# Tag Info

18

In order to generalize rings to structures with noncommutative addiiton, one cannot simply delete the axiom that addition is commutative, since, in fact, other (standard) ring axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right distributive law in different order to the term ...

11

The three basic properties of a ring are; The set under addition makes an abelian group, Multiplication is associative, and Left and right distributive laws hold. Thus, by definition of "abelian group", the addition must be commutative. Hopes that help.

8

The definition of a ring is that it has two binary operations, $+$ and $\cdot$. The $+$ operation forms an abelian group and $\cdot$ need only be associative. The distributive laws need hold. However, notice that the distributive laws force $+$ to be abelian when $R$ has $1$! $$(1+1)(x+y)=1(x+y)+1(x+y)=x+y+x+y$$ and $$(1+1)(x+y)=(1+1)x+(1+1)y=x+x+y+y$$ ...

5

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 ...

4

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 ...

3

Let's take $I=(a)$ and $J=(b)$. Then $IJ=(ab)$. $I\cap J= (\operatorname{lcm}(a,b))$ and $I+J=(\operatorname{gcd}(a,b))$. The fact that lcm and gcd exist is due to the fact that we're working in a UFD. Therefore $(I\cap J)(I+J)=(\operatorname{lcm}(a,b)\cdot\operatorname{gcd}(a,b))=(ab)$. This holds generally for principal ideals in a UFD, but since every ...

3

For $1$: Suppose $I$ is a proper finite ideal and $|I|=n$. Take $x\in I$, $x\neq 0$ and think about $\{x,x^2,\ldots,x^n\}\subseteq I$. Suppose none of these powers of $x$ in this set are zero: then $x^k=x^m\neq 0$ for some $m> k$, but then $(x^{m-k}-1)x^k=0$, but the left hand factor must be a unit. See the end? For $2$ look at $(x)$ in $\Bbb R[[x]]$. ...

3

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 $e \in A$ ...

2

Often function rings are not noetherian and these sorts of rings come up frequently in applications, though what one counts as an application in mathematics obviously depends on who you are talking to. For instance, the ring of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ is not noetherian. Even more frequently we see applications of non artinian ...

2

One way to approach this is to show directly that the intersection of all zero sets of functions in $I$ is nonempty. Denote the zero set of $f$ as $z(f)$. Of course $z(f)$ is always closed and nonempty for $f\in I$ (If it's empty, the function is a unit.) The next observation is that the intersection of two zero sets for functions $I$ is nonempty. If to ...

2

You know that $(1,0) \cdot (0,1)=(0,0)$ then $f(1,0) \cdot f(0,1)=f(0,0)=0$. This implies that $f(1,0)=(a,0)$ or $(0,a)$, for some $a\in\mathbb{Z}$ and similarly for $f(0,1)=(b,0)$ or $(0,b)$. Since $f(1,1)=(1,1)$ you get that $a=b=1$ and that only two of the four options are valid. In this way you get exactly your two functions.

2

In order to show that $\phi(N)$ is an ideal we must show that it is an additive group and closed under multiplication in $\phi(R)$ Showing $\phi(N)$ is an additive group Identity So as $0\in N$ and $\phi$ is a ring homomorphism we have that $\phi(0)=0\in \phi(N)$ Closure Now take $x,y\in \phi(N)$ then by definition of $\phi(N)$ there must exists ...

2

As Marc indicates, your reasoning is incorrect: $N(z)$ composite does not justify saying that the number $z$ is reducible. Consider Mark's example; $N(3)=3^2$ is composite but $3$ is irreducible. There are two ways you could go about this problem: factor $2$ (hint: it's associate to a perfect square of a Gaussian integer; find out the absolute value of this ...

2

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 ...

2

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 $$\frac{R\oplus ... 2 In a UFD, irreducibles are prime (indeed, this condition plus acc on principal ideals characterizes UFDs among domains), so one should look for a ring that is not a UFD. One example would be R = k[x,y,z,w]/(xy-zw). The element x \in R is irreducible, but (x) is not prime, as zw = xy \in (x), but neither z nor w is in (x). 2 I don't know your background in algebraic geometry, but one way of seeing this is the following: It is well-known that on a variety (the vanishing set of a set of polynomials) over an algebraically closed field, the singular points form a proper closed set of the original variety, where "closed" means closed in the Zariski topology. In your case, it is ... 1 Let S\subset R be the multiplicative set S=R\setminus \{0\} so that all R-modules and in particular all ideals I\subset R have a fraction vector space S^{-1}I over the field F=S^{-1}R. Your equality of R-modules I\oplus R^{(m)}\cong R^{(n)} implies after applying S^{-1} the equality of F-vector spaces S^{-1}I\oplus S^{-1}R^{(m)}\cong ... 1 Well, the idea is to assume that for a ring R and this special subset H, we want H to be an ideal of R, that is,$$ \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 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 ... 1 There is not a whole lot to say here. You are given that \{0\} is semiprime, and you are asking when it is prime. You could say that R is a domain iff \{0\} is primary. fpqc's example demonstrates this nicely since (xy) is a semiprime but not prime ideal of k[x,y] and also not in the localization at (x) or (y). Locality never really comes into ... 1 Think about R/M. The cosets of M are just the subcollections of functions that have the same output for the input 1/3. This output could be any real number, so there is an obvious map suggested: \theta:C([0,1])\to \Bbb R \theta(f):=f(1/3) Show that this is a ring homomorphism of C([0,1]) onto \Bbb R, with kernel M. Conclude by the first ... 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 ... 1 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... 1 Take any nonzero ring homomorphism g:M_2(\Bbb R)\to T for some nonzero ring T. Then S=\prod_{i=1}^\infty T obviously isn't Noetherian on either side, and it obviously isn't simple. Define f:M_2(\Bbb R)\to S by f(x)=(g(x),g(x),\ldots,g(x),\ldots), and you have a counterexample to the first three. There's hope for the fourth statement, though. As ... 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 OK, the discussion in the comments is getting a little confusing, so let me write a full answer instead. I'll try to spell everything out in detail. We are assuming statement (a) and trying to deduce (b). Of course, we can assume that C_1 \cap C_2 contains at least 4 points. Claim 1: No 3 points of C_1 \cap C_2 lie on a line. Proof of claim: 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

von Neumann regular rings are pretty important in functional analysis, and they are Noetherian iff they are semisimple rings, so the "interesting" ones are the non-Noetherian ones. I should also say that this class of examples is kind of orthogonal to the other (also very good) example chstan gave about rings of continuous functions. The ring of continuous ...

1

First of all, as Zhen Lin observed, a prime ideal is not necessary maximal. Think to the ideal $(p) \subseteq \mathbb{Z}[x]$. We can ask if this holds in some particular categories of rings. It is not true if we chose UFD rings, in fact $\mathbb{Z}[x]$ is UFD and we have ever seen a counterexample. The assert is true for the PID Rings with only one ...

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