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9

How to prove that $(3,\sqrt{15})$ is not principal. Assume $(3,\sqrt{15})$ is principal - that is $(3,\sqrt{15})=(a+b\sqrt{15})$ for some $a,b\in\mathbb Z$. Write $\alpha=a+b\sqrt{15}$. What is $N(\alpha)=a^2-15b^2$? It must be a divisor of $N(3)=9$ and $N(\sqrt{15})=15$. We know $N(\alpha)$ cannot be $\pm 1$, or $\alpha$ would be a unit, and ...


5

The definition of $I^m$ is recursive: $I^1=I$ and $I^{m+1}=II^m$, where the right side is the product of ideals. Hint: Note that $1=i+j$ for some $i\in I,j\in J$. Then expand $1=(i+j)^{2m-1}$ to see that $1\in I^m + J^m$.


4

Maybe I'm dyslexic, maybe I've got it all backwards, but I think these questions are best answered in the reverse of the order you have posed them. How do I show that $R$ is (not) a UFD? Given primes $p$ and $q$, if $\mathbb{Z}[\sqrt{pq}]$ is a unique factorization domain, then both $p$ and $q$ split in this domain. That's what happens to 2 and 3 ...


3

Set $I=\langle xy^3, x^2y^2, x^3y\rangle$ and $m=\langle x, y\rangle$. $\dim I/mI=3$ Just show that the residue classes of the generators of $I$ are linearly independent over $K$, that is, if $axy^3+bx^2y^2+cxy^3\in mI$ with $a,b,c\in K$ then $a=b=c=0$. This is obvious since the generators of $mI$ are homogeneous of degree $4$. If $\dim I/mI=3$ ...


3

No, this is incorrect. To say that $n$ is a unit means that there is another element $m$ in the ring for which $nm = mn = 1_R$, where $1_R$ is the identity of the ring. For example, every non-zero element in $\mathbb{Q}$ is a unit in $\mathbb{Q}$, but the equation $nk = k$ can only hold for exactly one value of $n$. So what you can conclude is that $1_R \in ...


3

A maximal ideal is by definition a proper ideal (which happens to not be strictly contained in any other proper ideal). If $I$ is an ideal and an element $x\in I$ is invertible, then $x^{-1}x=e\in I$ so $a=a\cdot e\in I$ for all $a\in A$, so $I=A$. So no proper ideal (and in particular, no maximal ideal) can contain an invertible element.


3

It's not entirely clear to me what you are looking for exactly, but here is a short proof of your statement: Consider the $R$-module $N= \bigoplus_{x \in X} R_x$, where each $R_x$ is just a copy of $R_R$. Denote the unit of $R_x$ by $1_x$. Then the map of sets $f: X \to N, x \mapsto 1_x$ gives a unique module homomorphism $M \to N$ extending $f$ (and which ...


3

when are ideals also rings with unity? Proposition: An ideal $I\lhd R$ will be a ring with identity iff there exists a central idempotent $e$ such that $eR=I$. Proof: ($\implies$) The identity of $I$, call it $e$, is an idempotent element of $R$ and satisfies $I=eI$. Then $I=eI\subseteq eR\subseteq I$, so $I=eR$. Since $e\in I$, we have ...


3

Yes, you have to prove exactly what you say. Note that $A=1A$ is a principal ideal.


2

Hint: For every $\alpha \in \mathbb C$ consider the ideal $$ (x-\alpha^3,y-\alpha^2) \subset \mathbb C[x,y].$$ Can you show that every ideal of this form contains $I$?


2

Continuing from Hilbert's Nullstellensatz: suppose $(x^2-y^3)$ is contained in a maximal ideal $(x-a,x-b)$. Then we can write $x^2-y^3=p (x-a)+q (y-b)$ for polynomials $p,q$ and evaluating at $(a,b)$ gives that $a^2=b^3$. So your task is the same as showing the zero locus of $x^2-y^3$ is infinite. This is true since $\Bbb C$ is infinite and every pair ...


2

The (nonzero) prime and hence maximal ideals of $\mathbf{C}[x]$ are precisely linear polynomials. One can see directly that $(x^2-2)$ is not maximal in several ways though. One is to note that $x^2 - 2 = (x+ \sqrt{2}) (x-\sqrt{2})$. This shows that it is not irreducible and therefore, not prime (hence not maximal).


2

For $r\in\Bbb C$ let $$I_r=\left\{\begin{bmatrix}a&br\\0&0\end{bmatrix}:a,b\in\Bbb Q\right\}\;.$$ If $$\begin{bmatrix}c&s\\0&t\end{bmatrix}\in R\quad\text{and}\quad\begin{bmatrix}a&br\\0&0\end{bmatrix}\in I_r\;,$$ then ...


2

You can say something more precise: $\overline S=A-\bigcup_{\mathfrak m\in\operatorname{Max}(A);\ \mathfrak a\subseteq\mathfrak m}\mathfrak m$.


2

Let $\;I\;$ be an ideal contained in both $\;\mathfrak a\,,\mathfrak b\;$ . Then it is also contained in their intersection $\;\mathfrak a\cap\mathfrak b\;$ , and this proves the maximality.


2

Suppose $x_1>\cdots>x_n$. Then $g_i=g_i(x_i,\dots,x_n)$ is a polynomial in $x_i,\dots,x_n$ for $i=1,\dots,n$. In particular, $g_n$ is a polynomial only in $x_n$. Let $\alpha_n$ be a root of $g_n$. (There are at most $d_n$ roots.) Then consider $g_{n-1}(x_{n-1},\alpha_n)$. This has at most $d_{n-1}$ roots, and let $\alpha_{n-1}$ be one of them. Now ...


2

Here are some hints: An ideal of a quotient ring $R/I$ has the form $J/I$ where $J$ is an ideal of $R$ containing $I$. The ideals of $\Bbb Z$ are of the form $n\Bbb Z$ for some $n≥0$. The third isomorphism theorem states that $(R/I) \; / \; (J/I) \cong R/J$ In a PID (such as $\Bbb Z$), an ideal $I$ is prime iff it is maximal. Answer:


2

do we require the ring to have an identity element You may define a principal ideal of a commutative ring (or a principal right ideal of a ring) that way, but it is not standard. The problem is that you have no guarantee $a$ is in $aR$ if that is your definition. This is what having an identity does for you, and it is supposed to be an expectation of ...


2

Remark That $Q[X]/I_1=\{a+xb, a,b\in Q, x^2=2\}$ $Q[X]/I_2=\{a+yb, a,b\in Q, y^2=-2\}$. Suppose $\phi$ exists, write $\phi(x)=a+by$, $\phi(x^2)=2=(a+by)^2=a^2-2b^2+2aby=2$. This implies $ab=0$. If $a=0$, $-2b^2=2$ impossible in $Q$. Suppose that $b=0$, $a^2=2$ impossible in $Q$.


2

In general, if $I=(n,m)$ then $I=(gdc(n,m))$. Therefore $I=(15,12)=(3)$ and the quotient is the field $\mathbb{Z}/(3)=\mathbb{Z}_{3}$.


2

Suppose that $P$ is a maximal ideal, and let $Q\subset B$ be a maximal ideal containing $PB$. Then $Q\cap A\supseteq P$, so $Q\cap A=P$. If $P$ is not maximal, then localize at $A\setminus P$ and find a finite ring extension $A_P\subseteq B_P$. Now use the previous result for the maximal ideal $PA_P$.


2

Put $S=H_1\cap\cdots\cap H_n$. To see that $S$ is an ideal, just verify that for any $a\in S$ and $r\in R$, we have (i) $ra\in S$ because $\forall i:a\in H_i$ by definition of $S$ and $\forall i:ra\in H_i$ by definition of ideal, hence $ra\in S$; we also have that (ii) $a,b\in S\Rightarrow a+b\in S$, since certainly $\forall i:a,b\in H_i$ by definition of ...


2

It's pretty clear that $a = (2, 0) \not \in 4\mathbb{Z}\times\mathbb{Z}$, while $a^2 = (4, 0) \in 4\mathbb{Z}\times\mathbb{Z}$. There's a general method of finding whether given ideal of $I \subset \mathbb{Z}^n$ is prime. Let $I = (a_1, \ldots, a_k)$. Then, $I$ is an image of an obvious map $\mathbb{Z}^k \to \mathbb{Z^n}$. This map is given by a matrix ...


2

a) Let $a \in A$. Then $a = ea + (1-e)a \in I_1 + I_2$, so $A = I_1 + I_2$. As for the intersection, assume that we have an element $f$ in the intersection, i.e. there exists an $a$ and a $b$ such that $f = (1-e)a = eb$. We then multiply by $e$ to get $$ ef = e(1-e)a = e^2b\\ ef = 0 = eb = f $$ so $0$ is the only element in the intersection. b) There is a ...


2

The key is the Lattice Isomorphism Theorem. In fact the correspondence between the ideals given in the theorem extends to prime and also to maximal ideals. The latter should be obvious, you may wish to think about why the former is also true. Now you are reduced to finding the prime/maximal ideals of $\mathbf{Z}$ that contain the ideal $(12)$.


2

Note that $\mathbb{Q}$ is a field, so $\mathbb{Q}[x]$ is a PID. In a PID $P$, if $p\in P$ then $\langle p\rangle$ is maximal iff $p$ is irreducible in $P$. Now simply observe that $x^2-2$ has no roots in $\mathbb{Q}$, thus $x^2-2$ is irreducible in $\mathbb{Q}[x]$, so $\langle x^2-2\rangle$ is maximal.


2

Note that if $f \in R$ with $f(0) \neq 0$ then $f$ is invertible in $R$ (as $\frac{1}{f} \in R$). If an ideal $I$ of $R$ contains a function $f$ with $f(0) \neq 0$ then it contains $\frac{1}{f} \cdot f = 1$ and so we must have $I = R$. This shows that $M$ is maximal and that any proper ideal of $R$ is contained in $M$ so $M$ is the only maximal ideal of $R$. ...


2

It's nice to see people studying non-UFDs besides $\textbf{Z}[\sqrt{-5}]$. If $\langle 3, \sqrt{15} \rangle$ was a principal ideal, we could find a number $n \in \textbf{Z}[\sqrt{15}]$ such that $\langle 3, \sqrt{15} \rangle = \langle n \rangle$. Then $\langle n \rangle$ contains numbers with a norm of $9$ and numbers with a norm of $15$. Since $\gcd(9, 15) ...


2

For $s\in U$ and $x\in M$ we have $sx=0\implies sx\in Q$, where $Q$ is a primary submodule of $M$ which appears in a primary decomposition of $(0)$ and $r_M(Q)\cap U=\emptyset$. Conclude that $x\in Q$. For the converse, $x\in\cap Q_j$ with $r_M(Q_j)\cap U=\emptyset$. On the other side, for some $Q_i$ such that $r_M(Q_i)\cap U\ne\emptyset$ we get an $s_i\in ...



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