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5

If it is not a field, it has exactly one prime element (up to associates, of course.) (explain why) It is a UFD, and apparently everything that is nonzero has a factorization which is a power of p times a unit. From this, observe the nontrivial ideals are just $(p^i)$ for integers $i>0$.


5

This situation is impossible. Indeed, suppose you have such ideals $A,B$ and $C$. Since $C\nsubseteq A$, there is an element $b\in C$ that is not in $A$; thus it must be in $B$. Similarly, there is an element $a\in C$ that is not in $B$, so it must be in $A$. Consider the element $a+b\in C$. This element is not in $A$, otherwise so would $b$. Neither ...


4

The ring $$R=\mathbb{C}[x_1,x_2,\ldots]/(x_1,x_2,\ldots)^2\cong\mathbb{C}[\epsilon_1,\epsilon_2,\ldots]$$ is non-noetherian (where $\epsilon_i$ denotes the image of $x_i$ in the quotient), and the radical of the (obviously finite) zero ideal $I=(0)$ in $R$ is equal to the ideal $(\epsilon_1,\epsilon_2,\ldots)$ which is not finitely generated, much less ...


3

Write $f(x,y)=(x^2+y^2-1)g(x,y)+a(x)y+b(x)$ with $a(x), b(x)\in\mathbb R[x]$. Since $f(x_0,y_0)=0$ for all pairs $(x_0,y_0)$ such that $x_0^2+y_0^2=1$, $x_0\ne0$ it follows $a(x_0)y_0+b(x_0)=0$ for all pairs $(x_0,y_0)$ such that $x_0^2+y_0^2=1$, $x_0\ne0$. In particular, $a(\sin t)\cos t+b(\sin t)=0$ for $t\ne k\pi$, so $a^2(\sin t)(1-\sin^2 t)=b^2(\sin t)$ ...


3

A polynomial $f\in \mathbb{R}[x,y]$, considered as a function $f:\mathbb{R}^2\to\mathbb{R}$, is continuous (where $\mathbb{R}^2$ and $\mathbb{R}^2$ are given their usual topologies). Therefore if $f\in I(\{(x,y): x^2+y^2=1, x \neq 0 \})$, we have $$f(0,1)=\lim_{t\to \pi/2}f(\cos(t),\sin(t))=\lim_{t\to 0}0=0\\ f(0,-1)=\lim_{t\to ...


2

What you've written isn't an ideal. What you presumably meant to write is $$I=(2,1+\sqrt{-6})=\{2\alpha+(1+\sqrt{-6})\beta:\alpha,\beta\in\mathbb{Z}[\sqrt{-6}]\}$$ Now, how to prove that $I$ is principal? I'll give a further hint there. Because $1+\sqrt{-6}\in I$ and $1-\sqrt{-6}\in\mathbb{Z}[\sqrt{-6}]$, we have (by the definition of ideal) ...


2

Consider the case of $(a_1,\ldots,a_n)=(0,\ldots,0)$ first, where it's obvious. Then observe that $$f(a_1,\ldots,a_n)=0\iff g(0,\ldots,0)=0$$ and $$f\in (x_1-a_1,\ldots,x_n-a_n)\iff g\in (x_1,\ldots,x_n)$$ where $$g(x_1,\ldots,x_n)=f(x_1+a_1,\ldots,x_n+a_n)$$


2

$I, J$ are each generated by a single element, say $x, y$ respectively. Every ring ideal is contained in a maximal ideal, in this case the (principal) maximal ideal, say, $(t)$. If $x$ is a unit, then $I=R$, so $J\subseteq I$. Similarly if $y$ is a unit. If not, then $x=ut^j$, $y=vt^k$ for some units $u,v$ and $j,k\in\mathbb{N}$. If $j\leq k$, then ...


2

By hypothesis, there are elements $i\in I$, $j\in J$ such that $i+j=1$. Observe that $j\equiv 1\mod I$ and $i\equiv 1\mod J$. Now take $u=ib+ja$. We have: $$u\equiv ja\equiv 1 a=a\mod I,\quad\text{and similarly}\quad u\equiv ib \equiv 1 b=b\mod J.$$


2

For any ring $R$, and any ideal $J$, $R/J$, as an $R$-module, is cyclic, meaning it is generated by a single element, namely the coset of $1$, $1+J$. But certainly there are $R$-modules that aren't cyclic. An ideal is cyclic if and only if it is principal, so, excluding rings where every ideal is principal, you have many examples of ideals for which this ...


2

As suggested in the comments, I will try to write an answer (although I'm a little against this). First of all, I may accidentally forget to write the word "compact" in "compact Riemann surface". So I'm assuming that all Riemann surfaces are compact. Actually everything fails if it's not compact. For instance, the correspondence between curves and Riemann ...


2

Is my answer to the first question on the right track? (Show that there is only one simple right R−module up to isomorphism) You made a comment about a chain and a minimal ideal, but this isn't going anywhere. The question has been asked a few times before, and you can find an explanation here: http://math.stackexchange.com/a/1011301/29335 How do I ...


2

Any Dedekind domain is dimension one, but not all Dedekind domains are PID's. For instance $\mathbb{Z}[\sqrt{-5}]$ is a standard example. The ideal $(2,1+\sqrt{-5})$ is maximal and therefore prime and is not the zero ideal, so it has rank $1$, but is not principal.


2

If $r\in R$ and $A$ is a minimal right ideal, then $rA$ is either $0$ or a minimal right ideal, because the mapping $R\to R$ defined by $x\mapsto rx$ is a homomorphism of right modules. In any case $rA$ is contained in the sum of the minimal right ideals.


2

$R$ is generated as a ring by any $2$ non-zero elements of the additive subgroup, $A$ say, with elements $\{0, x, y, x + y\}$. If $I$ is an ideal of $R$ with $4$ elements the quotient ring $R/I$ must be isomorphic to the ring $S = \mathbb{F}_2[z]/\langle z^2 \rangle$. The image of $A$ in $S$ generates $S$ and does not contain $1$ (since the elements of $A$ ...


2

It's not quite "self-evident" from what you wrote, because the finite basis of $i_j$'s you found are not necessarily a subset of $S$. (Also, you should be careful with your notation; it's not true that an ideal of $F[x_1,\ldots,x_n]$ necessarily can be generated by $n$ elements – take a look at this math.SE thread – but you wrote your finite ...


2

It is IMHO never a good idea to try and prove that some structure is not some other structure using the axioms only. After all, in many cases the said structure also IS that other structure, so a proof "generally speaking" is doomed to fail. Here you cannot rule out the possibility that, in addition to what you describe as a substitute to the associativity, ...


1

Here is an outline of the proof that I would write, with details for you to fill in: Show that $\operatorname{soc}(R)$ is equal to $\operatorname{ann}_l(N) = \{x \in R \mid xN = 0\}$, the left annihlator of $N$. Show that $\operatorname{ann}_l(N) = N^{r-1}$. Now that you know $R/S^1(R) = R/N^{r-1}$, finish the argument using induction on $r$.


1

Suppose $\phi$ is invertible. Then we have $\phi\psi=1$ for some $\psi$ and $(\phi)\supset (1)$. Now $\textbf{V}_V(\phi)\subset \textbf{V}_V(1)=\emptyset$. This proves one direction. For the converse, suppose $\textbf{V}_V(\phi)=\emptyset$. We also have $\textbf{I}_V(\textbf{V}_V(\phi))=\textbf{I}_V(\emptyset)$. Now $1$ vanishes nowhere so that $1\in ...


1

You can use the CRT to prove this. Recall that for co-prime ideals $I,J$, that is $I+J=R$, one has $R/IJ$ is isomorphic to $R/I \times R/J$ (as already mentioned in a comment) under the usual isomorphism $u+IJ \mapsto (u+ I , u+ J)$ (this is a slight abuse of notation). So you have some class modulo $IJ$, say $u+IJ$ that maps to $(a+ I, b+J)$. Takes ...


1

I don't know if this is what you mean by norms (you mean the absolute norm?). If this is what you already knew I apologize in advance! $$\begin{aligned}\mathcal{O}_L/\pi\mathcal{O}_L &\cong \mathcal{O}_K[T]/(p(T))/(\pi\mathcal{O}_K[T]/(p(T))\\ &\cong \mathcal{O}_K[T]/(p(t),\pi)\\ &\cong \left(\mathcal{O}_K/\pi\right)[T]/(p(T))\\ &= \prod_i ...


1

Why would the centre of $\mathfrak{M}_n(K)$ be a field? To commute with all of the unit matrices, an element of the center must be contained in the set $D=\{kI_n\mid k\in K\}$. This is already isomorphic to $K$. Then in order for an element to commute with all the elements of $D$, the scalar multiplier must be in the center of $K$ (which is sometimes ...


1

Yes. Another way to see this is by the following: We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=\varphi_0+\varphi_1X_n+\cdots + \varphi_kX_n^k$$ for some $\varphi\in R[X_1,...,X_{n-1}]$ with $\varphi_k(a_1,..,a_{n-1})\neq 0$. This reduces to the case you are okay with!


1

Since the first part has been dealt with in the comments: Assume $f(X,Y)=\sum_{i,j\ge 0}a_{i,j}X^iY^j\in\mathbb R[X,Y]$ is a polynomial with $f(x,y)=0$ whenever $x^2+y^2<1$. Let $\alpha\in\mathbb R$ and consider the polynomial $g(T)=f(T,\alpha T)\in\mathbb R[T]$. Then $g(t)=0$ whenever $t^2(1+\alpha^2)<1$. As there are infinitely many such $t$, $g$ ...


1

Since $I$ is a principal ideal, it is isomorphic to $R$ as an $R$-module. Hence $$I \otimes_R I \cong R \otimes_R R \cong R$$ which is obviously torsionfree.


1

1. "Yes" to the second question. More precisely, he is regarding $B$ as a $\left(B,A\right)$-bimodule, where the left $B$-module structure is obvious (i.e., given by multiplication) and where the right $A$-module structure is the one you define. "No" to the first question. He is tensoring a $\left(B,A\right)$-bimodule with a left $A$-module. This yields a ...



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