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7

No, $IJ\subseteq I\cap J\subseteq I\cup J\subseteq I+J$. The middle inclusion is true for all sets. So you just need to prove that $IJ\subseteq I\cap J$ and $I\cup J\subseteq I+J$. It's easy to find counter-examples in $\mathbb Z$ to your inclusions: $$(6\mathbb Z)(4\mathbb Z)=24\mathbb Z\subsetneq 12\mathbb Z=6\mathbb Z\cap 4\mathbb Z$$ $$6\mathbb ...


5

I am assuming that $R$ has a multiplicative neutral element $1$ (below I show that the claim is false otherwise). The definition of two ideals $I,J$ being coprime is (AFAIK) that for some elements $i\in I, j\in J$ we have $i+j=1$. This is equivalent to the requirement $I+J=R$, when $R$ is unital (see comments). So your assumptions imply that to each index ...


5

A silly example : Consider the field of real numbers, $\mathbb{R}$. Clearly $p\mathbb{R}=\mathbb{R}$, so $\mathbb{R}/p\mathbb{R} \cong \lbrace 0\rbrace$. More interesting: Let $R=\mathbb{Z}[x]$. $\mathbb{Z}[x]/p\mathbb{Z}[x] \cong (\mathbb{Z}/p\mathbb{Z})[x]$ which is not a field!


4

Try it like this. Let $p(x)= a_0 + a_1x + a_2x^2 + \dots + a_nx^n$ (with each $a_k \in \mathbb{Z}$) be any element that is in $J$ but not in $I$. Now note that $a_1x + a_2x^2 + \dots + a_nx^n$ is already in $I$ (because it is a multiple of $x$), hence it must be true that $a_0 \in J$ but $a_0 \notin I$. Now $a_0$ can't be an even integer -- if it were, it ...


3

No, $I^{\cdot2}=(2)$. Note that $I^{\cdot2}$ contains $(4,-10)=(2)$. On the other hand, every element of $I^{\cdot2}$ is clearly a multiple of $2$.


3

Note that $R/J\cong S/I$, hence if one is a domain or field so is the other.


2

Editted to remove reference to unity Continuing along the lines of your solution, we want to show that every $y \in R$ is also in $J'$. Note in particular, since the assumption is that $f(J')=S$, that there exists $x \in J'$ so that $f(x) = f(y)$. But then $$f(x-y)=f(x)-f(y)=0,$$ so $$x-y \in J = f^{-1}(I).$$ Can you see what to do from here?


2

According to WolframAlpha the gcd between $f$ and $g$ is indeed $x^2 - 1 = (x+1)(x-1)$. The easiest way to see this by hand is to use the Euclidean algorithm. Namely, by polynomial long division we have $$ \begin{gather} x^6-1 = (x^2-2 x+2) (\color{green}{x^4+2x^3+2x^2-2x-3}) + \color{red}{2 x^3-5 x^2-2 x+5} \\ \color{green}{x^4+2x^3+2x^2-2x-3} = ...


2

You can prove it is not maximal by finding a prime $p$ such that $x^2-5$ is irreducible modulo $p$. Thus we'll have $$(x^2-5)\varsubsetneq(p,x^2-5)$$ and this last one is indeed a maximal ideal since $\mathbf Z[x](p,x^2-5)\simeq\mathbf F_p[x]/(x^2-5)\simeq\mathbf F_{p^2}$. For instance $x^2-5$ is irreducible modulo $3$, since the only squares mod. $3$ ...


2

This is a good approach. You could carry on by showing that $(x^2 - 5)$ is the kernel of the homomorphism from $\mathbb{Z}[x]$ to $\mathbb{R}$ that maps $x$ to $\sqrt{5}$. Then show that the image of this homomorphism (which is isomorphic to $\mathbb{Z}[x]/(x^2-5)$) is the subring $A$ say of $\mathbb{R}$ comprising all numbers of the form $m + n\sqrt{5}$ ...


2

You already showed that $IJ\subset I\cap J$. So one side is obvious. That is, if $I\cap J\subset P$ then $I J\subset P$. For the other side, assume $I J\subset P$ and assume there is $x\in I\cap J$ such that $x\not \in P$. Then since $P$ is prime $x^2\not\in P$, but this is a contradiction since $x^2\in IJ\subset P$.


2

Let $I=\{x\in R:(x,0)\in M\}$ and $J=\{y\in S:(0,y)\in M\}$. If $(x,y)\in M$, then $(x,0)=(x,y)(1,0)\in M$ and $(0,y)=(x,y)(0,1)\in M$. This can be used to show one of the inclusions. The other one is easy.


2

In fact, $$\phi(R)/\phi(I)\simeq R/(I+\ker\phi)$$ and this follows easily by the fundamental theorem of isomorphism for rings. (Find the kernel of the surjective homomorphism $R\to\phi(R)\to\phi(R)/\phi(I)$.)


2

One might argue in this fashion, instead: Suppose that $J$ is an ideal such that $I\subsetneq J\subseteq\Bbb Z[x],$ and take any $y\in J\setminus I.$ By definition of $I,$ we know that $x$ doesn't divide $y,$ and so if $y=px+n$ for some $p\in\Bbb Z[x]$ and some $n\in\Bbb Z,$ then $n\ne 0.$ In fact, $n$ cannot be any even integer. (Why?) Thus, if $y=px+n$ for ...


2

If $I/I^2$ is generated by $r$ elements, you can pick $r$ elements in $I$ which generates $I/I^2$. Let $J$ be the ideal generated by these $r$ elements and then consider $S=R/J$ and $I'=I/J$ an ideal in $S$. Then $I'/I'^2=0$ and you have your argument for this case, showing $I'$ is generated by one element and thus $I$ is generated by $r+1$ elements. My ...


1

There are also plenty of counterexamples apart from fields. For example $p$ could be a unit even if $R$ is not a field, e.g. in $\mathbf Z [\frac{1}{p}]$, or $p$ could fail to be irreducible, e.g in $\mathbf Z[i]$ we have $(1+i)(1−i)=2$.


1

No, this argument fails at the point where $Ra_0b_0$ is replaced by $Ra_0$. In general $Ra_0$ is larger than $Ra_0b_0$. In particular, your claim $Rt = R$ is wrong, since $Rt$ is the ideal generated by $t$ in $R$.


1

Theorem-2.2.7[Cohen-macaulay ring by Bruns and Herzog] Let $(R,m,k)$ be a Noetherian local ring. Then the following are equivalent: (a) $R$ is regular (b) proj dim M is finite for every finite $R$ module $M$ (c) proj dim k is finite.


1

This is because a regular local ring has finite homological dimension (it is a characterisation of regularity). Indeed $\operatorname{dh}(A)$, the global homological dimension of $A$, is equal to the projective dimension $\operatorname{dp}(k)$ of its residue field $k=A/\mathfrak m$, and the latter has a finite free resolution of length $\dim A$ via the ...


1

Write $f = a_0 + ... + a_mX^m$ and $g = b_0 + ... + b_nX^n$. It suffices to show that for each maximal ideal $M$, there is a coefficient of $fg$ not in $M$. Let $k$ and $l$ be minimal with $a_k,b_l \notin M$. Check that the coefficient of $X^{k+l}$ is not in $M$.


1

Here is a slight modification of Thomas's example that is easy to prove. Let $R$ be a quasilocal ring with a nonzero idempotent maximal ideal $M$, e.g., the ring in Thomas's solution localized at his maximal ideal. Then $M$ has no minimal generating subset. Suppose to the contrary that it did have such a subset $\{a_i\}$. Then we have an expression $a_j ...


1

You shouldn't need to worry about PIDs or factorization or anything like that. The proof that $I\subset A$ is maximal if and only if $A/I$ is a field is a short one, and therefore it is easy to convert your proof that $\mathbb Z[x]/(x,2)$ is a field into a direct proof that $(x,2)$ is a maximal ideal in $\mathbb Z[x]$. Let me show you how that's done. ...


1

We prove that if $I\subsetneq J$ is an ideal, then $J= \mathbb Z[X]$. Let $P(x) \in J \backslash I$ be any polynomial. Then you can do long division by $X$ and get $$P(x)=XQ(X)+m$$ Now, $XQ(X) \in I \subset J$ which implies that $m \in J$. If $m$ is even, then $XQ(X) \in I, m \in I$ would imply that $P(X) \in I$. Therefore $m=2k+1$ for some $k$. This ...


1

for first question see page 27 of Gohberg, I. C, Krein, M. G. Introduction to the Theoryof Linear Non-Self-Adjoint Operators in Hilbert Space. For $A\in\mathscr{I}_1(H)$ and $B\in\mathcal{B}(H).$we have $s_{j}(AB)\leq{s_{j}{(A)}} \Vert{B}\Vert_\infty$ & $s_{j}(BA)\leq{s_{j}{(A)}} \Vert{B}\Vert_\infty$.


1

Just going on the first sentence - I would first mod out by $(x-1)$, which has the effect of setting $x =1$ (x becomes just another symbol for 1 so we forget about it). Now the quotient is (isomorphic) to the ring $Q[y]/(y^2 + 2)$. Since $(y^2 + 2)$ is irreducible over $Q$, the ideal it generates is maximal, hence the quotient is in fact a field. But since ...


1

Hint. Use the fundamental theorem of isomorphism for rings and notice that $\dim K[T^4, T^3U, TU^3, U^4]=2$.


1

The direction you are trying is possible. However, it is important to notice that the zero ideal is prime in every integral domain, and hence...


1

Yes and yes. You switch notation from $m$ to $m_0$, so I'm going to stick with $m\in M$ non-zero. You can deduce the maximality of the left ideal $A(m)$ from the second claim as follows. Define $\varphi:R\to M$ by $\varphi(r)=rm$. This is an $R$-module homomorphism whose image is a submodule of $M$ containing the non-zero element $m$, and hence must be all ...



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