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16

The answer is both yes and no, so this will take a bit of elaboration. There are two ways to define a ring. One of them require the existence of a $1$, the other does not. Let's start with the one that does. In this case, a subring is required to contain the $1$ from the larger ring, and hence no proper ideal can be a subring (as any ideal containing $1$ ...


6

In general, an ideal is a ring without unity - i.e. without a multiplicative identity - even if the ring it is an ideal of has unity. For example $\mathbb{2Z} \subset \mathbb{Z}$ is an ideal but $\mathbb{2Z}$ is not a ring with unity. So if you require your rings to have unity - and a lot of the time one does - then an ideal is in general not a ring. This is ...


4

This answer uses the information, given in OP, that $2, 1 + \sqrt{-5}$ are irreducible elements that are not associated. Suppose $I = (a)$. If $\{2, 1 + \sqrt{-5}\} \subset I $ then $a \mid 2$ and $a \mid 1 + \sqrt{-5}$. Thus as $2, 1 + \sqrt{-5}$ are irreducible: either $a$ is a unit, or $a$ is associated to $2$ and $a$ is associated to $1 + ...


4

Usually an ideal also has to be an additive subgroup of the additive structure of the ring. Groups are never empty (they at least contain an identity), so ideals cannot be empty.


4

Prime ideals do correspond under the correspondence theorem, so your argument suffices. To see this, Let $I\subset P\subset R$ be any prime in $R$ containing an ideal $I$, then $R/P \cong \frac{R/I}{P/I}$ by the 3rd isomorphism theorem. Since $P$ is prime, $R/P$ is an integral domain, hence so is $\frac{R/I}{P/I}$. Thus, $P/I$ is a prime ideal in $R/I$. ...


4

No, it is not the case that for any irreducible integer $c$ the principal ideal $(c)$ is properly contained in an $(c, a + b \sqrt{-5})$ that is properly contained in the full ring. If $p$ is a prime number number that is inert in $\mathbb Z[\sqrt{-5}]$ then essentially by definition $(p)$ is a prime ideal in $\mathbb Z[\sqrt{-5}]$. Now, as a Dedekind ...


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

Another nice example of a non-Euclidean PID is $ K[[x,y]][1/(x^2\!+\!y^3)]\,$ over a field $\,K,\,$ i.e. adjoin the inverse of $\,x^2\!+\!y^3$ to a bivariate power series ring over a field. The proof employs a general construction method, see D. D. Anderson. An existence theorem for non-euclidean PID’s, Communications in Algebra, $16:6, 1221\!-\!1229, ...


3

The hint really says it all. First note that $1, x,x^2$ are integral in your field. They satisfy $t-1=0,\, t^3-2=0,\, t^3-4=0$ which are monic, integral polynomials. They are also a $\Bbb Q$-basis for the number field, so if we have an integral basis, $\alpha_0,\alpha_1,\alpha_2$ for $K$, then you know there is an integer matrix, $M$ so that $M\alpha_i=x^i$. ...


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

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

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

Easy counterexample: consider $\langle x \rangle \subset \Bbb Z[x]$. Alternatively (following quid's idea), consider $\langle x \rangle \subset \Bbb C[x,y]$. Remember that an ideal is maximal in $R$ if and only if $R/I$ is a field, and that an ideal is prime in $R$ if and only if (it is non the entire ring and) $R/I$ is an integral domain.


3

First you might want to recall that in a domain $\{0\}$ is a prime ideal. Thus, if you have any non-zero prime ideals you have your examples. Perhaps, you however meant non-zero prime ideals that are not maximal. This is also not difficult to find. Take a polynomial ring, over som field, in several variables like $k[x,y,z]$. The ideals generated by ...


3

It's not true. Let $F$ be a field, $A:=F[X_1,X_2,\dots]$ the polynomial ring over $F$ in countably many indeterminates. Then taking $R:=A/(X_1,X_2^2,X_3^3,\dots)$ and $M:=(\bar{X}_1,\bar{X}_2,\bar{X}_3,\dots)$ gives a counterexample.


2

We want to prove that there is $c\in K^\times$ such that $c\mathfrak a+\mathfrak b=R$. Let $0\ne a\in\mathfrak a$ and $\mathfrak c$ an ideal of $R$ such that $\mathfrak c\mathfrak a=aR$. (In a Dedekind domain every non-zero ideal is invertible, so one can chose $\mathfrak c=a\mathfrak a^{-1}$.) Moreover, $R/\mathfrak c\mathfrak b$ is a principal ideal ring ...


2

Take $r\in R$ and $x\in M$ and suppose that $rx\in U(R)$. Then, there is $u\in R$ such that $rxu=urx=1$. So the element $e=xur$ is idempotent, since $ee=xurxur=xur=e$, we have $e\neq 1$ because $x$ is not invertible and $e\neq 0$ because $R\neq\{0\}$. So $e$ and $1-e$ are non-invertible idempotents, that is $e,1-e\in M$ and their sum is $1\notin M$, ...


2

No need to use Zorn's Lemma. The proof is quite elementary. Lemma 1. If $A,B,C$ are $R$-modules such that $A \oplus B$ is cyclic, then every bilinear map $\beta : A \times B \to C$ is trivial. Proof. Let $(u,v)$ be a generator of $A \oplus B$. Then $u$ is a generator of $A$ and $v$ is a generator of $B$. It suffices to prove $\beta(u,v)=0$. Choose some $r ...


2

If $\alpha\neq\beta$, then $x-\alpha$ and $x-\beta$ are comaximal in $\mathbb F_p[x]$ (that is $(x-\alpha)+(x-\beta)=(1)$. So Chinese Remainder Theorem implies $\frac{\mathbb F_p[x]}{(x-\alpha)(x-\beta)}\cong\frac{\mathbb F_p[x]}{(x-\alpha)}\times\frac{\mathbb F_p[x]}{(x-\beta)}\cong\mathbb F_p\times\mathbb F_p$.


2

To remove this from the unanswered questions; the problem is that $A\backslash S$ is not necessarily closed under addition and thus there's no reason to expect that it's an ideal. A counterexample would be $S=\{1\}$, where $A$ has other invertible elements. With thanks to Nate, mt_ and brick.


2

We have $3+-2=1 \in (3,2)$, hence the generated ideal contains $1$, hence it contains $\mathbb Z$


2

Let $R = \frac{k[x,y,z]}{(1-xy)z}.$ Let $P = (z)$ and $I = (x)$. Note that $P$ is prime because modding out by $z$ gives $k[x,y]$, but that $P^2 \ne P$. And $z \in (x)$ because $z = xyz$.


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

You have to prove that if $r\equiv s \mod I$, then $rm=sm$ for any $m\in $M$. That is equivalent to $(r-s)m=0$, which is by definition since $r\equiv s\mod I\iff r-s\in I\subseteq\operatorname{Ann}_AM$.


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

I just thought that I would recast ajotatxe’s argument in a slightly different manner. Observe that we have a multiplicative function $ N: \Bbb{Z}[\sqrt{-5}] \to \Bbb{N}_{0} $ defined by $$ \forall (a,b) \in \Bbb{Z}^{2}: \quad N(a + b \sqrt{-5}) \stackrel{\text{df}}{=} a^{2} + 5 b^{2}. $$ Suppose that $ \{ 2,1 + \sqrt{-5} \} \subseteq \langle x \rangle $ ...


2

Note that $(4, 2x, x^2)=(2,x)^2$. Set $J=(2,x)$. We have $\mathbb Z[x]/J\simeq\mathbb F_2$ (the field with two elements). Show that $4, 2x, x^2$ are linearly independent in $J^2/J^3$ over $\mathbb F_2$, so $\dim_{\mathbb F_2}J^2/J^3\ge3$. If $J^2$ is two-generated, then the quotient $J^2/J^3$ is also two-generated, a contradiction.


2

and I still can't seem to prove that if they do [correspond], $P$ can't be some non-principal ideal properly larger than $(x)$. Assuming you convince yourself of the correspondence of prime ideals (which is just fine) here's an elementary way to see that there can only be one prime ideal in this ring. Now $(x)$ is a maximal ideal of $\Bbb F[x]$ for any ...


2

Such rings are called residually finite, or rings with the finite norm property FNP). Below is an entry point into the literature. Levitz, Kathleen B.; Mott, Joe L. $ $ Rings with finite norm property. Canad. J. Math. 24 (1972), 557--565. Let $A$ be a ring with $A^2 \ne 0 ,$ and $A^+$ the additive group of $A$ . If each non-zero homomorphic ...



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