# Tag Info

4

Hint: Try to find an ideal such that the quotient is $\mathbb{Z_2}[x]$

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Hint for #1. Show that $2$ must be in all prime ideals. Also convince yourself of the fact that $\Bbb{Z}_8[x]/\langle 2\rangle\cong\Bbb{Z}_2[x]$. Can you find prime ideals/maximal ideals in the latter ring, and pull them back to the original ring? Hint for #2. Can you think of a field that contains $R$? What is the smallest such field? Can you show that it ...

4

Extending the comment of Warren Moore, we see that for $p=2$ we have $x^4+x^2+1\equiv (x^2+x+1)^2 \mod p$. Hence the polynomial is reducible over $\mathbb{F}_2$; and the quotient $\mathbb{Z}[x]/I$ is not a field, so that $I$ is not maximal. Regarding principal ideals, it could help reviewing the classical example $I=(2,x)$.

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Easier would be to identify the quotient $\mathcal{A}/I$ and show that this is a field. (Hint: this will be isomorphic to $\mathbb{R}$). Thus $I$ is a maximal ideal.

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Define $\phi : \mathcal{A} \to \mathbb{R}$ by $\phi (f)=f(0).$ Then clearly $\phi$ is a ring homomorphism. Also we see that $\forall r\in\mathbb{R}$ if we define $f:[0,1]\to\mathbb{R}$ by $f(x)=r\:\:, \forall x\in [0,1]$ then $f$ is continuous i.e $f \in \mathcal{A}$ and also $\phi(f)=f(0)=r$ So $\phi$ is onto. Now $\ker\phi=\{f\in \mathcal{A}| ... 3 Lots of questions here, so I'll focus on just one of them. One way of finding a maximal ideal in$\mathbb{Z}_8[x]$is using the fact that an ideal$I$is maximal if and only if$\mathbb{Z}_8[x]/I$is a field. First, recognize that$\mathbb{Z}_8/\langle 2 \rangle \cong \mathbb{Z}_2$, and so it follows that$\mathbb{Z}_8[x]/\langle 2 \rangle \cong ...

3

(1) Nudge: break the problem up using the remainder theorem. (2) If you want something, make it happen: $R=k[a,b,u,v]/(a-bu,b-av)$. (This is $\cong k[b,u,v]/(b-buv)\supset k[b,1-uv]/(b(1-uv))\cong k[x,y]/(xy)$ in which $y\ne 0$.)

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You did everything correctly. And I think the definition of nilradical does make sense (as a set) even in the noncommutative sense (you should be careful which ideals you are considering, left, right or two-sided). I guess the problem will be that $N(J)$ will no longer be an ideal. Indeed, take a (non-commutative) ring generated by two letters $a,b$ and with ...

3

You didn't do anything wrong. The definition given for the radical of an ideal does not actually yield an ideal for noncommutative rings, because nilpotent elements needn't be closed under multiplication. There are a couple of ways to fix this, at least for the nilradical (radical of the zero ideal), both achieved by replacing the usual definition with an ...

2

OK, I reviewed the Chinese remainder theorem in Dummit and Foote. I will first provide a quick summary and then apply it to this problem. Chinese Remainder Theorem Let $R$ be a commutative ring with $1$, and let $I_1, \ldots I_m$ be distinct ideals of $R$ with the property that $I_j + I_k = R$ for all $j \neq k$. Then: $I_1 I_2 \ldots I_m = I_1 \cap I_2 ... 2 Answer to the title question: Yes,$(a) = (b)$in$\mathbb{Z}/(n)$iff$a, b$are associates in$\mathbb{Z}/(n)$. Hint: by Chinese Remainder it suffices to consider the case$n = p^k$is a prime power. Then show that every ideal of$\mathbb{Z}/(p^k)$is of the form$(p^n + (p^k))$for some$n = 0, \ldots, k$. Then check directly that$p^n + (p^k)$,$p^m + ...

2

The definition requires $I$ to be a proper ideal. Why? Because if it didn't, "maximal ideal" would be a really boring concept. Here are two ways to show $p\mathbb{Z}$ is maximal in $\mathbb{Z}$. First is direct. Say there exists something larger, $p\mathbb{Z} \subset I$. Pick $a \in I \setminus p\mathbb{Z}$. Because $p$ is prime, and $a$ is not a multiple ...

2

You cannot "establish this statement" because it is false! For example $(X^2)\subset k[X]$ is irreducible but not prime. Remark The correct implications (valid in any noetherian ring) are:$$\text {prime} \implies\text {irreducible} \implies \text {primary}$$

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The first option looks good, and recall that here "maximal" means that the generating element, that is the polynomial, is irreducible over $\mathbb{Z}/3\mathbb{Z}$. Now, you are left with the task of deciding which of the polynomials are irreducible. I cannot know which means you have for this, but if nothing else you can note that since the polynomials ...

2

We could just as easily write $K'=K(\sqrt[h]{x})$. One issue in writing $K'=K[X]/(X^h-x)$ is that the class group ${\rm Cl}(K)$ needn't be simple, so there can be elements with order properly dividing the class number $h$, in which case if $I^r=(y)$ with $r< h$ then $I^h=(y^{h/r})$ implying $X^h-y^{h/r}$ is reducible over $K$ since it's expressible as ...

2

Maximal ideals in a number ring (the ring of integers of a number field) need not be principal, contrary to what you say. But I'm guessing you don't realize "ring of integers" has a more general meaning in algebraic number theory, and you're just using the phrase to refer to $\Bbb Z$. In $\Bbb Z$, you say that $(2,3)$ will include multiples of $2$ and $3$, ...

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The claim is not true. If $A$ is an integral domain which has a non-trivial Picard group, there is some invertible $A$-module $M$ which is not free. But then $M$ has no quotient isomorphic to $A$, since any epimorphism $M \to A$ is an isomorphism (using that $M$ is locally free of rank $1$). However, we can prove the following: Let $M$ be a finitely ...

1

$\DeclareMathOperator{\Hom}{\operatorname{Hom}}$There is a well-known criterion for $\Hom$ to vanish: Proposition: Let $A$ be a Noetherian ring, $M, N$ f.g. $A$-modules. Then $\Hom_A(M, N) = 0$ iff $\text{ann}_A(M)$ contains a nonzerodivisor on $N$. Taking $N = A/p$ for $p \in V(\text{ann}_A(M))$ gives that $\Hom_A(M, A/p) \ne 0$, i.e. some nonzero ...

1

Actually $J$ is the kernel of the map $K[x,y,z,t]\to K[u,v]$ given by $x\mapsto u^3$, $y\mapsto u^2v$, $z\mapsto uv^2$, $t\to v^3$. (See also here.) Now you find that $I=(x,y)\cap J$, an intersection of two prime ideals.

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This is not true, even if $n = 1$: consider $R = \mathbb{Z}$, $P_1 = (2)$. Then $\pi(S_1) = \{ \frac{n}{1} : n \text{ odd}\}$, whereas $S_1^{-1}R - P_1 \cdot S_1^{-1}R = \{\frac{n}{s} : n, s \text{ odd}\}$, which contains e.g. $\frac{1}{3}$, but $\frac{1}{3}$ is not of the form $\frac{m}{1}$ for $m$ odd (otherwise $s(3m - 1) = 0$ for some odd $s$, ...

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Note that an ideal is closed under addition and subtraction, and $3-2=1$. Since the gcd of two integers is an integer linear combination of the two integers, any non-zero ideal is generated by the gcd of the elements of the ideal. If this gcd is $1$, the ideal is all of $\mathbb{Z}$. If this \gcd is composite, say $ab$ where $a$ and $b$ are greater than ...

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With your help, here is my solution: Let $p(x) := x^4+x^2+1$ a polynomial in $(\mathbb{Z}/2\mathbb{Z})[x]$. $\mathbb{Z}[x]/I\cong(\mathbb{Z}/2\mathbb{Z})[x]/(p(x))$ $p(x)=(x^2+x+1)(x^2+x+1) \Rightarrow \mathbb{Z}[x]/I$ is not integral $\Rightarrow \mathbb{Z}[x]/I$ is not a field. $\Rightarrow I$ is not maximal. By contradiction, if ...

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We have an isomorphism of rings $$\mathbb{Z}/20 [x] \cong (\mathbb{Z}/4 \times \mathbb{Z}/5)[x] \cong \mathbb{Z}/4[x] \times \mathbb{Z}/5[x]$$ It is well-known how to describe (prime) ideals in products in terms of the factors, so that we only have to look at $\mathbb{Z}/4[x]$ and $\mathbb{Z}/5[x]$. Well, $\mathbb{Z}/5$ is a field, hence $\mathbb{Z}/5[x]$ is ...

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You can do it using your approach. Suppose $I\subset J$ and that inclusion is proper. Then there is a continuous function $f\in J$ with $f(0)\not= 0$. Clearly $f(x)-f(0)\in I\subset J$. Thus $f(x)-\Big(f(x)-f(0)\Big)\in J$ and so $f(0)\in J$. which is to say $J$ contains $1.\:\:$

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Take $S=k[x,y],\quad R=k[x,y,z]/(yz-x)$. Take $I=(x),J=(y)$. Note that $z\in(I^e:J^e)\setminus(I:J)^e$. Or an alternative formulation of the same counterexample: Take $k[x,y]\subset k[y,\frac{x}{y}]$, with the same ideals - $I=(x), J=(y)$, and observe $x/y$.

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I'm really curious why you would want to consider bounded linear operators over a real space. Besides making everything more awkward, what is it there to gain? Regarding your second question: $$|UA|=((UA)^*UA)^{1/2}=(A^*U^*UA)^{1/2}=(A^*A)^{1/2}=|A|.$$ And $$|AU|=((AU)^*AU)^{1/2}=(U^*A^*AU)^{1/2}=U^*(A^*A)^{1/2}U=U^*|A|U.$$ The last equality is ...

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Also you can use Chinese Remainder Theorem. Since $(x-1)$ and $(x-2)$ are comaxal so, $\mathbb{R}[x]/(x^2-3x+2)$ isomorphic to $\mathbb{R}[x]/(x-1) \times \mathbb{R}[x]/(x-2)=\mathbb{R} \times \mathbb{R}$. Now ideals of $\mathbb{R} \times \mathbb{ R}$ are only $\mathbb{R}\times {0},{0}\times \mathbb{R}$ and itself. Whose pre-image are respectively ...

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Yes. If $S=R/K$, then $I=I'/K$ and $J=J'/K$. If $I', J'$ are not coprime, then there is a prime ideal $P$ such that $I'+J'\subseteq P$, and thus $I+J\subseteq P/K$. (Note that $P\supseteq K$ since $I',J'\supseteq K$.)

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Write $1=i+j$ with $i \in I, j \in J$. Find preimages $u,v$ in $f^{-1}(I)$ resp. $f^{-1}(J)$. Then $1 \equiv u+v \bmod \ker(f)$. Since $\ker(f) \subseteq f^{-1}(I)$, it follows that $1 \in f^{-1}(I) + f^{-1}(J)$. (Notice that the proof by user26857 uses the existence of enough prime ideals, which is independent from ZF.)

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An ideal $I$ maximal if it is proper and there are no other ideals other than the ring itself that properly contain $I$. In other words, you won't have another ideal $J$ such that $I \subset J \subset R$. Another way to say this is that if we have $I \subseteq J$, then $I = J$ or $J = R$. By definition, any maximal ideal $I$ must be properly contained in ...

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