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8

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


5

So $\phi$ is defined by $\phi(g) = (1,\omega)$. Hence we must have $$ \phi(1) = \phi(g^3) = (1, \omega)^3 = (1,1) $$ (Perhaps $(1,0)$ is a typo). For your second objection, note that $1+g$ mapsto $(2, 1+\omega)$, and $$ \phi(1 + g)^2 = (2,1+\omega)^2 = (4, 1 + 2\omega + \bar\omega) $$ On the other hand \begin{align*} \phi(1 + 2g + g^2) &= (1, 1) + ...


5

Not necessarily. The classical example of PID which is not Euclidean, $R=\mathbb Z[\frac{1+\sqrt{-19}}{2}]$ has the property that $R/I$ is finite for all $I\ne(0)$. (In order to prove this use a similar argument to the one given in this answer.)


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) ...


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 ...


4

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


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 ...


3

Suppose $3+2\sqrt{-2}=xy$. Then, using conjugates, you get $$ (x\bar{x})(y\bar{y})=(3+2\sqrt{-2})(3-2\sqrt{-2})=17 $$ Then… Note that an element $z$ is irreducible if it is not $0$, it is not invertible and, if $z=xy$, then either $x$ or $y$ is invertible.


3

$N(3+2\sqrt{-2})=(3+2\sqrt{-2})(3-2\sqrt{-2})=17.$ If $3+2\sqrt{-2}=a\cdot b$, then $N(a)N(b) = N(ab) = 17$, thus wlog $N(a)=1$, hence $a$ is invertible.


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 $S+T$.


3

It is an if and only if, prove both ways. If the characteristic is $2$, then $2 = 0$, so $(a + b)^2 = a^2 + 2 a b + b^2 = a^2 + b^2$ If $a, b \ne 0$ are such that $(a + b)^2 = a^2 + b^2$, then $2 a b = (a + b)^2 - a^2 - b^2 = 0$. As $a b \ne 0$, and it is an integral domain, it has no zero divisors, so $2 = 0$.


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 $i\sqrt{2}$.


2

Take $R$ to be the following subring of $\mathbb{Q}[x]$: $$ R = \{ P\in \mathbb{Q}[x] \ | \ P(0)\in \mathbb{Z} \}.$$ Then $2$ is a universal side divisor. Indeed, the only units of $R$ are $\pm 1$, and an element $P$ of $R$ is divisible by $2$ if and only if $P(0)$ is even, so $P$ or $P-1$ is divisible by $2$. However, $R$ is not Euclidean. In fact, it is ...


2

Hint: For the second portion, try a proof by contradiction: For non-zero divisors $a, b \in R$, suppose $(ab)c=0$ for nonzero $c$, then by associativity $a(bc)=0$...


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 ...


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 ...


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.


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 ...


2

An ideal will be a ring with identity iff it is a two sided ideal of the form $eR$ where $e$ is an idempotent and $eR(1-e)=0$. Clearly $e$ is the identity element, and in fact $eR=eRe$, the "corner ring" of e. This is all straightforward to see: the identity element is an idempotent, and acts as a left identity on $eR$, so we want it to be a right identity ...


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

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 $\Bbb R^2$. ...


1

The isomorphism you have written down is just the Chinese remainder theorem. You have an isomorphism $$\phi:\mathbb{Q}[x]/(x^3-1)\to\mathbb{Q}[x]/(x-1)\oplus\mathbb{Q}[x]/(x^2+x+1),\;\;\; f+(x^3-1)\mapsto (f+(x-1),f+(x^2+x+1)).$$ I will omit the coset notation for convenience. Under this map, we have $$\phi(x^2+x+1)=(3,0)$$ since $x^2+x+1\equiv 1^2+1+1=3$ ...


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

The document only says that $$\phi(g)=(1, \omega)$$ As you observed you must have $$\phi(1)=(1,1)$$ Fixing this, the second relation becomes: $$((1,1) + (1,\omega))((1,1)+(1,\omega)) = (2,1+\omega)(2,1+\omega) = (4,1+2\omega+\bar{w}) \\ =(1,0) + 2(1,\omega) + (1,\overline \omega)$$


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

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

$u^2 - u + 1 = (u^3+1)/(u+1)$, 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 $u = -g^{(p-1)/3}$.


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

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 ...



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