# Tag Info

16

You can think of ideals as subsets that behave similarly to zero. For example, if you will add $0$ to itself, it is still $0$, or if you multiply $0$ with any other element, you still get $0$. So ideals are like "zeros with several elements". It is a subset $I$ of a ring $R$ that is closed under addition, and also if $a\in I, b\in R$, then $ab\in I$. In ...

7

An ideal is a generalization of a number that makes factoring work better if you allow square roots of some numbers. What numbers does $n$ divide? When studying how numbers factor, we can take a specific number, $n$, and look at all the numbers it divides, $(n) = \{ nk : k \in R \}$ where $R$ is the universe of numbers we are examining. If you've worked ...

5

You can write down the factorization into prime ideals of the principal ideal $(\pi)$. This will contain exactly one factor of $\mathfrak{p}$ by construction, and similarly for powers of $\pi$. If you quotient by $\mathfrak{p}^n$, only ideals including this, thus dividing it remain. Hence the stated equality and the result for $\mathfrak{a}$ prime. The ...

4

In general, if $A$ is a finite $k$-algebra, then $A$ is necessarily $A$ is Artinian. This implies that $\text{Spec}(A)=\text{MaxSpec}(A)$ and that $A$ decomposes as a finite product of local artinian rings. In fact, let $A$ be a finite type $k$-algebra. Then, the following are equivalent: $\text{Spec}(A)$ is finite. $\text{Spec}(A)$ is discrete. ...

4

Consider the ring $\mathbb{Z}\times\mathbb{Z}$ (which is arguably the simplest ring having non-trivial idempotents, so it should be one of the first examples we look at). Its non-trivial idempotents are $(1,0)$ and $(0,1)$, and its prime ideals are those ideals of the form $\mathbb{Z}\times P$ and $P\times\mathbb{Z}$ where $P\subset\mathbb{Z}$ is a prime ...

4

The method of @DonAntonio is best, but here’s a rough and ready argument: To factor out by the ideal $(x,3)$ is to treat computations in the original ring $\mathbb Z[x]$ as if both $x$ and $3$ are zero. So polynomials immediately lose all their nonconstant terms, and even these behave as integers modulo $3$. Alternatively, just construct a ring ...

4

Notice that $(y^2+x^3-17)\subset (y^2,x^3-17)$. Therefore, the quotient of the ring $\mathbb{C}[x,y]/I$ by the ideal, generated by $(y^2,x^3-17)$ is actually isomorphic to $\mathbb{C}[x,y]/(y^2,x^3-17)$. All I am using here is that for any ring $A$ and ideals $I\subset J$, we have an isomorphism $(A/I)/(J/I)\simeq A/J$. So the question boils down to ...

3

The general strategy here is to take the exact sequence $$0\to \ker \pi\to R^2\overset{\pi}\to I\to 0$$ and find a splitting for $\pi$. Then we'd have $R^2\cong \ker\pi\oplus I$, so $I$ is projective as a direct summand of a free module. You have to implicitly use fractional ideals of $R$, so if you know what that is do some Googling, but if you don't then ...

3

Apply the isomorphism theorems for rings to justify the following $$\left(\Bbb Z[x]\right)/\left(\langle x\,,\,3\rangle\right)\cong\left(\Bbb Z[x]/\langle x\rangle\right)/\left(\langle x,3\rangle/\langle x\rangle\right)\cong\Bbb Z/\langle 3\rangle=:\Bbb F_3$$ So... Added as punishment for myself for being uncareful: $$\left(\Bbb Z[x]\right)/\left(\langle ... 3 You can easy answer to the 1st request using the isomorphism$$R/I \simeq (R/J) / (I/J)$$that holds if I and J are ideals such that J \subseteq I. Then you obtain$$\mathbb{k}[x_1,\dots, x_n]/(x_1-c_1, \dots, x_n-c_n) \simeq \simeq \mathbb{k}[x_1,\dots, x_n]/(x_n-c_n)/(x_1-c_1, \dots x_{n-1}-c_{n-1})^* \simeq \simeq ...

3

Your general statement is unfortunately false: Consider the irreducible polynomial $p_1(x)=p_2(x)=x^2+1\in \mathbb R[x]$. Then the ideal $\langle x^2+1, y^2+1\rangle \subset \mathbb R[x,y]$ generated by $p_1(x)$ and $p_2(y)$ is not maximal. Indeed $$\mathbb R[x,y]/\langle x^2+1, y^2+1\rangle\cong (\mathbb R[x]/\langle x^2+1\rangle)[y] /\langle ... 3 Let's take I=(a) and J=(b). Then IJ=(ab). I\cap J= (\operatorname{lcm}(a,b)) and I+J=(\operatorname{gcd}(a,b)). The fact that lcm and gcd exist is due to the fact that we're working in a UFD. Therefore (I\cap J)(I+J)=(\operatorname{lcm}(a,b)\cdot\operatorname{gcd}(a,b))=(ab). This holds generally for principal ideals in a UFD, but since every ... 3 I think your work is good, but I might take a few points off if I were grading this on a homework because I don't think you justified entirely why$$\mathbb{Q}[X,Y]/(X^2+Y,Y^2-2)\cong\mathbb{Q}[X]/(X^4-2).$$I suppose I'm not convinced (by reading your argument) that there may not have been other relations between X and Y that were forgotten after you ... 3 You can do it by hand very easily: two elements a+b\sqrt{2} and c+d\sqrt{2} lie in the same coset iff their difference belongs to 17R. Since 17R=\{17x+17y\sqrt{2}\colon x,y\in \mathbb Z\}, you have that a+b\sqrt{2} and c+d\sqrt{2} lie in the same coset iff a\equiv b\bmod 17 and c\equiv d\bmod 17. So you see that R/17R has 17^2 elements ... 3 This is how I look at such problems. First of all, \Bbb Z[\sqrt{2}]\cong \Bbb Z[x]/(x^2-2). Then using isomorphism theorems, \Bbb Z[\sqrt{2}]/(17)\cong \Bbb Z[x]/(x^2-2,17)\cong(\Bbb Z/(17))[x]/(x^2-2)=\Bbb F_{17}[x]/(x^2-2). (I'm taking some liberties with the notation: the parenthesis around elements denote the ideal generated in the ring in context. ... 3$$J_1\cap J_2=\emptyset \iff I_1+I_2=\Bbb{C}[x_1,\cdots, x_n] J_1\cup J_2=\Bbb{C}^n \iff I_1=(0)\; \text {or} \; I_2=(0)$$Edit (answer to Jack's request in the comments) The second equivalence relies on \mathbb C^n being irreducible in the Zariski topology. This means that the union of two closed subsets J_1,J_2\subset \mathbb C^n is the whole ... 3 Indeed, J is an ideal of A_0. Any product of two elements of A_0 belongs to J, since (f\cdot g)'(0) = f'(0)g(0) + f(0)g'(0) = f'(0)\cdot 0 + 0\cdot g'(0) = 0, and hence J is non-modular, as we have au\in J for all a,u\in A_0, and hence a-au = a-ua \in J \iff a\in J. Also, J is a 1-codimensional (closed) linear subspace of A_0, hence ... 3 Considering that the list of justifications hasn't been fully populated, I will provide an answer so as to remove this from the unanswered queue. (1) k is not an ideal, since it is not closed under multiplication using elements from outside k. For instance 1x_1\not\in k. (2) This is basically the same as (1). (3) This is actually an ideal! (4) You are ... 3 The distributive property states (a,b)(c,d)=(ac,ad,bc,bd). So we compute$$\begin{array}{ll} (2,1+\sqrt{-5})^2 & =(4,2+2\sqrt{-5},-4+2\sqrt{-5}) \\ & =(2)(2,1+\sqrt{-5},-2+\sqrt{-5}) \\ & = (2)(2,3,-2+\sqrt{-5}) \\ & = (2)(1)=(2). \end{array} by $3=(1+\sqrt{-5})-(-2+\sqrt{-5})$, valid since $(a,b)=(a-b,b)$. Thus $I^2=(\sqrt{2})^2$. By ...

3

The first thing to note is that the units of $k[[x_1,\dots,x_n]]$ are precisely the elements that have a non-zero constant term. I will let the proof of that to you (hint: try to directly compute an inverse). Hence if an element is not a unit, then it must have zero constant term. But then that means that the element must be inside the ideal ...

3

The maximal ideals are related to ultrafilters. Let $\def\U{\mathcal U}\U$ be a ultrafilter in $\mathbb N$, and let $I_\U$ be the set of sequences in $\mathbb R^\mathbb N$ such that set indices on which one of them vanishes is an element of $\U$. That is a maximal ideal, and this way you get all of them.

2

Below I show that the proof is an ideal-theoretic form of a well-known proof of  Euclid's Lemma  for integers. To highlight the analogy I give three proofs: first the common proof for integers using Bezout's Identity for the gcd; second, the translation in terms of gcds; third, the ideal translation. $\begin{eqnarray} Ax\!+\!ay&=&1,\ \ A ... 2 One characterisation that is implicit in some of the other answers, especially Sasha's, is the following: Ideals are exactly those subsets which can arise as the kernel of a ring homomorphism. That kernels are ideals is immediate (and can hence motivate the definition of an ideal), and the quotient construction (modding out) yields an epimorphism ... 2 Another place where you find ideals quite naturally is in polynomial rings of various kinds. Working with ideals helps to generalise ideas about dimension. And it is easy to give simple examples of ideals which are generated by more than one element. So for example, if there are three variables and integer coefficients we have$\mathbb Z[x,y,z]$with ... 2 Clearly$\phi^{-1}(N)$is nonempty. Suppose$x,y\in \phi^{-1}(N)$. Then$\phi(x),\phi(y)\in N$so$\phi(x)-\phi(y)=\phi(x-y)\in N$. So$x-y\in\phi^{-1}(N)$. If$a\in R, x\in \phi^{-1}(N)$. Then$\phi(ax)=\phi(a)\phi(x) \in N$, since$\phi(a)\in S$and$\phi(x)\in N$. So$ax\in\phi^{-1}(N)$, and similarly for the reverse multiplication. 2 It is enough to show that$R/\mathfrak{m} \cong R_\mathfrak{m}/\mathfrak{m}_\mathfrak{m}$, since$\mathfrak{m}/\mathfrak{m}^2$and$\mathfrak{m}_\mathfrak{m}/\mathfrak{m}_\mathfrak{m}^2$are just the base changes of the$R$-module$\mathfrak{m}$to these respective rings. This is straightforward with universal properties.$R\to R/\mathfrak{m}$and$R\to ...

2

Here is another possibility. EDIT: Thanks to Bruno for pointing out that I was being silly :) Because of this, the material between the %%%%%%%% is good motivation for the final argument, but no longer directly applicable. %%%%%%%% As a warm up, prove the following basic result: every Dedekind domain with finitely many prime ideals is a PID. This is ...

2

A surjective endomorphism of a noetherian ring is an automorphism. Take $f:A\to A$ a surjective endomorphism and the ascending chain of ideals $\ker f\subseteq\ker f^2\subseteq\cdots$. Then there exists $n\ge 1$ such that $\ker f^n=\ker f^{n+1}=\cdots$. Replace $f$ by $f^n$ and assume that $\ker f=\ker f^2$. Now take $a\in\ker f$. Since $f$ is surjective ...

2

If $R$ is an integral domain with fraction field $K$, then an overring of $R$ is a ring $T$ with $R \subset T \subset K$. Easy examples of overrings are given by localization at a multiplicatively closed subset $S$ of $R \setminus \{0\}$. For a general integral domain there are overrings which are not obtained by localization: this comes down to the fact ...

2

As already noted in the comments, embedding of noncommutative domains in general into division rings is fraught with peril: The domain may not "densely" embed into a division ring (this is what the condition "everything in the field is of the form $rs^{-1}$ for some $r,s\in D$" amounts to in commutative fields of fractions) The domain may not embed into ...

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