New answers tagged

0

Here is a different geometric proof. We have two Gaussian integers $a$ and $b$, and we have to prove that there exists a Gaussian integer $z$ such that $$|az-b|<|a|$$ Well, let's consider the set $A=\{az\mid z\in\mathbb Z[i]\}$. What does it look like? Writing $z=x+yi$, we see that $az=xa+y(ai)$. But $ai$ is just $a$, rotated clockwise by $90$ degrees, ...


1

In these cases, considering the conjugate can help. The conjugate of $a+b\sqrt{2}$ is $a-b\sqrt{2}$. Now, if $(a+b\sqrt{2})(c+d\sqrt{2})=0$, also $$ (a+b\sqrt{2})(a-b\sqrt{2})(c+d\sqrt{2})(c-d\sqrt{2})=0 $$ and therefore $$ (a^2-2b^2)(c^2-2d^2)=0 $$ Since the integers form a domain, we conclude $a^2-2b^2=0$ or $c^2-2d^2=0$. The irrationality of $\sqrt{2}$ ...


1

Hint Suppose we have $$(a + b \sqrt{2})(c + d \sqrt{2}) = 0$$ from some $a, b, c, d \in \Bbb Z$. Expanding gives $$(ac + 2 bd) + (ad + bc) \sqrt{2} = 0,$$ and since $\sqrt{2}$ is irrational, the coefficients must vanish separately $$ac + 2bd = ad + bc = 0 .$$ Substituting $0$ for any of the four parameters gives quickly that $a = b = c = d = 0$, so we may ...


1

The main proof you mention is the easiest and the best since it generalizes very well. If you really want something more intrinsic, note $$(a+b\sqrt 2)(c+d\sqrt 2)=0\implies (a^2-2b^2)(c^2-2d^2)=0$$ But then if so, either $a^2=2b^2$ or $c^2=2d^2$, WLOG assume the former. Then $a$ is even, but then if the prime factorization of $a$ is $2^kp_1^{e_1}\ldots ...


3

Here is a solution that isn't satisfying because it isn't really deduced from the givens of the problem. Nevertheless, it completely explains this and many more weird operations of the same type. If you have any bijection of $f:\Bbb Z\to\Bbb Z$ (or from any ring into itself, for that matter) then you can define two new operations $$ a\oplus b := ...


0

Yes, it can be isomorphic to a “ring of numbers”; more precisely, to a subring of the complex numbers. It's easy to see that the ring $R$ is generated by $$ J=\begin{bmatrix}0&1\\-3&0\end{bmatrix} $$ in the sense that every element of $R$ can be written (in a unique way) as $aI+bJ$, for $a,b\in\mathbb{Z}$. Notice that $R$ is indeed a ring, because ...


0

I suppose $R$ and $S$ are local rings with maximal ideals $\mathfrak m$, respectively $\mathfrak n$. If $r_1,\dots,r_n$ is a sop for $R$, then the ideal $(r_1,\dots,r_n)$ is $\mathfrak m$-primary. We wonder if $(r_1,\dots,r_n)S$ is $\mathfrak n$-primary. First note that $\sqrt{(r_1,\dots,r_n)S}=\sqrt{\mathfrak mS}$. But $\mathfrak mS$ is $\mathfrak ...


0

Is it possible for a ring of matrices to isomorphic a ring of numbers? What hampers me the most here is that I have no idea what a "ring of numbers" is supposed to be. If it is supposed to be one of $\Bbb C$, $\Bbb R$, $\Bbb Q$ or $\Bbb Z$, or even $\Bbb H$, then yes, it is possible. For example, ...


1

1) $S$ is also a left ideal, in fact, if $r\in R $ and $I$ is a right ideal of $R$ so is $rI$ (Evident!). Let $x\in I$, we first observe that $(xr)^k=0$ for some integer $k$. Therefore, we have $(rx)^{k+1}=0$, showing that $rI$ is a nil right ideal of $R$. Now, if $y\in S$ we can write $y=y_1+...y_n$, where each $y_j$ belongs to some nil right ideal $I_j$ ...


0

As Mathematician 42 has already answered, the cardinality argument shows that $R$ cannot be isomorphic to $\mathbb{C}$. In fact, $R$ is isomorphic to $\mathbb{Z}[i\sqrt{3}]$ via the map $a + bi\sqrt{3} \mapsto \begin{pmatrix} a & b \\ -3b & a \end{pmatrix}$. It can be easily shown that the multiplication is just the same: ...


1

Hint: $\mathbb{Z},+$ is generated by $1$. Hence a group morphism $f:\mathbb{Z},+\rightarrow \mathbb{Z}^*,\circ$ is completely determined by $f(1)$.


1

No, matrices in $\mathbb{Z}$ are countable whereas $\mathbb{C}$ is not. If you allow matrices with entries in $\mathbb{C}$ then the $1\times 1$-matrices do the job.


1

Yes, $d(n)$ can attain any positive integer for some $n$, consider $n=2^k$. Hence all ring numbers are of the form $d(n)$.


1

Take $R = \mathbb{F}_2[X]/(X^n)$. Then $R$ is finite and has exactly $n+1$ ideals. Indeed, ideals of $R$ are in canonical bijection with ideals of $\mathbb{F}_2[X]$ containing $X^n$, ie with polynomials dividing $X^n$ : these are the $X^k$ for $0\leqslant k\leqslant n$.


0

a) $6$ is a zero. $(x-6) = (x+1)$. So $\frac {x^3 +4x^2 +5x +2}{x+1} = (x+1)(x+2)$ b) $x^2 + x +8 = x^2 -9x + 8 = (x-1)(x-8) = (x+9)(x+2)$.


5

We're in luck because this is a very small field. We can just test out all the values. Obviously $0$ is not a root. $f(1) = 1 + 3 + 5 = 9 = 2 \neq 0$. $f(2) = 8 + 6 + 5 = 19 = 5 \neq 0$. $f(3) = 27 + 9 + 5 = 41 = 6 \neq 0$. $f(4) = 64 + 12 + 5 = 81 = 4 \neq 0$ $f(5) = 125 + 15 + 5 = 145 = 5 \neq 0$ $f(6) = 216 + 18 + 5 = 239 = 1 \neq 0$. So there are ...


2

No, $x$ is not a unit; there is no polynomial $p(x)$ such that $xp(x)=1$. This is because $\deg xp(x)=1+\deg p(x)\ge 1$. Actually, one shows that, for a polynomial of positive degree to be a units, it is necessary that its leading coefficient to be a zero-divisor, which implies $n$ to be composite. A simple example: in $\mathbf Z/4\mathbf Z[x]$, we have: ...


2

Much of this would be covered in "basic ring theory". Let's say that $R$ is a commutative ring with unity (although more general settings can be accommodated, it seems like a good place to start). As you studied "polynomials" in calculus you were asked to consider them as functions of one or more variables. In abstract algebra there is more of an emphasis ...


1

Yes, I'm pretty sure that's definition is correct. Basically, this means that the ring has commutative addition, commutative multiplication, and there's no way you can multiply two non-zero polynomials and get $0$. Any polynomial that has an $x$ term in it can not be a unit. There is no way you can multiply something with an $x$ in it and then get $1$. It ...


1

Well if $f$ is an isomorphism, let $u\in R^*, v\in S^*$. Then $$1_S = f(1) = f(uu^{-1})=f(u)f(u^{-1})$$ so $f(u)$ is invertible with inverse $f(u^{-1})$ and so $f(R^*)\subseteq S^*$. But then $$1_R=f^{-1}(1_S)=f^{-1}(vv^{-1})=f^{-1}(v)f^{-1}(v^{-1})$$ so $S^*\subseteq f(R^*)$ proving the result.


0

Use that for the complex conjugate we have $$(1)\;\overline{z+w}=\overline z+\overline w\;,\;\;(2) ;\overline{wz}=\overline w\overline z\,,\;\;(3)\;\;\overline z=z\iff z\in\Bbb R$$ So that $$p(\overline z)=\overline{p(z)}=\overline0=0$$


1

Yes, $x$ being a zero of $f$ just means that $f(x) = 0$. As for what a polynomial ring $R[x]$ is, it's simply creating a ring (in the usual way) by taking a base ring $R$, and then using it to create polynomial in $x$ with coefficients from $R$. So the elements of $R[x]$ are simply polynomials $f(x) = \sum_{i=0}^{n} a_i .x^i$, where $n$ is some finite ...


8

Lemma 1: The only unit is the identity. If $e$ is the leftmost idempotent in a factorization of a unit $u$, then $(1-e)u=0$ implies $1-e=0$ and $e=1$. After finitely many steps we have proven $u=1$. Lemma 2: The ring is reduced, that is, it has no nonzero nilpotent elements. The only nilpotent element is $0$. If $x$ is nilpotent, then $1-x$ is a ...


0

The map certainly defines an endomorphism $\phi$ of $R[x]$. And $x\mapsto a^{-1}(x-b)$ is also an endomorphism $\psi$ which apparently is the two-sided inverse of the given automorphism. However, in general $\phi$ is not idempotent. Instead $\phi^n(x)$ will be of the form $a^n x+b_n$, and there is no reason event that $a^n=1$.


1

Let $R$ be a commutative ring and $f\in R[X]$. Then $f$ is a zerodivisor if and only if there exists $a\in R$, $a\neq 0$ such that $af=0$. Let $f=a_0+a_1X+\dots +a_nX^n$ with $a_n\neq 0$. If $f$ is a zerodivizor, then there exists $g\in R[X]$, $g\neq 0$ with $fg=0$. Choose $g$ of minimal degree with this property. Set $g(X)=b_0+b_1X+\cdots+b_mX^m$, ...


1

The most important property of polynomial rings is the following: given a ring homomorphism $\varphi\colon K\to R$ and $r_1,\dots,r_n\in R$, there exists a unique ring homomorphism $\hat{h}\colon K[X_1,\dots,X_n]$ such that $\hat{\varphi}(k)=\varphi(k)$, for $k\in K$, and $\hat{\varphi}(X_i)=r_i$, for $i=1,\dots,n$. Here $R$ is any commutative ring. Now ...


1

If for $\mathbf b=(b_1,\ldots, b_n)\in K^n$ we denote with $\phi_{\mathbf b}$ the endomorphism of $K[X_1,\ldots,X_n]$ given by $\phi_{\mathbf b}(X_i)=X_i+b_i$, then we see that $\phi_{-\mathbf b}\circ\phi_{\mathbf b}=\phi_{\mathbf b}\circ\phi_{-\mathbf b}=\operatorname{id}$ because these compositions map $X_i\mapsto X_i+b_i\mapsto X_i$ (resp. $X_i\mapsto ...


3

The most modern term for this is $R$ satisfies the $SI$ condition This is seen in papers by Greg Marks, which I have found to be the most thoughtful and comprehensive recent papers discussing these things. For example , see A taxonomy of 2-primal rings. Earlier works used the following terms: $R$ is zero-insertive (zi) $R$ satisfies the ...


1

A finite commutative ring (indeed, any artinian commutative ring) is a finite direct product of local rings, and a finite direct product of principal ideal rings is a principal ideal ring, so it's enough to prove that a finite local commutative ring with all ideals of different sizes is a principal ideal ring. Let $R$ be such a ring. Since it is local, up ...


1

A right ideal of a ring with identity which is a direct summand is always idempotent: if it is $eR$, then $e=e^2\in (eR)^2$, so $eR\subseteq (eR)^2$, and $(eR)^2\subseteq eR$ trivially. (Are you really asking about PI rings without identities?)


0

Let $F[x]=\mathbb{F}_p[x]/(p(x)) = \{h(x): \deg(h) \leq p(x)\}$. If we let $n = deg(p)$ then we have all polynomials of the form; $$a_{n}x^{n}+ a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \in F[x]$$ Now its just a simple counting argument to say that you have $q^n$ polynomials.


1

Yes, I think that argument works. I should add that in the general case, every finite field is $\mathbb F_{q^n}$ for some $n$, so as soon as you knew as it was a field and contained $\mathbb F_q$ you were in principal done. So you don't even need $p$ to be irreducible here! (You wouldn't necessarily have that $n = \deg p$, in that case.)


2

I have seen these rings discussed, but never a never with a name attached. The reason is probably because this class of rings is a disjoint union of the following two classes of rings which do not need complicated names and which have rather divergent properties compared to each other: Rings of Krull dimension 0 Domains of Krull dimension 1


1

In a PID, $R$, every ideal is principal. That is every ideal is of the form $(a)$ for some $a \in R$. So let $(a)$ be a proper ideal in $R$. PIDs are unique factorization domains, so we can write $a = p_1p_2 \cdots p_n$ for some prime/irreducible elements $p_1,p_2,\dots, p_n \in R$. The ideals $P_i = (p_i)$ are prime ideals. Claim: $(a) = P_1P_2\dots ...


1

The book "Introduction to commutative algebra" by "Atiyah-Macdonald" is good terse text. But since you want a Book which is easy to understand, the book "Steps in Commutative Algebra" by "R.Y.Sharp" would be better for you, (it has more explanation). Maybe the book Monomial Ideals by "Herzog-Hibi" is the best for you, if you want to see examples of ...


1

Your answer is right. You only have to observe that $<x -a_i>$ is maximal or, equivalently, that $x-a_i$ is irreducible. But it follows, since $\frac{\mathbb{C}[x]}{(x-a_i)}$ is isomorphic to $\mathbb{C}$


4

The first three lines of your attempt are perfectly fine. The fourth line is - strictly speaking - also correct (since we will have $n=1$ anyway, and then it is correct a posteriori), but you lack an argument. Notice that $M=\langle p(x) \rangle \subset \langle x-a_1 \rangle$. But $M$ was assumed to be maximal, hence we must have equality, so $M=\langle ...


0

$\newcommand{\Ann}{\operatorname{Ann}}$ The point here is that you need to see you can take two representatives of the same class, $r+\Ann(A)=s+\Ann(A)$ and see that their product by $a$ is the same. The important fact to note here is that for an ideal $I$, $r+I=s+I$ if and only if $r-s\in I$. So in our case, $r+\Ann(A)=s+\Ann(A)$ means $r-s\in\Ann(A)$, so ...


0

More generally, a $R$-module $A$ is also an $R/I$-module in the natural way defined above, iff $IA=0.$ Certainly $\operatorname{Ann}(A)A=0$, hence we can let $I=\operatorname{Ann}(A)$.


1

If you assume that the operation from $R/Ann(A) \times A$ to $ A$ is well defined, then A will be an $R/Ann(A) $ module (it follows from the fact that $A$ is an $R $ module, do you see why?). Now, you need to prove that the operation is well defined. Take two elements $r_1 , r_2 \in R$ that lie in the same equivalence class in $R/Ann(A)$. Then we must have ...


1

Hint. Compute $(re-ere)^2$ for $e,r\in R$, $e$ idempotent.


0

Given a trilinear map $\mu:M_1\times M_2\times M_3\to T$, for each $m_3\in M_3$ the map $(m_1,m_2)\mapsto \mu(m_1,m_2,m_3)$ is an $(A_0,A_1,k)$-bilinear map $M_1\times M_2\to T$. There is thus a unique map $\nu:(M_1\otimes_{A_1} M_2)\times M_3\to T$ such that $\nu(m_1\otimes m_2,m_3)=\mu(m_1,m_2,m_3)$ and $\nu$ is $A_0$-linear in the first variable. It is ...


0

Do you know these facts? $A_p$ is local ring with $p A_p$, maximal. $aA_p$ is an ideal of $A_p$ which can be generated by $n$ elements. $p A_p$ is minimal over $aA_p$. $\operatorname {ht} pA_p= \operatorname {ht} p.$ In a local ring $(R,m)$, if $m$ is minimal over an ideal $J$, then $\sqrt J = m$. Now in order to prove that $\operatorname {ht} p ...


4

Recall the ideal correspondence for quotient rings: The ideals of a ring of the form $R/I$ are ideals of $R$ which contain $I$. This is just like the subgroup correspondence for quotient groups. So you are right to start off by noting the maximal ideals of $k[x,y]$ are of the form $(x-a, y-b)$ by the Nullstellensatz. Now any maximal ideal $m$ of your ring ...


0

We don't say that an ideal under multiplication or under addition. An ideal $I$ of a ring $R$ is a subring of $R$ which satisfying $ri\in I$ and $ir\in I$ for all $r\in R$ and $i\in I$. In general, the principal ideal generated by $a\in R$ is defined by $$\langle a\rangle=\left\{ra+as+ma+\sum_{i=1}^{n}x_i a y_i\mid r,s,x_i,y_i\in R, n\in \Bbb{N}^+, m\in ...


2

Hint $$(-x)^3 - (-x) - 1 = -(x^3 - x + 1) .$$


0

You can do it as you suggest. Consider say $(2,0)$ and note that after whatever multiplication the second coordinate will still be $0$. Thus, you can never get $(1,1)$ the identity with respect to multiplication. Note that the subset $\{(q,0) \colon q \in \mathbb{Q}\}$ would be a field, yet with a different identity element namely $(1,0)$. (This however ...


2

You know that in a field $\;ab=0\iff a=0\;\;or\;\;b=0\;$ . Now try with $\;(1,0)\;,\;\;(0,1)\;$ in your case


5

$(1, 0) \ne (0, 0) \ne (0, 1)$, but $(1,0) \cdot (0,1) = \dots$


0

A source of radical but not prime: Let $K$ be a field, and let $S = K[x_1, . . . , x_n]$ be the polynomial ring in $n$ variables over $K$. The monomial prime ideals are all of the form $(x_{i_1}, . . . , x_{i_k} )$. But A monomial ideal $I$ is a radical ideal, iff $I$ is a squarefree monomial ideal (This is because: $\{\sqrt u : u ∈ G(I)\}$ is a set of ...



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