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15

There are proper-class-many groups and rings - in fact, there are proper class many $X$s, for any reasonable kind of mathematical structure $X$. One way to prove this is the following: given a cardinal $\kappa$, the direct product of $\kappa$-many rings is still a ring, and ditto for groups. So is there a sense in which there are "more" groups than rings? ...


8

The collection of all groups and the collection of all rings are, like the collection of all sets, proper classes (colloquially, they are "too big to be a set"). For any set $S$ whatsoever, you can form the free group $F_S$ and the commutative ring $\mathbb{Z}[S]$, and if you have distinct sets $S\neq T$ then $F_S\neq F_T$ and ...


8

In $2\mathbb Z$, there exists a generator $a$ of the additive group such that $a+a=x^2$ has a solution. This is not the case for $3\mathbb Z$.


6

Hint A ring isomorphism is in particular an isomorphism of the underlying abelian group under addition, but the abelian groups $n \Bbb Z$, $n > 0$, each have only two automorphisms; using this one can show that there are only two isomorphisms (of abelian groups) $n \Bbb Z \to m \Bbb Z$ for any $n, m > 0$.


6

If $A$ is any ring, then $A$ and $M_n(A)$ have equivalent categories of modules, and usually $A$ and $M_n(A)$ are not isomorphic. This is the simplest example of a Morita equivalence.


6

Automatically if $I=IJ$ then $I=IJ\subseteq J$ i.e. $I\subseteq J$. But equality isn't necessarily true. Consider $R=F\times F\times F$ and the ideals $I$ and $J$ generated by $(1,0,0)$ and $(1,1,0)$. Moreover, equality can occur when $I=I^2$ but can't occur if $I\ne I^2$.


5

A proper and nonzero ideal $I$ in $\mathbb{C}[x]$ is generated by a single non constant monic polynomial $f$. Decompose $f$ as $$ f(x)=(x-a_1)^{m_1}(x-a_2)^{m_2}\dots(x-a_k)^{m_k} $$ with $a_1,\dots,a_k$ pairwise distinct and $m_i>0$ (for $i=1,2,\dots,k$). Then the Chinese remainder theorem applies and $$ \mathbb{C}[x]/I\cong ...


4

You can see here more generally: Theorem. If the coefficient ring $R$ is an integral domain, then so is also its polynomial ring $R[X]$. Let $f(X)$ and $g(X)$ be two non-zero polynomials in $R[X]$ and let $a_f$ and $b_g$ be their leading coefficients, respectively.  Thus  $a_f≠0,  b_g≠0$,  and because $R$ has no zero divisors,  $a_fb_g≠0$.  But the ...


4

No, there is no such example, because if the polynomial ring $R[x]$ is a PID, then $R$ is a field, so that $R[x]$ is also Euclidean. For references see here.


3

Let $m,n\in\Bbb Z$. We wish to show that $\langle m\rangle\subset\langle n\rangle$ if and only if $n\mid m$. First, suppose that $\langle m\rangle\subset\langle n\rangle$. Since $m\in\langle m\rangle$ it follows that $m\in\langle n\rangle$. That is, there exists a $k\in\Bbb Z$ such that $nk=m$. Hence $n\mid m$. Conversely, suppose that $n\mid m$. To show ...


3

Let $J$ be an ideal of $R_P$. Consider $$ I=\{x\in R: x/1\in J\} $$ It's easy to prove that $I$ is an ideal in $R$: obviously $0\in I$; if $x,y\in I$, then $$ \frac{x+y}{1}=\frac{x}{1}+\frac{y}{1}\in J $$ and therefore $x+y\in I$; if $x\in I$ and $r\in R$, then $$ \frac{xr}{1}=\frac{x}{1}\frac{r}{1}\in J $$ and thus $xr\in I$. Suppose $x\in I$ and $x\notin ...


3

By definition an element $f$ of a ring $R$ is irreducible if whenever $f=uv$ for $u,v\in R$, either $u$ or $v$ is a unit. By definition if $f$ is nonprimitive in $\Bbb{Z}[x]$, $f=ng(x)$ where $n$ is the gcd of it's coefficients, which is not $\pm 1$, and hence not a unit, and $g(x)$ is a primitive polynomial in $\Bbb{Z}[x]$ of positive degree and hence not ...


3

Yes, as a DVR is a local ring there is only one prime number that is not invertible. (Note that a nontrivial ideal cannot include two distinct prime numbers and each non-unit is contained in some maximal ideal.) In more detail: Every non-zero subring of $\mathbb{Q}$ must contain $\mathbb{Z}$ and since a field is not a DVR the ring must not equal ...


3

You seem to be focusing on understanding the elements of a single coset. But really, that is not so important. The important thing to understand the quotient is to understand the cosets all together. A good way to do this is to realize (in this case) that each coset has a unique element of degree at most $1$. Now you can play around with multiplying and ...


2

A variant on this way of seeing things is that when you do operations on representative of cosets, each time you meet some $x^2+1$, you can replace it by $0$ ($x^2+1$ is ‘killed’ in the quotient. In other words, each time you have an $x^2$, you replace it with $-1$, an $x^3$ is replaced with $-1$, $x^4$ with $1$, &c.


2

Yes, $f\notin m^2$ for any maximal ideal $m$ is sufficient for proving the regularity. For showing $f\notin m^2$ you need to check that not all the partial derivatives of $f$ at $(\alpha,\beta,\gamma)$ are $0$. (If $f(\alpha,\beta,\gamma)\ne 0$ there is no need to go further.)


2

Assume $\varphi(\frac rs)=0$. Then $0=\varphi(s\cdot \frac rs)=\varphi(r)=\varphi'(r)$.


2

Well, actually your proof relies on the fact that $(-1)\cdot x=-x$. This does not follow immediately the definition, but it's a lemma that is usually proven as a corollary of $0\cdot x=0$. You are not giving an alternative proof, though. Moreover, $(-1)\cdot x=-x$ doesn't hold for rngs (rings without $1$), while $0\cdot x=0$ does. So I see little point in ...


2

I would say this is not correct. You see, the proof looks fine because you assume that $(-1)x=-x$. However to show that this is true you most likely have to use the fact that $0x=0$. If you found a way to show $(-1)x=-x$, without using that $0x=0$ then you're in the clear. I doubt that you can find one though. But if you (or anyone) can, I would love to ...


2

Let $x\in S$. Then there are two chances: $x\in A$, or $x\notin A$. If the latter happens then $x^{-1}\in A$ and then $x^{-1}\in\mathfrak m$, so $x^{-1}\in\mathfrak m_S$. But $x\in S$, so $1=xx^{-1}\in\mathfrak m_S$, a contradiction.


2

Your seem to have the main idea, but let me try to clarify. The map $\pi : R \to R/I$ is given by $\pi(r) = r + I$. In order to show $\pi$ is surjective, we need to show that given any element $ y \in R/I$, there exists an element $x \in R$ such that $\pi(x) = y$ (this is just the definition of surjectivity of a map of sets). Now, as $R/I = \{r + I \mid r ...


1

One way to describe the algebraic closure is that it is in some sense a "maximal" algebraic extension: it's an algebraic extension into which every other extension embeds. So it seems to me like the following question is a more basic one that should be answered first: What's an algebraic extension of commutative rings? There are various ways to answer ...


1

I think the famous theorem of Jacobson about such a ring being commutative holds, even if there is no identity. The hypothesis implies the ring is von Neumann regular. Therefore every ideal is idempotent, including the entire ring. Since the ring is finite, by Nakayama's lemma, we have that R has an identity. It would follow that $R$ is a finite product ...


1

When you quotient by $x-1$ you are saying that $x-1 \equiv 0$ (or $x \equiv 1)$. With this in mind \begin{align*} a_nx^n+a_{n-1}x^{n-1}+ \dotsb +a_1x+a_0 & \equiv a_n(1)^n+a_{n-1}(1)^{n-1}+ \dotsb + a_1(1)+a_0 \pmod{x-1}\\ & \equiv a_n+a_{n-1}+ \dotsb + a_1+a_0 \pmod{x-1}\\ \end{align*} Thus every polynomial is equivalent to its value at $x=1$. Thus ...


1

The ring in question is $\{\frac{a}{b}\in\mathbb{Q} \mid b\text{ is odd}\}$, also known as $\mathbb{Z}_{(2)}$. The key property of this ring is that $R^\times = \{\frac{a}{b}\in\mathbb{Q} \mid a\text{ is odd, }b\text{ is odd}\} = R\setminus 2R$, from which we can conclude that $2R$ is the unique maximal ideal. Presumably, you have identified all the ...


1

Indeed, an alternative definition of Noetherian ring is a ring where every ideal is finitely generated. As you noticed, a ring $R$ admits an infinitely generated ideal if and only if there is an infinite ascending chain of ideals. Indeed, suppose there is an ideal of $R$ which can only be generated by an infinite set $\{e_i\}_{i \in I}$ and extract a ...


1

I'll assume what Pavel Čoupek indicated in his comment to the question. The desired statement then essentially follows from the following lemma (this is Lemma 3.32 from the book Algebraic Geometry 1 by Görtz and Wedhorn where you will also find a more complete proof on p. 79): Let $B$ be a ring and let $A$ be a $B$-algebra. Assume that there are finitely ...


1

I assume what they mean is that this homomorphism makes $M$ into an $R$-module and the kernel is the annihilator of $M$ over $R$.



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