# Tag Info

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$4 You have,$(a-b)(a^2+b^2)=a^3+ab^2-ba^2-b^3=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$p\ne2we have \begin{align}p\mid u^2-u+1\ \hbox{for someu$}\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$p}. \end{align} Using Legendre symbols and supposingp\equiv1\pmod4$, 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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