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5

Hint: as you noted, a necessary condition is that the constant term be equal to $\pm1$. Conversely, if $P$ has constant term equal to $\pm 1$, then $\pm P=1+XQ$ is a unit, since it is the sum of a unit and a nilpotent element. As for the group structure, you have a bijection $$\phi:\lbrace -1,+1\rbrace\times\Bbb Z^2\to A^\times, ...


4

See M. Harper, ${\bf Z}[\sqrt{14}]$ is Euclidean, Canad. J. Math. Vol. 56 (2004), 55–70. It's also done in Bernhard Lutzmann, Quadratic number fields that are Euclidean but not norm-Euclidean, available here. Also worth a look is Malcolm Harper and M. Ram Murty, Euclidean rings of algebraic integers, Canad. J. Math. Vol. 56 (1), 2004 pp. 71–76, available ...


4

Maybe you should have read the paragraph after Theorem $3.5$: This last theorem is probably why some people do not insist that rings contain 1. Kernels of ring homomorphisms have all the properties of a subring except they almost never contain the multiplicative identity. So if we want ring theory to mimic group theory by having kernels of ring ...


4

Use distribution for all it's worth, together with $1\cdot a = a \cdot 1 = a$: $$ a + (-1) \cdot a \\= 1\cdot a + (-1)\cdot a \\= (1 + (-1)) \cdot a \\= 0\cdot a \\= 0$$ and thus, by the uniqueness of additive inverses (remember, if we forget multiplication, any ring becomes an abelian group), we get that $(-1) \cdot a = -a$. By the same reasoning, we get $$ ...


3

We have $\space[1+(-1)]=0$. Therefore $a\cdot [1+(-1)]=a\cdot 0=0$. By left distributive law, $$a\cdot 1+a\cdot (-1)=0$$ $$a+a\cdot (-1)=0.$$ Now $-a \in R$. Adding $-a$ to both sides, we get $$(-a)+[a+a\cdot (-1)]=(-a)+0,$$ or $$[(-a)+a]+a\cdot (-1)=-a\quad [\text{associative property}]$$ or $$0+a\cdot (-1)=-a,$$ or $$a\cdot (-1)=-a.$$ Similarly we can ...


3

No. A counterexample is the ring ${\cal O}(D)$ of the holomorphic functions defined on a domain $D\subset\Bbb C$. The maximal ideals are the ideals $\{(z-a){\cal O}(D)\}$ for $a\in D$ (which are principal), but there are ideals which are not finitely generated. For instance, the ideal $I=\{\sin(nz)\}_{n\in\Bbb N}$ in ${\cal O}(\Bbb C)$ is proper (it is ...


3

For general sets, $A^B$ denotes the set of functions from $B$ into $A$. In your case, $\Bbb R^{\Bbb R}$ denotes the set of functions from $\Bbb R$ to $\Bbb R$. A way to see that this makes sense is that $\Bbb R^3$ is usually viewed as a vector space with 3 basis vectors. We write a general element as $(x_1,x_2,x_3)$. We could instead view this as a function ...


3

There is no injective ring homomorphism $\phi:M_3(\mathbb R)\to\mathbb R$ for simple reasons: consider the matrix $A=\pmatrix{0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0}$. Notice that $A^3=0$. Then $\phi(A)^3=0$, so $\phi(A)=0$. Since $\phi$ is injective we get $A=0$, false.


3

$I$ is not prime since $(2+i)(2-i) = 5 \in I$, and $2+i,2-i \notin I$. In particular, $I$ is not a maximal ideal. Finally, note that $\Bbb{Z}[i]/I = \{ a+bi : a,b \in \{ 0,1,2,3,4\} \}$, so it has $25$ elements.


3

An infinite direct sum of nonzero modules is never finitely generated. Since $R$, as a left module over itself, is finitely generated, the result follows. However, in this case it's even easier: you can write $1=\sum_{\lambda\in\Lambda}x_\lambda$, with $x_\lambda\in I_\lambda$ (I changed slightly the notation, with $\Lambda$ as index set) and all but a ...


3

There is no special notation for powers which indicates specifically that composition is the operation. You will have to depend on context. For example, the set of continuous functions from $\Bbb R\to \Bbb R$ can be made into a ringmonoid with composition or pointwise multiplication, and in both cases you would use ordinary power notation for its elements. ...


3

Well in $\frac{F[t]}{t^2}$ you have that $t^2=0$ so $t^n$ with $n\geq 2$ will be null. Hence, the most general element of this ring is : $$a_0+a_1t\text{ where } a_0,a_1\in F $$ Hence there are $p^2$ elements in it. If I want to be complete, you can show (I think) that : $$GL_2(\frac{F[t]}{t^2}) \text{ is isomorphic to } M_2(F)\rtimes_{Ad}GL_2(F) $$ ...


3

I hesitate to say it -- Take any functor $\mathsf{Int} \times \mathsf{Int} \to \mathsf{Set}$ which doesn't factor through a projection (e.g. take the underlying set of the product ring), and then follow it with the free ring functor $\mathsf{Set} \to \mathsf{Int}$ (the free ring on a set is a domain). Ugh. I hope there's something nicer. If you want to ...


3

Note that we have a surjective morphism of rings ${\rm ev}_0:\Bbb R[X]\to \Bbb R$ with ${\rm ev}_0(p(X))=p(0)$. What is the kernel of this morphism? Since $\Bbb R$ is a field, what does this give?


3

A ring $R$ such that the intersection of any two principal ideals of $R$ is principal is exactly a GCD domain. And a GCD domain need not be an UFD.


3

Jason in his comment has completely answered your specific question. I'd like to say that if you stick to positive rationals, unique factorization theorem can be interpreted as: $\mathbf{Q_{>0}^*}$ is a free abelian group on the set of prime numbers. Your map $\varphi$ is simply a permutation(actually a transposition) of two basis elements and so leads ...


3

Actually, what you've showed already is closer to a proof of the other inclusion. To show that $Im(\phi)\subseteq\Bbb Z[i],$ you should start with an arbitrary $f(x)\in\Bbb Z[x],$ then show that $f(i)\in\Bbb Z[i].$ It will probably help to first write $f(x)=\sum_{k=0}^na_kx^k$ where $a_0,\dots,a_n\in\Bbb Z.$ For the other inclusion, you will instead start ...


2

Given a ring $A$, $A^\star$ is the group of invertible elements of $A$ (that is, the group of units). Here, they are invertible matrices, that is matrices $M=\left(\begin{matrix} a & b\\c & d\end{matrix}\right)$ with $ad-bc\neq0$, because in a field, every nonzero element has an inverse. The inverse of $M$ is then ...


2

The first implication : $$p=a^2+b^2 \implies (ab^{-1})^2\equiv -1 \mod p $$ The other implication: if $x^2\equiv -1 \mod p$ then $x$ has order $4$ in $\Bbb Z_p$ and hence $4$ divides $p-1$ i,e $p$ is of the form $p=4k+1$ and use the implication you have in the book


2

Let $R$ be an ordered ring. I assume you are asking for the set of positive elements to be well-ordered. If so, let $\varepsilon > 0$ be the smallest positive element. If $\varepsilon$ is invertible, then $\varepsilon^2 \neq 0$. So $0 < \varepsilon^2 \leq \varepsilon \cdot 1 = \varepsilon$ and minimality of $\varepsilon$ imply that $\varepsilon^2 ...


2

No, this is not true. A counterexample is the ring of holomorphic functions $R=Hol(\Bbb{C})$. $R$ is not a UFD because irreducible (and prime as well) elements of $R$ are functions with exactly one zero with multiplicity 1. However, there are some functions with infinite zeroes (e.g. $\sin z$), and such functions cannot be factorized as a product of ...


2

$<5>$ is a prime ideal of $\mathbb{Z}$ and therefore is a maximal ideal of $\mathbb{Z}$. (Because $\mathbb{Z}$ is a PID). $\mathbb{Z}/<5>$ is a field. (Result well known) Now $$\begin{align}\mathbb{Z}[X]/<X,5>&=\mathbb{Z}[X]/<X>/<5>\end{align}$$ and $\mathbb{Z}[X]/<X>=\mathbb{Z}$ so ...


2

We can the 3rd isomorphism theorem: $$\mathbf Z[x]/(5, x)\simeq \mathbf Z[x]/5\mathbf Z[x]\color{red}/(5,x)\mathbf Z[x]/5\mathbf Z[x]\simeq \mathbf F_5[x]/x\mathbf F_5[x]\simeq \mathbf F_5. $$


2

In $\mathbb{Z}[\sqrt{-5}],$ $2$ and $3$ are irreducible, since $6 = 2 \cdot 3$ and $6 = (1 + \sqrt{-5})(1 - \sqrt{-5}).$ Observe that $U(\mathbb{Z}[\sqrt{-5}]) = \{\pm1\}$ (group of units). Therefore, the factorization of $6$ are truly different $\Longrightarrow \mathbb{Z}[\sqrt{-5}]$ is not a unique factorization domain. Since all principal ideal domains ...


2

If $A$ were a maximal proper subgroup of $\mathbb Q$, then the quotient group $\mathbb Q/A$ would be nontrivial (because $A$ is a proper subgroup of $\mathbb Q$), divisible (because $\mathbb Q$ is divisible), and therefore not cyclic. So there's a nontrivial proper subgroup $G$ in $\mathbb Q/A$. The inverse image of $G$ in $\mathbb Q$ is a subgroup ...


2

Hint: the equation $x^2 \equiv x$ is true mod $mn$ if and only if it is true mod $m$ and mod $n$.


2

$(2x^3−3x^2+2x−3)+(2x^2−x−3)=\{a(2x^3−3x^2+2x−3)+b(2x^2−x−3)| a,b\in R\}$ By definition. Every ideal is an additive subgroup by definition so yes $0$ is contained in it. (Note $0\in R$ so set $a=b=0$).


2

Let $x$ be a left zero divisior but not a right zero divisior. Then set $f:R\to R$ by $f(r)=rx$. Notice that $f$ is one to one as, $r_1x=r_2x\implies (r_1-r_2)x=0\implies r_1-r_2=0$ as $x$ is not a right zero divisior. Then $f$ must be onto as $R$ is finite which means there exist $r$ s.t. $1=rx$. Since $x$ is left zero divisior, $xy=0\implies rxy=0 ...


2

Your lattice consists of the points $(a/2,b/2)$ in $\Bbb{R}^2$ with $a,b$ integer and $a\equiv b\mod 2$. Then you can show that for any point $(x,y)$ in $\Bbb{R}^2$, (hence the same holds for points in $\Bbb{Q}^2$), you can find a point in the lattice with $|x-a/2|\le 1/2$ and $|y-b/2|\le 1/4$. Now, if you are working in $O_3$, you take the multiplicative ...


2

The $(\Rightarrow)$ part is good. The $(\Leftarrow)$ part is wrong, unfortunately. You can't say that $abx=bx$. Assume $ab\mid c$; then, for some $x\in R$, $(ab)x=c$ and so $a(bx)=c$. Then $b(ax)=c$ and therefore $b\mid c$. Note that this part doesn't require $a$ is a unit.



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