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That's not usually the most helpful unless you know that the element has prime norm, since unique factorization of numbers doesn't hold in general. Generally you look at norms then try and find ideals above a given ideal. In the case of $(10)$, we know the disciminant of the field is $40$ so $5$ and $2$ ramify, and since the degree of the extension is $2$ ...

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Let $E\subseteq X$. We set $J_E:=\{f\in C(X):\text{$f$vanishes on a neighbourhood of$E$}\}$. It is easy to see that $J_E$ is an ideal in $C(X)$. Take $E=\{x\}$ for some $x\in X$. Then each function in $I_{\{x\}}$ can be approximated uniformly by functions in $J_{\{x\}}$, but not every function such that $f(x)=0$ is zero on a neighbourhood of $x$. Thus ...

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Since $\mathbb{C}[x]/(x-a) \cong \mathbb{C}$, the ideal generated by $x-a$ is even maximal. If a power of a polynomial in $\mathbb{C}[x]$ lies is the ideal generated by $(x-a)(x-b)$, then both factors appear in the prime factorization of that polynomial (by uniqueness of the factorization) if $a \neq b$, i.e. the ideal generated by $(x-a)(x-b)$ in ...

3

Call $R= \Bbb{Z}+ x \Bbb{Q}[x]$. I highly suspect that $R$ is a Bezout domain, (i.e. every finitely generated ideal is principal), so I give you a non finitely generated ideal. Consider the ideal $$I=(x, x/2 , x/4 , x/8 , \dots) = \bigcup_{k \ge 1} \left( \frac{1}{2^k}x \right)$$ Clearly, for all $k \ge 1$ we have $$\left( \frac{1}{2^k}x \right) \subsetneq ... 3 You can see all this done by looking up (or Googling) "determinantal variety." For instance, it's done on the Wikipedia page, and there's a more comprehensive section in Harris's Algebraic Geometry: A First Course. 2 Let (R,m) be a local (not necessarily noetherian) ring, a\in R a non-zero divisor, and I,J\subset R ideals such that IJ=(a). Then I and J are principal. Write a=\sum_{i}a_ib_i with a_i\in I and b_i\in J. Then I(a^{-1}J)=R. (Note that a^{-1}J is an R-submodule of Q(R), the total ring of fractions of R, and this is the frame ... 2 Let’s call f=x^6+x^3+1. You want three linearly independent elements of the ideal (f) of the ring R=\Bbb F_2[x]/(x^9+1). Since (f) is just the set of multiples of f, you certainly have 1\cdot f, xf, and x^2f. Notice that x^3f=x^9+x^6+x^3=1+x^6+x^3=1\cdot f, already counted. I’ll leave it to you to show that those three polynomials are \Bbb ... 2 The proof I know for the second lemma also assumes that A is integrally closed: since 1\in \mathfrak{p}^{-1} we have \mathfrak{a}\subseteq \mathfrak{ap}^{-1}. So we have to show that the inclusion is strict. Assume that \mathfrak{a}= \mathfrak{ap}^{-1}. Let a\in \mathfrak{p}^{-1}. Then a is integral over A iff there is a finitely-generated ... 2 If \mathfrak a\mathfrak p^{-1}=\mathfrak a, then \mathfrak a\mathfrak p=\mathfrak a. Moreover (\mathfrak aA_{\mathfrak p})(\mathfrak pA_{\mathfrak p})=\mathfrak aA_{\mathfrak p}. By Nakayama lemma we get \mathfrak aA_{\mathfrak p}=0, so \mathfrak a=0 which I suppose you don't want. 2 For the other direction, let x+y\in S where s\in \langle a\rangle, y\in \langle b \rangle. Then there are h,k\in\mathbb{Z} such that x = ha, y = kb. Then with d = \gcd(a,b), d is a divisor of a and b, so there are n and m such that a = nd and b = md. So x + y = ha + kb = hnd + kmd = (hn + km)d \in d\mathbb{Z}. 2 Consider the ideal I = (x y + 1) of the ring k[x,y]. Then the ideal J consists only of g(x) \in k[x] which are divisible by x. This follows, as every h \in I is of the form$$h = (x y + 1) (g_m(x)y^m + \cdots + g_1(x) y + g_0(x))$$In general J is an ideal: if f_1 = g_1 y^{m_1} + r_1 \in I where every power of y in r is less than m_1 ... 2 We consider the following condition. Condition. There exists A \in \operatorname{GL}_2(R) such that$$\begin{pmatrix} a \\ b\end{pmatrix} = A \begin{pmatrix} a' \\ b'\end{pmatrix}.$$The sufficiency of the condition has already been noted, so I will only consider its necessity. Lemma. If R is a Dedekind domain, then the condition is ... 1 I'm skipping the questions about irreducibles. Too many questions here. If you take the typical example: R=\mathbb Z[\sqrt{-5}] where:$$2\cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})\tag{1}$$Then the ideal, I=(2,1+\sqrt{-5}) is not principle, and it is prime - it is easy to show that R/I\cong \mathbb Z/2\mathbb Z, which is a field, so I is maximal, hence ... 1 Hints: (b) See if you can show that \mathbb{R}[x]/(x-1) \cong \mathbb{R} in a similar manner to (a). From this, you conclude... (c) Is (x^{2}+1) reducible over \mathbb{R}[x]? What do you know about irreducible elements of PIDs? (d) Is (x^{2}+1) reducible over \mathbb{C}[x]? 1 Actually this is a finite extension of fields (hence algebraic, hence integral), by Zariski's Lemma. 1 "\Rightarrow" Let S=A\setminus\mathfrak p. Then S^{-1}\tilde A=\tilde A_{\mathfrak p}=A_{\mathfrak p}. If prime ideal \tilde{\mathfrak p}\subset\tilde A lies over \mathfrak p the it survives in A_{\mathfrak p}, so there is only one with this property. Then \mathfrak p\tilde A=\tilde{\mathfrak p}^e, so S^{-1}(\mathfrak p\tilde ... 1 It is so much easier to think about that in terms of P(S) \cong \operatorname{Maps}(S,\mathbb F_2) :=R with the identification$$P(S) \ni A \longleftrightarrow \chi_A \in \operatorname{Maps}(S,\mathbb F_2) The first statement then says: Take a map, which vanishes almost anywhere. If you multiply it with any map, it still vanishes almost anywhere. The ...

1

When you square an ideal you're really looking at products of two things coming from your ideal. In you case $\langle 2, 1+\sqrt{-5}\rangle^2$ means take something in $\langle 2, 1+\sqrt{-5}\rangle$ and multiply it by something else in the ideal; the set of all the things you get by doing this is the ideal $\langle 2, 1+\sqrt{-5}\rangle^2$. In order to ...

1

For a P. I. D., yes. For the general case, the product of two ideals $I=(a_1,\dots, a_n)$ and $J=(b_1,\dots, b_p)$ is the ideal generated by all possible products $a_ib_j$. Hence the square of $I=(2, 1+\sqrt{-5})$ is the ideal $I^2=(4,2+2\sqrt{-5},-4+2\sqrt{-5})$.

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Note that given any ideal $I$ and elements $a,b \in I$ we have $a-b \in I$. Also note that if $1 \in I$ then $I = R$. Using these two facts I'd continue your proof as follows: So, $2\mathbb Z$ is in the form $2k$ for $k$ integer. Therefore, an ideal greater than $2\mathbb Z$ must have an element of the form $2k+1$. Hence $1 = (2k + 1) - 2k \in 2 ... 1 Subrings are discussed a lot, though perhaps not explicitly. When you are first learning about subrings, you may be asked to do some exercises that have you prove that a certain subset of the ring is in fact a subring. Beyond that, when you're learning ever more advanced concepts, the fact that a certain subset of a ring is a subring is usually ... 1 I can't think of any easier way to prove the statement, but the proof should not require very much effort (which is why the author was so casual about it). First, I claim that it is very obvious that$N_R(U/T)$is an additive subgroup of$R$:$U0=0\subset T$If$Ur\subset T$, then$U(-r)\subset T$If$Ur, Us \subset T$, then$U(r+s)=Ur+Us\subset T+T=T$... 1 Another way to look at it is that$N_R(U/T)$is the annihilator of the$R$-module$\frac{U+T}{T}$. This is useful if you believe the annihilator of an$R$-module is always an ideal in$R$(which is an easy exercise you may or may not have already done.) 1 Identifying the image of$\Bbb Z$in$R$as$Z$, if$I$is some ideal such that$aZ=I$, then trivially$I=aZ\subseteq aR\subseteq I$. If you already get all elements of the ideal with addition, then multiplication won't give anything new. 1 We just submitted a paper, http://arxiv.org/pdf/1510.05699v1.pdf, in Section 2 you can find examples of quite natural coanalytic ideals (and more). 1 In a local domain, every invertible ideal is principal. The proof of this fact can be found in Kaplansky's book "Commutative rings", page 37, Theorem 59. 1 A height one primary ideal in a UFD is principal and generated by a power of a prime. Let$\mathfrak q$be a$\mathfrak p$-primary ideal of height one. Then$\mathfrak p=(p)$with$p\in R$a prime element. We have$\mathfrak q\subseteq (p)$. Suppose$\mathfrak q\subseteq (p^n)$, and$\mathfrak q\nsubseteq(p^{n+1})$. Since$\sqrt{\mathfrak q}=(p)$, there ... 1 Hint: First do the trivial case$S=\{0\}$. Then move onto the non-trivial case. Let$T \subseteq S$, where$T=\{a \in S \, | \, a > 0\}$. Now use well-ordering principle to claim that$T$has a smallest element. Call it$m$, use that to show that$S=m\mathbb{Z}$. 1 Let$m$be the minimal positive in$S$. Then by the second rule$km\in S$for each$k\in\Bbb Z\$.

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