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5

The ring $\mathbf Z[w]$ is isomorphic to the quotient ring $A=\mathbf Z[x]/(x^2+x+1)$. To show $2$ is prime, you have to show $A/2A$ is an integral domain. We have: $$A/2A \simeq \mathbf Z[x]/(2,x^2+x+1)\simeq \mathbf Z/2\mathbf Z[x]/(x^2+x+1)$$ so it is enough to show the polynomial $x^2+x+1$ is irreducible over $\mathbf Z/2\mathbf Z$, which means it has ...


5

Hint: Use the fact that you can mod out in any order, and use the isomorphism $\mathbb{Z}[\sqrt{2}][x]/(5,x^2 + 1)\cong\mathbb{Z}[x,y]/(5,x^2 + 1,y^2 - 2)$. I'll start you off: \begin{align*} \mathbb{Z}[\sqrt{2}][x]/(5,x^2 + 1)&\cong\mathbb{Z}[x,y]/(5,x^2 + 1,y^2 - 2)\\ &\cong\mathbb{F}_5[x,y]/(x^2 + 1,y^2 - 2) \end{align*} Now look at these ...


5

Are you familiar with prime and maximal ideals? The assumption is that $I$ is prime. $R/I$ is a field iff $I$ is maximal. So you'd like to show that nonzero prime ideals in a PID are maximal. Can you do that?


4

Yes, you are on the right track. All your reasoning makes sense to me. On your question about the associates By the properties of the norm, associates have the same norm. So the only possible associates in your list are $1 + \sqrt{-5}$ and $1 - \sqrt{-5}$. Now determine all units of $\mathbb{Z}[\sqrt{-5}]$ by finding all integer solutions of $N(a + ...


4

I've seen these same arguments in quite a few books, it all looks pretty standard issue to me. But what you seem to be missing is the motivation for all of this, the why do we care. That motivation is this famous fact: $$6 = 2 \times 3 = (1 - \sqrt{-5})(1 + \sqrt{-5}).$$ It's also true that $$6 = -2 \times -3 = (-1 - \sqrt{-5})(-1 + \sqrt{-5}).$$ We've got ...


4

If an element is not a unit, the ideal it generates is not all of $R$. Now, every proper ideal is contained in a maximal ideal.


3

The structure of the ring $\mathbb{F}_p(x)/(q(x))$ obviously depends on how $q(x)$ splits over $\mathbb{F}_p$. If $q(x)$ is irreducible over $\mathbb{F}_p$, then the quotient ring is a field, hence the only zero divisor is zero. In our case, over $\mathbb{F}_3$: $$ x^4+2 = x^4-1 = (x-1)(x+1)(x^2+1), $$ so the quotient ring is isomorphic to ...


3

HINT: Whatever $a$ and $b$, the ring will be a $\Bbb F_p$ vector-space of dimension $2$. As ring we have the following three possibilities for the quotient: A ring with a unique maximal ideal $\frak m$ such that $\frak m^2=(0)$. The ring $\Bbb F_p\times\Bbb F_p$. A field with $p^2$ elements. These situations should be linked to the behaviour of the roots ...


3

No. Search the site (or better yet, rack your brain) for the examples of rings which have elements $a,b$ such that $ab=1$ and $ba\neq 1$, and you'll have an example. If you are silently assuming the ring is commutative, then yes. Just look at what it means for $xR$ not being a proper ideal.


3

No, an element $r$ can be left invertible, but not right invertible (example follows). Any two-sided ideal that contains $r$ is the whole ring. Example: consider the endomorphism ring of an infinite dimensional vector space with basis $\{e_n:n\in\mathbb{N}\}$. Let $f$ be defined by $f(e_n)=e_{n+1}$. Then $f$ has a left inverse, the map defined by ...


3

First, to construct the quotient space, you want the relation $\sim$ defined by $x\sim y\iff x-y\in I$ to be an equivalence relation. Then you want to be able to define operations for the cosets so that $[x]+[y]=[x+y]$ and $[xy]=[x][y]$ are well defined and give $A/I$ the structure of a ring. From these requirements it follows that $I$ is an ideal. That is, ...


2

It is simply a statement about the maximum additive order of something in the ring. Quotients of $\Bbb Z$ are a natural source of rings with different characteristics, of course. The most physical analogy that comes to mind is modular arithmetic. If you're familiar with any sort of cyclic behavior that repeats after finitely many steps, you can view the ...


2

In order to better distinguish between elements and ideals I'll use Fraktur letters for the latter: $\mathfrak{b}$ and $\mathfrak{c}$. With $\mathfrak{c}/\mathfrak{b}$ the ideal $$ \{x+\mathfrak{b}:x\in \mathfrak{c}\} $$ is meant. You have to prove that $\mathfrak{c}/\mathfrak{b}$ is an ideal of $A/\mathfrak{b}$ Every ideal of $A/\mathfrak{b}$ is of this ...


2

$S^{-1}R$ is noetherian, reduced ($N(S^{-1}R)=0$), and every element of $S^{-1}R$ is a zerodivisor or invertible. Now your question follows from the following result: Let $A$ be a reduced noetherian ring with the property that every element of $A$ is a zerodivisor or invertible. Then $A$ is artinian. Let $\{\mathfrak p_1,\dots,\mathfrak p_n\}$ be the ...


2

Hint $\, \alpha,\alpha'$ are roots of $\,0 = (x-\alpha)(x-\alpha') = x^2 - (\alpha+\alpha') x + \alpha\alpha' = x^2-\color{#c00}k\,x+\color{#c00}n,\,\ \color{#c00}{k,n\in\Bbb Z},\, $ where $\,\alpha' = \frac{1}{2}(1-\sqrt{-19})\,$ is the conjugate of $\,\alpha.$ Therefore $\, \alpha^2 =\, k\,\alpha - n.\ $ Use this equation to show closure under ...


2

Any odd number is congruent to $1,3,5,7$ mod $8$. Each of these when squared gives remainder $1$ mod $8$.


2

Consider $\mathbb{Z} \rightarrow \mathbb{Z}$ multiplication by 2.


2

$$\varphi:Z\rightarrow Z_8; \quad \varphi(n)=\bar n $$ $J(Z)=0$. so $\varphi(J(Z))=0$ but $J(Z_8)= \langle 2 \rangle$


2

Hint: (1). $\mathbb Z[X]/A \cong \dfrac{\mathbb Z[X]/(5)}{A/(5)} \cong \mathbb Z_5[X]/(X^3 - 3X + 2)$ and $(X^3 - 3X + 2)$ is reducible in $\mathbb Z_5$ as it has a root in $\mathbb Z_5.$ (2). $\mathbb Z[X]/B \cong \dfrac{\mathbb Z[X]/(X-1)}{B/(X-1)} \cong \mathbb Z_5.$


2

A general nonzero element of $R$ can be uniquely written in the form $t^a f(t)$ where $a \in \mathbb{Z}$ and $f(t)$ is a polynomial with $f(0) \ne 0$. This shows that the invertible elements of $R$ are precisely the monomials $ct^a$, with $c \ne 0$. Since $R$ is the localization of $F_3(t)$ which inverts the element $t$, an endomorphism of $R$ corresponds ...


2

Starting with a ring $A$ having no multiplicative unit. Consider the direct sum $\mathbb{Z}\oplus A$ with the following addition $(m+\alpha)+(n+\beta)=(m+n)+(\alpha+\beta)$ and multiplication $(m+\alpha).(n+\beta)=mn +m\beta+\alpha n+\alpha\beta$ You can check it is a ring with unit $(1,0)$ and identifying $(0,\alpha)$ with $\alpha$ (and not $(0,\alpha)$ as ...


2

There are only two units in $\mathbb{Z}[\sqrt{-5}]$: 1 and $-1$. To obtain the associate of a number, you multiply that number by a unit other than 1, and in this domain there is only one choice: $-1$. So, for example, the associate of 2 is $-2$, the associate of 3 is $-3$, the associate of $1 - \sqrt{-5}$ is $-1 + \sqrt{-5}$ and the associate of $1 + ...


2

First prove that $I$ is prime: let $ab\in I$ for $a,b\in R$ then. $ab+I=I\implies (a+I)(b+I)=I \implies a+I\in I\ or\ b+I\in I$ since $R/I$ has no zero divisors Thus $I$ is prime and in a PID every non zero prime ideal is maximal


2

There is no paradox. The set of integers is a subset of the set of rationals, and the set of all fields is a subset of the set of all integral domains. (I don't want to irritate set theorists, so you can say the second fact just means every field is an integral domain.) The latter does not mean the set of rationals is a subset of the set of integers.


1

The "paradox" you have discovered is because the integers are a subset of the rationals when you consider the elements themselves. Clearly all the members of the integers belong in the rationals, and the rationals are an extension of the integers themselves (a fraction field). But as an algebraic structure, a field is less general than an integral domain. ...


1

In general, any integral domain can be extended to a field of fractions containing an embedding of the original integral domain. This is due to the fact, that what an integral domain is possibly missing is multiplicative inverses before it becomes a field. So we actually have to supply those two an integral domain that is not already a field in order to ...


1

$I$ is generated by a single element $p$. The fact that $R/I$ has no zero divisors mean that if $ab\in R$ is divisible by $p$, then either $a$ is divisible by $p$ or $b$ is. Thus $p$ is prime. The maximal nonzero ideals in a PID are exactly the ideals generated by a single prime element. Thus $I$ is a maximal ideal, hence $R/I$ is a field.


1

You can write it as $$\mathbf F_5[x,y]/(x^2+1,y^2-2).$$ As $y^2-2$ has no root in $\mathbf F_5$, it is irreducible over $\mathbf F_5$ and $\mathbf F_5[y]/(y^2-2)$ is a field, isomorphic to $\mathbf F_{25}$. As for $x^2+1$, it has roots in $\mathbf F_5$ ($\pm 2$), so the quotient $\mathbf F_5[x,y]/(x^2+1,y^2-2)$ is isomorphic with $\mathbf ...


1

(Caveat: this is all done assuming you mean domain where you wrote ring. This is the natural environment for unique factorization.) If all (non-zero and non-unitary) irreducible elements of a ring are prime elements, we have a unique factorization ring. This is incorrect. In addition to irreducibles being prime, you also need the ascending chain ...


1

2 is a prime number so the ring $\mathbb{Z}_2$ is actually a field.



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