),\quad c = r + ...

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I suggest you R. Y. Sharp's book Steps in Commutative Algebra. By THEOREM 15.4 (KRULL'S GENERALIZED PRINCIPAL IDEAL THEOREM), $\dim R$ is less than or equal to "the number of elements in each minimal generating set for $m$", which is finite since $R$ is Noetherian. And by PROPOSITION 9.3 this is equal to $\dim_k m/m^2$.

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You need to prove that if $c^3=(ab^{-1})^3=1$ then $c=1$. Note that the multiplicative group of the finite field of order $8$ has order $7$, and the order of any element divides the order of the group. What you have at the moment is simply a reformulation of the problem, and you aren't referring to any properties of the field itself. Since there are fields ...

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In a UFD, irreducibles are prime (indeed, this condition plus acc on principal ideals characterizes UFDs among domains), so one should look for a ring that is not a UFD. One example would be $R = k[x,y,z,w]/(xy-zw)$. The element $x \in R$ is irreducible, but $(x)$ is not prime, as $zw = xy \in (x)$, but neither $z$ nor $w$ is in $(x)$.

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The canonical example is $\mathbb{Z}[\sqrt{-5}]$. Consider: $$2 \cdot 3 = 6 = (1 + \sqrt{-5})(1 - \sqrt{-5})$$ One can verify that $2$ is irreducible, but $2$ does not divide $(1 + \sqrt{-5})$ or $(1 - \sqrt{-5})$. Therefore, $(2)$ is not prime.

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Hint: define $$\phi:RG\to R\;\;,\;\;\phi\left(\sum_{k=1}^na_kg_k\right)=\sum_{k=1}^na_k$$ Show that $\;I=\ker\phi\;,\;\;\phi$ is onto, and then use the first isomorphism theorem

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Put $x=\sum_{i=1}^n a_ig_i \in I$, i.e. $\sum_{i=1}^n a_i=0_R$. Recall that the additive identity of $R(G)$ is $0_R e$. Then $0_R e=(\sum_{i=1}^n a_i)e=\sum_{i=1}^n a_i e$.($*$) Thus, $x=\sum_{i=1}^n a_i(g_i-e)$ by substituting the equation ($*$).

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If by $e$, you mean the multiplicative identity, then you can see that $I=\{\sum_{i=1}^n a_ig_i\in R(G)|\sum_{i=1}^na_i=0_R \}$ meaning $e$ is a root of $\sum_{i=1}^n a_ig_i$. So $I$ can be written as $\{<g-e>|g \in G\}$ because whatever belongs to $I$ will definitely have $e$ as its root.

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You haven't introduced a base-ring for the various algebras in play; let me denote it by $A_0$. (In the case of varieties, we would take $A_0$ to be a field, but that doesn't affect anything.) For a finitely presented $A_0$-algebra $A$, formal smoothness is equivalent to smoothness. (And if $A_0$ is Noetherian, e.g. a field, then f.p. is equivalent to ...

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Notice that $(y^2+x^3-17)\subset (y^2,x^3-17)$. Therefore, the quotient of the ring $\mathbb{C}[x,y]/I$ by the ideal, generated by $(y^2,x^3-17)$ is actually isomorphic to $\mathbb{C}[x,y]/(y^2,x^3-17)$. All I am using here is that for any ring $A$ and ideals $I\subset J$, we have an isomorphism $(A/I)/(J/I)\simeq A/J$. So the question boils down to ...

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It appears that the category of finitely cogenerated modules is not closed under extensions in general. Researching this problem led me to this article by Wisbauer. You need to see theorem 6.5 which I believe, when modified for this situation, says that the finitely cogenerated modules of $Mod-R$ are closed under extensions iff $R$ is right ...

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R/p is semiprime and right artinian, so it is semisimple. Since R/p is in fact prime, it can have only one simple component. Therefore, R/p is simple, so p is a maximal ideal.

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I think you're overthinking it. You're already at the answer :) All minimal right ideals of the same isotype sum to $I_S$ (and are hence contained in $I_S$.) You've just said that $rR_i\subseteq I_S$ for each $R_i$ in the isoclass of $S$. So, $rI_S=r(\sum R_i)=\sum rR_i\subseteq I_S$. I think I see a little confusion in the second paragraph of the answer ...

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Consider the homomorphism $f : M_n(\Bbb{Z}) \to M_n(\Bbb{Z}/2\Bbb{Z})$ which sends a matrix $M$ to the same matrix, but with entries modulo $2$. Is $f$ surjective? What is the kernel of $f$?

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Let $S\subset R$ be the multiplicative set $S=R\setminus \{0\}$ so that all $R$-modules and in particular all ideals $I\subset R$ have a fraction vector space $S^{-1}I$ over the field $F=S^{-1}R$. Your equality of $R$-modules $I\oplus R^{(m)}\cong R^{(n)}$ implies after applying $S^{-1}$ the equality of $F$-vector spaces $S^{-1}I\oplus S^{-1}R^{(m)}\cong ... 0 Yes: as you probably we'll know, the ideals of$R$and those of$M_n(R)$are in correspondence via the map$I\to M_n(I)$. Moreover this map preserves containment and intersections. If U is a right primitive ideal, say the annihilator of the simple R module S, then it's easy to show that$\oplus_{i=1}^n S$Is a simple right$M_n(R)\$ module with annihilator ...

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