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10

You can easily construct an isomorphism using the universal properties of quotient and polynomial rings. Start as follows: We have $A \to A/q \to A/q [x] \to A/q[x]/(p)$ and the element $x$ in $A/q[x]/(p)$. Hence, we get $A[x] \to A/q[x]/(p)$. Clearly $q$ and $p$ lie in the kernel, so that we get $A[x]/(p,q) \to A/q[x]/(p)$. Now construct the inverse ...


7

To complement Martin's answer in accordance with the wishes expressed in the comments of the OP, why is the kernel of the canonical map $A[x]/(q,p)\to (A/q)[x]/(\bar{p})$ zero? Suppose $f=a_nx^n+\cdots+a_1x+a_0\in A[x]$ is in the kernel. This means that if we reduce the coefficients modulo $(q)$, to get $\bar{f}=\bar{a}_nx^n+\cdots+\bar{a}_1x+\bar{a}_0$, ...


6

Hint : Enough to prove $R/P$ is a field.. See that $R/P$ is already an integral domain and then see what would $r^n=r$ imply? I repeat $R/P$ is integral domain... Spoiler : $r^n=r\Rightarrow r^n-r=0\Rightarrow r(r^n-1)=0\Rightarrow???$


4

I'm not sure trying to show every ideal is principal will work (though I can't verify a counterexample off the top of my head!) however I'll start you off a different way: First assume $I \not\subseteq J$, then we can take some $x\in I\smallsetminus J$ and now consider the principal ideal $Rx$, what can we say about this ideal?


3

The ideal $f(I)A$ is the ideal of the scheme-theoretic fiber $\text{Spec}(A) \times_f \text{Spec}(B/I)$, that is, the restriction of the morphism $\text{Spec}(A) \to \text{Spec}(B)$ to the subscheme $\text{Spec}(B/I)$. So algebraically, $f(I)A$ is the kernel of the map of $B$-algebras $$A \to A \otimes_B B/I \cong A/f(I)A.$$ So it is the (ideal of the) ...


3

Your proof is basically correct, but perhaps it's helpful to look at a more general situation (indeed it's irrelevant that $R$ is local, keeping in mind that, over a field, a module being finitely generated is the same as it being finite-dimensional). Let $R$ be a ring and $M$ a finitely generated $R$-module. Then for any ideal $I$ of $R$, $M/IM$ is a ...


3

My comment above is already a hint. And here the definition along with a further hint are listed. Definition: A subset $I$ of a ring $R$ is called an ideal if it satisfies the following two conditions: I. $$\forall a, b\in I, a-b\in I.$$ II. $$\forall a\in I, r\in R, ra\in I.$$ Notice that the first implies that $I$ is a subgroup under the addition in ...


2

Just write down what $\mathcal L \cap \mathcal L^*$ is. You have $$ \mathcal L^* = \{x: \phi(xx^*)=0 \}, $$ as $x \in \mathcal L^*$ if and only if $x^* \in \mathcal L$ (and remembering that $(x^*)^*=x$). Then $$ \mathcal B = \{x: \phi(xx^*)=0 = \phi(x^*x) \}. $$


2

Instead of commenting on this proof, which is too complicated in my opinion, here is a fast proof. Assume that $a$ is not nilpotent. Then the localization $A_a$ is non-zero (otherwise $1/1=0$ in $A_a$, which would mean $a^n \cdot 1 =0$ for some $n$). Hence, it has a prime ideal $\mathfrak{p}$. The preimage in $A$ is a prime which doesn't contain $a$ ...


2

If $\sqrt{0}^n=0$, then in particular $(2 e_n)^{n-1}=0$, i.e. $2^{n-1} = 0 \bmod 2^n$, a contradiction.


2

I'll explain Praphulla Koushik's argument in more detail so that you can see what he is getting at. Let $R$ be a ring, and $\ast$ be the property you mentioned: namely that for any $a \in R$ there exists an $n > 1$ such that $a^n = a$. Lemma 1: Let $S$ be an integral domain. If $S$ satisfies property $\ast$, then $S$ is a field. Proof: $S$ is ...


2

Suppose $I\not\subseteq J$ and $J\not\subseteq I$ you have $x\in I\setminus J$ and $y\in J\setminus I$ Now consider Principal ideals $Rx$ and $Ry$


2

Yes. Let $R$ be ring of endomorphisms of $\mathbb{R}^2$, $I$ and $J$ be annihilators of subspaces spanned by $(1,0)$ and $(0,1)$, respectively. Let $\theta\in R$ be given by $\theta(x,y)=(y,x)$, and $\phi(f)=f\circ \theta$ for $f\in R$. Then $\phi$ is an automorphism of $R$ as a module over itself, and a bijection between $I$ and $J$. Thus, it induces ...


2

I believe a slightly more elementary version of the example of Marcin Łoś consists in taking the ring $R$ of $2 \times 2$ matrices over $\mathbb{R}$, say, and the left ideals $$ I = \left\{ \begin{bmatrix}a & 0\\ b & 0\end{bmatrix} : a, b \in \mathbb{R} \right\} , \qquad J = \left\{ \begin{bmatrix} 0 & a\\0 & b\\\end{bmatrix} : a, b \in ...


2

When I try to find (counter)examples, I don't consider random examples first, but rather try to simplify the general case so that, in the end, examples pop out automatically without any effort. This also works here: By the universal properties of $R$ and quotient modules, a homomorphism of left $R$-modules $R/I \to R/J$ corresponds to some element $[x] \in ...


2

Take $R=\mathbb Z / 9 \mathbb Z$ and $I=3R$. Then $I=\{0, 3, 6\}$, which are exactly the non-units. More generally, $R=\mathbb Z / p^2 \mathbb Z$ and $I=pR$, where $p$ is a prime.


1

Chose $g \in I\setminus P$ and let $f^{\prime}=fg$


1

Let $I$ be a fractional ideal according to definition $1$ which I believe includes also $I\neq(0)$. Pick any $0\neq z\in I$ then $z\in K^\times$ and $zR\subset I$. Also, from $rI\subset R$ you get $I\subset\frac1rR$ and $u=\frac1r\in K^\times$. If, on the other hand, you start from definition $2$, from $zR\subset I$ you get $z\cdot1\in I$ so that ...


1

If $h: R \rightarrow \mathbb{Z}$ is a ring homomorphism, then $h$ is also a homomorphism of (additive) abelian groups. If $P$ is the kernel of $h$, then $R/P$ is isomorphic to the image of $h$, which is a subgroup of $\mathbb{Z}$. And a subgroup of $\mathbb{Z}$ is either trivial or isomorphic to $\mathbb{Z}$. So if $P$ is a prime ideal of $R$ (prime ...


1

Let $x\in I$ and $r \in R$ and $e$ the unity of $I$ then $re \in J$ and so $rex \in I$ but $rex=rx$


1

To prove this we use that $PID$ are UFDs. Then Let $\mathfrak{p}=(p)$ be a prime ideal of $R$. Assume there is an ideal $\mathfrak{p}\subseteq\mathfrak{m}=(m)\subseteq R$. Then we would have that $m|p$, but then by the defintion of a prime element of a UFD this means that $m=p$ or $m=u$, a unit. Hence $\mathfrak{m}=(p)$ or $R$, proving maximality. Edit: ...


1

Hint $ $ Notice that for principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e. $(a)\supset (b)\iff a\mid b.\,$ Therefore $\qquad\quad\begin{eqnarray} (p)\,\text{ is maximal} &\iff&\!\!\!\! (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (a)\\ &\iff& p\ \ \text{ has no proper}\,\ ...


1

Your approach can't work. The ring $\mathbb{Q}[x]$, where $\mathbb{Q}$ is the field of rational numbers, is a PID. However, for the prime ideal $P=\langle x\rangle$ the quotient is $\mathbb{Q}[x]/I\cong\mathbb{Q}$, which is infinite. Suppose $P$ is a nonzero prime ideal and $P=\langle p\rangle$. Now, let $I=\langle a\rangle$ be an ideal such that $\langle ...


1

You can find this in the classical books on Lie algebras, e.g. N. Jacobson's book. Since $\mathfrak{gl}_n(K\simeq \mathfrak{sl}_n(K)\oplus K$ is a reductive Lie algebra, and $\mathfrak{sl}_n(K)$ is simple (if the characteristic of $K$ is zero, or not dividing $n$), we know all the ideals. Recall that a simple Lie algebra has only the trivial ideals. The ...



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