# Tag Info

1

When I perform a Google books search for "Boolean ring", a majority of the top two pages of hits explicitly do not assume identity. The same is still true when restricting to hits with publication dates in the past decade. So it seems like a standard usage of boolean ring fits your requirements quite well already. Naturally you are going to find category ...

0

For a given $f\colon R \to S$ one can define a homomorphism $\tilde f\colon R \to S/I$ which sends $r \in R$ to $f(r) + I$. The kernel of $\tilde f$ is precisely those elements whose image under $f$ is in the ideal $I$, thus $\ker \tilde f = f^{-1}(I)$. Since $f^{-1}(I)$ is the kernel of a homomorphism from $R$, this shows that $f^{-1}(I)$ is an ideal of ...

1

Let $x\in R$ and $y\in f^{–1}(I)$ be, then $f(y)\in I$. But $f(xy)=f(x)f(y)$ then $f(xy)\in I$, and hence $xy\in f^{-1}(I)$. This complete the proof of the claim.

2

Since "rng" is standard for "ring without assuming an identity", "Boolean rng" seems a reasonable name that's likely to be understood, even though it doesn't seem to have been used much in the literature.

2

I figured it out --- all my dreams were true. $\Phi$ is injective Since $e$ is full we have $ReR=R$. Now for any ideal $I$ of $R$ we have $$I=RIR=(ReR)I(ReR) = (Re)I(eR) = R(eIe)R$$ thus observing that $eIe$ completely determines $I$. So $I\mapsto eIe$ is injective. $\Phi$ is multiplicative We want to show that $eIJe =(eIe)(eJe)$. But this is again ...

0

You can also prove this in the following way: let $T = \{ a \in A \hspace{1mm} | \hspace{1mm} a \text{ of finite order}\}$ be the so-called torsion subgroup of $A$. Observe that the set $I$ that you describe is precisely the kernel of the following ring homomorphism $\phi: \text{End}(A) \rightarrow \text{End}(T), \hspace{1mm} f \mapsto f|_{T}$, where ...

1

Here's a different approach. Since prime elements are irreducible, it suffices to show that $4+i$ is prime. (In fact, since $\mathbb{Z}[i]$ is a UFD, then an element is irreducible iff it is prime.) Recall an ideal $I$ of a commutative ring $R$ is prime iff $R/I$ is a domain. Since $\mathbb{Z}[i] = \mathbb{Z}[x]/\langle x^2 + 1 \rangle$, then ...

0

For a ring to be a subring it does not need to have the unit element. It only has to have the zero element, closure under addition and multiplication and an additive inverse, meaning a+(some element)=(zero element). If k={0} it has the zero element of the integers, it is closed under Addition and multiplication and 0+(-0)=0.

3

That is not the way one usually addresses this kind of questions. The key tool is the norm map $N\colon\mathbb Z[i]\to \mathbb Z$ which maps $a+bi\mapsto a^2+b^2$. This has the property that $N(xy)=N(x)N(y)$ and it allows you to translate your question into a question about integers. So if there are $A,B\in\mathbb Z[i]$ s.t. $4+i=AB$ then ...

3

This is not necessarily the case. For example, $\mathbb{Q}(\sqrt{17})$ forms a star ring with absolute value under the conjugation $(a+b\sqrt{17})^*=a-b\sqrt{17}$ and the standard (Euclidean) absolute value, but certainly $|1+\sqrt{17}|\neq |1-\sqrt{17}|$.

0

Units in $\mathbf Z/4\mathbf Z[x]$ remain units modulo the maximal ideal of $\mathbf Z/4\mathbf Z$, i.e. in the ring of polynomials over the field $\mathbf Z/2\mathbf Z$. Over any integral domain, a polynomial is a unit if and only if it is a unit constant. So they must be non-zero constants modulo $2$. As all elements of the maximal ideal $1\mathbf ... 1 It suffices to show that$(a^{-1}+b^{-1})^{-1}\in V\cap W$if$a\in V\setminus W$and$b\in W\setminus V$. If, say,$(a^{-1}+b^{-1})^{-1}\not\in V$, then$a^{-1}+b^{-1}$is in the maximal ideal of$V$. But since$b\not\in V$,$b^{-1}$is also in the maximal ideal of$V$, so this would imply that$a^{-1}$is in the maximal ideal of$V$. This contradicts ... 4 The key here is that the real numbers form a commutative ring. Therefore $$|xy||x^{-1} - y^{-1}| = |x||y||x^{-1} - y^{-1}| = |x| |x^{-1} - y^{-1}||y| = |xx^{-1}y - xy^{-1}y| = |y-x|.$$ (P.S. I assumed that by normed ring you meant Banach algebra, which puts an inequality ; I only put equality because you did. Was I correct? If the answer is not helpful ... 9 Yes, this is a monad. Much more generally, if$\mathcal{C}$is a monoidal category and$M$is a monoid object in$\mathcal{C}$, then the functor$P(R)=M\otimes R$is a monad using the monoid structure of$M$. In this case,$\mathcal{C}=\mathbf{Ring}$, the monoidal structure is$\otimes_\mathbb{Z}$, and$M=\mathbb{Z}[x]$(note that a monoid object in ... 0 Let$x\notin Rad(I)$. By definition of$Rad(I)$you know$S=\{1,x,x^2,x^3,\ldots\}$is a multiplicatively closed set disjoint from$I$. At this point, I highly recommend you get your head around the idea in item 1 of this post. It is a standard bit of commutative algebra that you should not avoid. It is literally Theorem 1 on page 1 of Kaplansky's ... 3 Let$B$be the set of monomials in$x_1,\ldots,x_n$.$R$is merely a free$\Bbb Z$-module with basis$B$. The action of$G$corresponds to permutations on$B$, let$B/G$be the set of orbits of$B$under$G$. There is an obvious map$B/G \to R$(obtained by summing the monomials of the orbits together). If you remove from$B$one representant of each orbit ... 1 Just to add to completeness of the other answers let me find minimal polynomial of$\alpha = \sqrt 2 + \sqrt 3$over$\mathbb Qdirectly. \begin{align} \alpha = \sqrt 2 + \sqrt 3 &\implies \alpha - \sqrt 2 = \sqrt 3\\ &\implies \alpha^2 -2\alpha\sqrt 2 + 2 = 3\\ &\implies \alpha^2 - 1 = 2\alpha\sqrt 2\\ &\implies \alpha^4 - 2\alpha^2 + 1 = ... 3 Letx=\sqrt 2+\sqrt 3$. Evaluate$\frac{x+x^{-1}}2$. 5 Your belief that $$\def\Q{\mathbb{Q}} \Q(\sqrt{2}+\sqrt{3})=\{a+b(\sqrt{2}+ \sqrt{3}):a,b\in \Q\}$$ is wrong, as well as $$\Q(\sqrt{2}+\sqrt{3})=\{a+b\sqrt{2}+ c\sqrt{3}:a,b,c\in \Q\}$$ The second claim can be immediately dismissed, because this would mean that$[\Q(\sqrt{2}+\sqrt{3}):\Q]=3$, but clearly ... 5 The "usual" proof that$\Bbb Q(\sqrt{2},\sqrt{3}) \subseteq \Bbb Q(\sqrt{2}+ \sqrt{3})$), the other inclusion is obvious. It suffices to show that$\sqrt{2} \in \Bbb Q(\sqrt{2}+ \sqrt{3})$, since then$\sqrt{3} = (\sqrt{2} + \sqrt{3}) - \sqrt{2}$must be as well. Now$(\sqrt{2} + \sqrt{3})^3 = 2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3}= 11\sqrt{2} ...

0

A ring homomorphism from a field to any ring is necessarily injective (assuming that ring homomorphisms map $1$ to $1$), because its kernel is a proper ideal and a field has only $\{0\}$ as proper ideal. In this case it's also easy to verify it directly, because if $\phi(a+bi)$ is the null matrix, then necessarily $a=b=0$. The verification that $\phi$ is a ...

1

The canonical morphism is defined by $\phi(\overline p)=(\overline p,\overline p)$. (Sorry for using the same notation for the residue classes modulo $(f)$, $(g)$, and $(h)$.) To show that $\phi$ is well defined suppose that $\overline p=\overline q$ in $k[x]/(f)$, so $f\mid p-q$. Then $g\mid p-q$ and $h\mid p-q$ and you are done. The kernel of $\phi$ is ...

1

Typically I have seen it denoted $\mathbb{Q}[\sqrt[3]{2}]$. If it is an extension over $\mathbb{Q}$, then we consider it a vector space with scalars being elements of $\mathbb{Q}$. Given that it needs to remain a field it must be closed under multiplication, and so $(a+b\sqrt[3]{2})(c+d\sqrt[3]{2})$ must be able to be generated by the basis. Since ...

3

You're making things too complicate. ;-) Let $f\colon R/I\to R/J$ be a homomorphism. Consider the composition map $g=f\circ \pi\colon R\to R/J$ and write $g(1)=x+J$. Then, for every $r\in R$, $g(r)=g(1r)=g(1)r=xr+J$. Since $g(r)=0$, for every $r\in I$, we know that $xI\subseteq J$, so that $x\in (J\mathbin:I)$. Now, ...

0

Hint: This is a ''canonical '' isomorphism between $\mathbb{C}$ and a subring of $M_2(\mathbb{R})$ (here). to show this fact the first step is to prove that the set $\mathcal{C}$ of matrices of the form $$\begin{bmatrix} a&b\\ -b&a \end{bmatrix} \qquad a,b \in \mathbb{R}$$ is a field. The associative and distributive properties of sum and ...

0

I think I have solved it myself. We know that we must find $3$ prime ideals with norm $3$ and one with norm $5$. We start with norm $5$, and see that $3\cdot 5 + 1\cdot(-4 + \sqrt{-14}) = 11- \sqrt{-14}$ which means that $(5,-4 + \sqrt{-14}) \supset (11- \sqrt{-14})$ and therefore $(5,-4 + \sqrt{-14}) | (11- \sqrt{-14})$ which covers the norm $5$ part. ...

0

Indeed, after performing your division, you may write $$2x^2+3x+2=2\times(x^2+1)+3x$$ Hence taking modulus $(x^2+1)$, you will have $2x^2+3x+2 = 3x\pmod {x^2+1}$.

0

In $A=\mathbb{R}[x]/(x^2+1)$ you consider polynomial of form $f (x) + (x^2+1)$. In particular 2 polynomials $f, g\in \mathbb{R}[x]$ are equal in $A$ if and only if $x^2-1$ divide $f (x)-g (x)$ , in this case you write $(x^2+1)+ f (x)= (x^2+1)+ g (x)$. For example, the polynomials $x$ and $1$ are not equivalent in $A$.

1

By def'n of quotient rings, (x+A)(y+A) = xy+A = 0+A = A Since D is an integral domain, ab=0 implies a=0 or b=0. This step doesn't work. You know that $xy+A=0+A$, but that doesn't imply $xy=0$; it just implies that $xy\in A$. In fact, the result is not true. To find a counterexample, I suggest looking at various choices of ideals $A$ in the ring ...

1

Here's an answer if you meant $Z/24Z$. That's where you need the fact that 4 and 6 aren't relatively prime. For $Z/12Z$ the other answers are just fine - the sets aren't even the same size. Now $Z/24Z$ is cyclic of order 24, while $Z/4Z \times Z/6Z$ doesn't have an element of order 24. (Can you prove that?)

0

An isomorphism between groups is a bijective homomorphism. Without even stopping to consider the homomorphism part, a prerequisite condition for finite sets (as mentioned by John above) is that they must be of the same cardinality/size. In this case, $\mathbb{Z}/12\mathbb{Z}$ has $12$ elements, while $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$ has ...

0

If there are 12 elements in the first group and 24 in the other, how could there exist a bijection?

0

Yes of course. This is by definition of "isomorphism".

3

No, this is not necessarily true. For instance, suppose $K=\mathbb{Z}$, $R=2\mathbb{Z}$, and $[\cdot]_1=[\cdot]_2=[\cdot]_3=2\mathbb{Z}$. Then $[\cdot]_1\cdot[\cdot]_2=[\cdot]_3$, but if $a\in[\cdot]_1$ and $b\in[\cdot]_2$, then $ab$ must be divisible by $4$, not just by $2$. In particular, $2\in[\cdot]_3$ but cannot be written as such a product $ab$.

1

Yes. More generally, if $D$ is a UFD then for all $a,b\in D$ you have $\langle a\rangle\cap\langle b\rangle=\langle\operatorname{lcm}(a,b)\rangle$.

1

Let $\mathfrak m$ be the maximal ideal of $A$. Since $\operatorname{height}\mathfrak m=2$ we have $\mathfrak m\notin\operatorname{Ass}_A(A/(a))$, so $\mathfrak m$ contains a non-zero divisor on $A/(a)$, in other words, $\operatorname{grade}\mathfrak m\ge2$. (In particular, this shows that $A$ is Cohen-Macaulay.) Now let $\mathfrak ... 2 It seems like, by$K$-module, you are really talking about vector spaces, and they are isomorphic if and only if they each have a basis of the same cardinality. The size of the basis for$K[x]$is easily seen to be the cardinality of the natural numbers. Can the basis of$K[[x]]$be the cardinality of the natural numbers? It's not clear why you are ... 0 As darij grinberg showed this statement is generally not true. However we can prove it, if we require that$7\not| \operatorname{Char}(A)$. For in this case we can do the following: $$-7=7ab=7ba\implies7(ab-ba)=0$$ Now say$\alpha:=ab-ba$. We want to prove that$\alpha=0$. Now since$7\not|\operatorname{Char}(A)$we can find$x,y\in\mathbb Z$such that ... 2 Here is my answer. Assume that$A$is not artinian. Then there exists a maximal ideal not contained in the union of all minimal primes. Hence there exists an element$a$not invertible and not contained in any minimal prime. Consequently if$ab=0$then$b$is contained in all minimal primes and is therefore nilpotent, contradicting the hypothesis. 0 If$n$is not a prime, say$n=ab$with$a,b>1$, then$q_a(x)=\frac{x^a-1}{x-1}$is a divisor of$p(x)=\frac{x^n-1}{x-1}$, since every root of$q_a$is a root of$p$, and there are no repeated roots. 3 The key is the repeated application of the geometric summation formula $$\sum_{i=0}^{z-1} q^i = \frac{q^z - 1}{q-1}.$$ Plugging in$z=n, q=x$and assuming$n = kl$, we get $$p(x) = x^{n-1} + x^{n-2} + \ldots + 1 = \frac{x^n - 1}{x-1} = \frac{x^{kl} - 1}{x-1}.$$ Geometric summation with$z=l$and$q = x^k$yields $$\ldots = \frac{x^k - 1}{x-1} ... 1 Why is f_1(x)=x^3+1 not exactly the same as f_2(x)=x+1? f_1(0)=1=f_2(0),f_1(1)=0=f_2(1)[...] Then the only irreducible polynomials in \Bbb F_2[x] are x,x+1,0,1? You have discovered that the two polynomial expressions produce identical functions, but the 'correct' definition of equality of polynomial so is 'the ordered list of coefficients ... 2 The ring \mathbb{F}_{2}[x] does not contain functions; you cannot really argue that objects in this ring are equal because they define the same function. The element x is to be considered as an indeterminate, that is, it is a transcendental element (satisfies no algebraic equations). If you want to consider the polynomials as functions, you can work in ... 1 In the ring \mathbb{F}_2[x] (or even any polynomial ring), much part of the game lies in playing with "coefficients". Thus it is 3x+1=x+1 because the "coefficients" 3 and 1 are same in \mathbb{F}_2; but not x^3+1=x+1. 3 I think it's a typo, it should be x\mapsto x+1, as you said. This is an homomorphism: if p(x),q(x)\in \Bbb Q[x], then \phi(p(x)q(x)) is just the product pq evaluated in x+1, which is the same as p(x+1)q(x+1), i.e., \phi(p(x))\phi(q(x)). This is essentially the fact that the evaluation map is a ring homomorphism (whether you're ... 1 It turns out that all such rings have a multiplicative unit, so there are no further examples besides the ones already listed in Matthew Pressland's answer. Proof: Suppose R is a commutative ring (with or without multiplicative identity) of four elements, with a unique non-zero zero divisor x. Then there exists an element y\in R such that xy = 0 ... 1 A polynomial is of the form$$ f = \sum_{i = 0}^n a_i x^i$$So if f(1) = 1, we have$$1 = f(1) = \sum_{i = 0}^n a_i$$But now$$f(11) = \sum_{i = 0}^n a_i 11^i = \sum_{i = 0}^n a_i (10+1)^i $$If we look at the binomial expansion of (10+1)^i \mod{100} we see that$$(10+1)^i = \sum_{j = 0}^i {i \choose j} 10^j = \sum_{j = 0}^1 {i \choose j} 10^j = 1 + ... 0 Hint$a$is a double root of a polynomial if and only if$a$is a root for$gcd( P(X), P'(X))$. This gives you an algorithm to find if the are any... 4 A double root is a root of the derivative, i.e. a root of$X^6+1$. But if$X^6=-1$,$X^7+7X+1 = 6X+1$, hence there are no double roots. 2 Suppose there exists an$f\in Z_{100}[x]$such that$f(1)=1, f(11)=17$. We have:$\mod \rm x-y: x-y\equiv 0 \iff x\equiv y$. By the polynomial congruence rule (credit to @Bill Dubuque) we have: $$\rm x\equiv y \pmod r \implies f(x)\equiv f(y) \pmod r$$ Taking$\rm x=11,y=1,r=x-y$we get:$\$ 11\equiv 1\pmod{10}\implies f(11)\equiv f(1)\pmod{10}\iff ...

Top 50 recent answers are included