New answers tagged

2

If evaluation $\text{ev}_r : R[x] \to R$ were a homomorphism, then it would have to satisfy $$\text{ev}_r(x \cdot a) = \text{ev}_r(ax) = ar = \text{ev}_r(x) \text{ev}_r(a) = ra$$ for every $a \in R$: that is, $r$ would have to be central. And in fact $x$ is central in $R[x]$, so there is an evaluation homomorphism $\text{ev}_r : R[x] \to R$ precisely when ...


0

We can factor: $f(x)=(x^6-1)(x^2+2x+6)$. We are looking for an $n$ such that $\phi(n)=6k$ and such that $-5$ is a quadratic residue $\pmod n$. $21$ makes the job. In fact $\phi(21)=12$ and thus $x^{12} \equiv 1 \pmod {21}$ has 12 solutions, in particular $6$ are good for us! Solving the polynomial of degree $2$ we see that we would love to have a square ...


2

I think n=43 works. The roots are 1, 6, 7, 36, 37, 42, 8, and 33. The first observation is that this polynomial factors over $\mathbb{Z}$ as $(x^6-1)(x^2+2x+6)$. We can actually go further and factor $(x^6-1)$ more, but I don't think it's necessary. The next simplification is to just look for $n$ prime, in particular then the multiplicative group ...


1

Hint We can proceed naively but efficiently. It is plausibly useful to factor $f(x)$ over $\Bbb Q$: $$f(x) = (x^6 - 1)(x^2 + 2 x + 6),$$ and recalling our cyclotomic polynomials, we can easily factor the first factor, giving, respectively, $$f(x) = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1)(x^2 + 2 x + 6).$$ Obviously, we need $n \geq 7$. On the other hand, ...


1

Use the best definition of prime and maximal ideals: an ideal $I$ is prime if $R/I$ is an integral domain, and maximal if $R/I$ is a field. The natural bijection between ideals containing $I$ and ideals of $R/I$ respects taking quotients. (The best definition is, among other things, the shortest way to see why preimages of prime ideals are prime ideals, ...


2

Let $\;I\;$ be an ideal contained in both $\;\mathfrak a\,,\mathfrak b\;$ . Then it is also contained in their intersection $\;\mathfrak a\cap\mathfrak b\;$ , and this proves the maximality.


0

Hint: Use the fact that if $\lim_{x\to 1^{-}} g(x)=a$ and $\lim_{x\to 1^{-}} f(x)=b$, then you have $\lim_{x\to 1^{-}} (g\cdot f)(x)=ab$. Also note that in the ring of functions from $[0,1]\to \mathbb{R}$, there are problematic functions.


0

Indeed, let $f(x)=1-x$ and $g(x)=\begin{cases}0,&x\in\Bbb Q\\\frac1{1-x},&x\notin\Bbb Q\end{cases}$. Then $f\in I$ but $g\cdot f\notin I$.


0

A model-categorical enhancement of D(k-filt) is constructed in arXiv:1602.01515, see Corollary 3.58 there. It also shows that this model category is combinatorial and compactly generated, as requested. A morphism f: k→k' of Z-filtered rings induces a Quillen adjunction, as shown in Theorem 5.5 there. In fact, if f is a graded equivalence, then the Quillen ...


1

We can assume $f$ has degree $\ge 2$. Suppose that $f$ and $f'$ are relatively prime as polynomial over $K$. Then (Bezout) there exist polynomials $u(t)$ and $v(t)$, with coefficients in $K$ such that $uf+vf'=1$. But then $f$ and $f'$ cannot have a common root in any field extension of $K$, for such a root would have to be a root of the polynomial $1$. In ...


1

Given a root $\alpha$ of $f(x)$, $f(x)$ (divided by his director coefficient but who cares, we have coefficients in a field) is the minimal polynomial of $\alpha$. Thus there cannot be another polynomial with lower degree such that $p(\alpha)=0$. If $\alpha$ is a double root for $f(x)$, then $f'(x)$ is again a polynomial in $K[x]$ and of course ...


1

As $f$ and $f'$ are not relatively prime, there exists a non-constant prime polynomial $p$ such that $p \mid f$ and $p \mid f'$. Since $p \mid f$, there exists a polynomial $h$ such that $f=ph$. Thus, by product rule $f'=p'h+ph'$. Since $p \mid f'$ and $p \mid ph'$, $p \mid p'h$. By definition of the derivative, $p'$ has a lesser degree than $p$ and thus $p ...


1

You are right that for $A,B\in T$, we need $C\in T$ with $A\cup B\subset C$. But we get this automatically because of the total order, since either $A\leq B$ or $B\leq A$ and this order is defined by inclusion.


1

Well, first you have to see that if $x,y\in J$, then $\exists I'\in T, x,y \in I'$, this is true becasuse T is a totally ordered subset of K. Therefore, $x+y\in I'\subset J$. Then take $x \in J$, then $\exists I' \in T, x\in I'$, and $I'$ is an ideal, thus you have that $\forall a \in R, ax \in I'$ and $xa \in I'$. So, as $I'\subset J$, $\forall a \in R, ax ...


1

Well, yes, if you mean a local noetherian ring: the localisation of any (commutative) ring at a prime ideal $\mathfrak p$ is a local ring, with maximal ideal $\mathfrak pA_{\mathfrak p}$, and any ring of fractions of a noetherian ring is noetherian.


4

You can't, without the stronger hypothesis that $A$ is also a domain. For any Noetherian ring $A$ and ideal $I \subseteq A$, $I^{2} = I$ if and only if $I$ is a principal ideal generated by an idempotent. This is a consequence of Nakayama's lemma, and has been covered before on this site: see Georges Elencwajg's excellent answer here, for instance. The only ...


3

when are ideals also rings with unity? Proposition: An ideal $I\lhd R$ will be a ring with identity iff there exists a central idempotent $e$ such that $eR=I$. Proof: ($\implies$) The identity of $I$, call it $e$, is an idempotent element of $R$ and satisfies $I=eI$. Then $I=eI\subseteq eR\subseteq I$, so $I=eR$. Since $e\in I$, we have ...


1

If $R$ is a ring with unity everything works fine. You have an action of $\mathbb{Z}$ on $R$ namely if $m\in \mathbb{Z}$ we define $mr$ as you said above. This is because every ring has an abelian group structure with the sum. Returning to your question, if you avoid a morphism that sends everything to zero the rest of homomorphisms are determined by the ...


1

If $R$ is a ring then clearly is generated as a $R$ Module by $1\in R$. Considering $R^n$ the direct sum of n copies of $R$ we know that there are canonical injections of $i_i: R \hookrightarrow R^n$ sending $r \mapsto (0,0,..r,..)$. We denote $e_i=i_i(1)$ with this is clear that the set $\{e_i\}_{i=1}^n$ generates $R^{n}$ For your question, since you have ...


2

You're absolutely right that (ii) does not obviously follow from (i) as stated. What (i) should say is not just that $I_i$ is ring-isomorphic to $M(n_i,F)$ but that it is isomorphic to $M(n_i,F)$ as an $F$-algebra. Concretely, this means that if you take an element $a\in F$, consider it as an element of $FG$, project it to $I_i$, and then map it to ...


4

Yes, such rings are exactly the integral domains, i.e. the rings which are commutative and in which $xy=0$ implies either $x=0$ or $y=0$. Any field is an integral domain (if $xy=0$ and $x\neq 0$, you can multiply by $x^{-1}$ to get $y=0$) and clearly any subring of an integral domain is an integral domain, so any ring that embeds in a field is an integral ...


0

Set $I=(x,y)^d$ in $K[x,y]$. Then $|V(I)|=1$ and $\dim_KR/I=\frac{d(d+1)}{2}$.


1

I've already mentioned, if $S/P$ is finite for all prime ideals $P\subset S$ then the same holds for $R$ since the extension is integral and hence it has the "lying over" property. Let $\mathfrak p\subset R$ be a prime ideal such that $|R/\mathfrak p|<D$, where $D>0$ is fixed. Then there is a prime ideal $P\subset S$ such that $P\cap R=\mathfrak p$. ...


1

We always have $y=1\cdot y\in (y)\subseteq (x)$. This proves that $(y)\subseteq (x)\implies y\in (x).$


4

(1) If $\mathbb{Z} \to \mathbb{Z}_p$ was an epimorphism, this would imply that $\mathbb{Q} \to \mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Q} = \mathbb{Q}_p$ is an epimorphism. But $\mathbb{Q}$ is a field and $\mathbb{Q} \to \mathbb{Q}_p$ is not surjective, so this is a contradiction. (2) I think that $R \to \widehat{R}_I$ is almost never an epimorphism. (3) ...


1

Here are two examples in the ring $\mathbb C[x,y]$: Consider $I=(x-1)$ and $J=(x-1,y-1)$. Then $I:J=(x-1)$. Moreover $$Z(I:J)=Z(x-1)=Z(I)=\{(1,\alpha)|\alpha \in \mathbb C\},$$ while $$Z(I)\smallsetminus Z(J)=\{(1,\alpha)|\alpha \in \mathbb C\}\smallsetminus \{(1,1)\}.$$ Consider $I=((x-1)(x-2))$ and $J=(x-1)$. Then $I:J=(x-2)$. We have ...


1

a) For $I=(y), J=(x)\subset \mathbb C[x,y]$ we have $$(I:J)=I\quad \quad Z(I)=Z(I:J)=\mathbb C\times \{0\}\quad \quad Z(J)=\{0\}\times \mathbb C$$ and thus we get the equalities and strict inclusion $$Z(I:J)=\mathbb C\times \{0\}=\overline {\mathbb C^*\times \{0\}}=\overline {Z(I)\setminus Z(J)}\supsetneq\mathbb C^*\times \{0\}=Z(I)\setminus Z(J)$$ 2) ...


0

Let's show that $R$ contains the identity. Indeed, $y(1) = y(1 \cdot 1) = y(1) + y(1) \Rightarrow y(1) = 0$. Let now $\alpha, \beta \in R$ be arbitrary. First of all, $y(\alpha - \beta) \ge \min\{y(\alpha), y(-\beta)\}$. Furthermore, $y(-\beta) = y(-1) + y(\beta)$. Further, $y(1) = y((-1)(-1)) = y(-1) + y(-1)$ and hence $y(-1) = 0$. It follows $\alpha - ...


1

We have $$\mathbb{Z}[\sqrt{2}]/(1+3\sqrt{2})\simeq\mathbb Z[X]/(X^2-2,1+3X).$$ But $X+6\in(X^2-2,1+3X)$ since $X+6=X(1+3X)-3(X^2-2)$, so $$\mathbb Z[X]/(X^2-2,1+3X)\simeq\frac{\mathbb Z[X]/(X+6)}{(X^2-2,1+3X)/(X+6)}\simeq\mathbb Z/17\mathbb Z.$$


1

The rest is straightforward. $\phi$ is a surjection: If $(x,y)\in \Bbb R^2$, define $f\in \mathscr{F}$ by $$\begin{align} f(0) &= x \\ f(1) &= y \\ f(z) &= 0 \text{ if $z\ne x, z\ne y$}. \\ \end{align}$$ Then $\phi(f) = (x,y)$. The kernel of $\phi$ is all $f$ such that $\phi(f) = (0,0)$, as $(0,0)$ is the $\mathbf{0}$ of the ring $\Bbb R^2$. ...


2

Let $(x,y)\in\mathbb{R}^2$. We want to show that there exists a function such that $f(0)=x$ and $f(1)=y$. Define $f:\mathbb{R}\to\mathbb{R}$ piece-wise by these two expression, and let $f(z)=0$ at all other points. This is a function, so it's subjective. The kernel is exactly the set of functions that satisfy $f(0)=f(1)=0$


1

It's better to go the other way around: define the map $$ \chi\colon\mathbb{Z}\to R, \qquad k\mapsto k1_R $$ and prove that $\ker\chi=n\mathbb{Z}$. Since the image of $\chi$ is $\mathbb{Z}1_R$, the homomorphism theorem allows to finish up.


2

Suppose $x_1>\cdots>x_n$. Then $g_i=g_i(x_i,\dots,x_n)$ is a polynomial in $x_i,\dots,x_n$ for $i=1,\dots,n$. In particular, $g_n$ is a polynomial only in $x_n$. Let $\alpha_n$ be a root of $g_n$. (There are at most $d_n$ roots.) Then consider $g_{n-1}(x_{n-1},\alpha_n)$. This has at most $d_{n-1}$ roots, and let $\alpha_{n-1}$ be one of them. Now ...


1

You just need to apply the first isomorphism theorem now.


1

Suppose $S+T$ is a subring. Then, for every $s\in S$ and $t\in T$, $$ (s+0)(0+t)=st\in S+T $$ Thus you only need to find a counterexample to this situation. A simple one is given by $S=\mathbb{Q}(\sqrt{2})$ and $T=\mathbb{Q}(\sqrt{3})$ as subrings of $\mathbb{R}$. The sum is the $\mathbb{Q}$-subspace spanned by $1$, $\sqrt{2}$ and $\sqrt{3}$, which does not ...


2

Take $R = \mathbb{C}$ and consider the subfields $S = \mathbb{Q}[\sqrt{-1}] = \{a+bi \mid a,b \in \mathbb{Q}\}$ and $T = \mathbb{R}$. Then $S + T = \{a + bi \mid a \in \mathbb{R}, b \in \mathbb{Q}\}$. This contains $i$ and $\sqrt{2}$, but not $i\sqrt{2}$.


3

Ring $S$: the rationals, viewed as a subring of the reals; Ring $T$: the reals of the form $a+b\sqrt{2}$ where $a$ and $b$ are integers. Note that $\frac{1}{2}\cdot \sqrt{2}$ is not in $S+T$.


0

Let's be clear about what the bijection is between. There is a 1-1 correspondence $$ \{\mbox{Ideals $J$ s.t. $I\subseteq J\subseteq A$}\}\leftrightarrow\{\mbox{Ideals in $A/I$}\} $$ This bijection is induced by the canonical projection $\pi:A\to A/I$. Note that if $J$ is an ideal containing $I$, then $\pi(J)$ is an ideal in $A/I$. Also, if $K\lhd A/I$ is ...


1

They do indeed mean the map $\phi: A \mapsto A/I$ with inverse $A/I \mapsto A$ when they talk about this correspondence. They're saying that it's a bijection. The fact that it's inclusion preserving means that if $A \leq B$ in $R$ and both contain $I$, then $A/I \leq B/I$, and vice versa.


2

Is there any requirement that the two operations of a ring have to be related to each other, excluding the requirement of distributivity? The only requirements on what a ring is are given in the axioms for a ring, and if something does not appear there, there is no requirement for it. Are there any other examples of rings whose second operation is ...


0

The problem is the "only if" part of the latter statement. Consider the polynomial $x^4 + 2x^2 + 1 = (x^2+1)^2 \in \Bbb R[x]$. It has no root in $\Bbb R$, but still it is reducible. You are completely correct, though: The "if" part of the statement remains true. Any polynomial which has a root and degree greater than one is reducible over any field.


1

Here’s another argument: If $p\equiv1\pmod3$, then $\Bbb F_p^\times$ is a cyclic group of order divisible by $3$, and so has three cube roots of unity, call them $1$, $a$, and $a^2$. Then $(X+1)(X^2-X+1)=X^3+1=(X+1)(X+a)(X+a^2)$, and so $X^2-X+1=(X+a)(X+a^2)$.


1

$u^2 - u + 1 = (u^3+1)/(u+1)$, so $p \mid u^2-u+1$ iff $u$ is a non-trivial cube root of $-1$ modulo $p$, in other words $-u$ is a primitive cube root of unity. Let $g$ be any primitive root mod $p$, and then let $u = -g^{(p-1)/3}$.


0

Since $p\ne2$ we have \begin{align}p\mid u^2-u+1\ \hbox{for some $u$}\quad &\hbox{iff}\quad p\mid4u^2-4u+4\ \hbox{for some $u$}\cr &\hbox{iff}\quad p\mid(2u-1)^2+3\ \hbox{for some $u$}\cr &\hbox{iff}\quad \hbox{$-3$ is a square modulo $p$}. \end{align} Using Legendre symbols and supposing $p\equiv1\pmod4$, we have $$\Bigl(\frac{-3}p\Bigr) ...


1

Consider internal decomposition $\mathbb{Q}[G]=I\oplus J$. In $I$ there is an element $e=\frac{1+g+g^2}{3}$ (it is a Primitive central idempotent), and $I$ is two-sided ideal generated by this element. We can write $I=Ie$. Then $Ie$ becomes an algebra, in which additive identity is $0$ but multiplicative identity element is $e$; we can show that $Ie$ is ...


1

Write $S = R/J$ and $T = I/J$, then the Third Isomorphism theorem for rings immediately tells you: $S/T \cong R/I$ from which it follows that the quotient rings are integral domains together or fields together so that $I$ is a prime (maximal) ideal of $R$ if and only if $T$ is a prime (maximal) ideal of $S$.


0

Since $i^2=-1$, if we pass to a quotient where $i+1=0$, we must have $-1=(-1)^2=1$, so such a quotient contains $\mathbb{Z}/2$. In fact, such a quotient is exactly $\mathbb{Z}/2$, because $1\mapsto 1$ and $i\mapsto 1$, so any $a+bi$ is mapped to $a+b\in\mathbb{Z}/2$.


4

You could use $\mathbb{Z}[i] \cong \mathbb{Z}[x]/(x^2+1)$. Then $$ \mathbb{Z}[i]/(i+1) \cong \mathbb{Z}[x]/(x^2+1, x+1) \cong \mathbb{Z}[x]/(-x+1,x+1) \cong \mathbb{Z}[x]/(2,x+1) \cong \mathbb{Z}/2\mathbb{Z}[x]/(x+1)\cong \mathbb{Z}/2\mathbb{Z}. $$


8

Consider $a+bi$: we can write $a-b=2q+r$, where $q$ is an integer and $r$ is $0$ or $1$. Then $$a+bi=(b+2q+r)+bi=r+b(1+i)+q(1-i)(1+i)\equiv r\ .$$



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