Tag Info

New answers tagged

1

If $\alpha/F$ has deg $n$ then $\alpha$ cannot exist in an extension of $F$ of deg $<n$. One could argue this using vector space dimensions (if $F(\alpha)\subseteq L$ then $n\le \dim_FL$), although you can also employ the transitivity of field degrees if you want (seems only slightly stronger than necessary). Either way, we may conclude that since ...


0

There's an obvious difference insofar as $J$ of part 1 is defined in terms of a ring element $z$, whereas $I$ of part 2 is defined in terms of a ring element $z$ and an ideal $J$ (which should not be confused with the ideal $J$ defined in the first part of the problem). Actually if you do take the $J$ as defined in part 1 and also the same $z$, then you ...


1

Suppose $K$ is any subfield of $E$ containing $\alpha$ and $F$. Let $f(x)$ be any element of $F[x]$, say: $f(x) = a_0 + a_1x + \cdots + a_nx^n$. Now $\phi_{\alpha}(f(x)) = a_0 + a_1\alpha + \cdots + a_n\alpha^n$. Since $F \subset K$, we have that $a_0,\dots,a_n \in K$. Since $\alpha \in K$, we have that: $\alpha,\cdots,\alpha^n \in K$, by closure of ...


1

It's because if $K$ contains $\alpha$, it also must contain any power of $\alpha$, and any linear combination of these powers with coefficients in $F$. Isn't that the definition of the image of $F[x]$ under $\phi_\alpha$?


1

Not too difficult: let $\alpha \in K\subset E$ with $F\subset K$. Set $F(\alpha) = \phi_{\alpha}(F[x])$. Then, since $F[x]$ is generated by $F$ and $x$, we see that $F(\alpha)$ is generated by $F$ and $\phi_{\alpha}(x) = \alpha$. Since $F\subset K$ and $\alpha\in K$, then the field generated by $F$ and $\alpha$ is also in $K$. But this is just ...


2

You have to prove that if $r\equiv s \mod I$, then $rm=sm$ for any $m\in $M$. That is equivalent to $(r-s)m=0$, which is by definition since $r\equiv s\mod I\iff r-s\in I\subseteq\operatorname{Ann}_AM$.


1

If $S$ is a set (for example $S=\{X_1,\ldots,X_n\}$) and $A$ a ring, the polynomial ring $A[S]$ (more usually written $A[X_1,\ldots,X_n]$ if $S=\{X_1,\ldots,X_n\}$) can be defined as a ring $B$ together with a ring homomorphism $\iota\colon A\to B$ as well as a map $i\colon S\to B$ such that for any other ring $C$ and ring homomoprhism $\phi\colon A\to C$ ...


0

In $\mathbb Q[x]/(x^5-3)$, clearly $x^2+(x^5-3)$ is a root of the polynomial $X^5-9$. On the other hand, in $\mathbb Q[x]/(x^5-9)$, $x^3+(x^5-9)$ is a root of $X^5-9^3$ so that $\frac13x^3+(x^5-9)$ is a root of $X^5-3$. We conclude that $x\mapsto x^2$ and $x\mapsto \frac13x^3$ induce field homomorphisms in both directions, which must be isomorphisms.


3

Hint: Calculate $(\root5\of 3)^2$ and $(\root5\of 9)^3$.


4

Let's see some of the numbers in $\mathbb{Q}(\sqrt[5]3)$. We clearly have $\sqrt[5] {3}$. So we must have its powers $$ \sqrt[5] {3}, \sqrt[5] {9},\sqrt[5] {27},\sqrt[5] {81}.$$ And so in particular, we can see that $$ \mathbb{Q}(\sqrt[5]{9}) \subset \mathbb{Q}(\sqrt[5] 3).$$ Since both are degree five extensions, this subset relation must actually be an ...


0

In $\mathbb Z$, every ideal is principal.


0

Let $d=gcd(a,b)$. Then $a\mathbb{Z}+b\mathbb{Z}=d\mathbb{Z}$, sine the Euclidean algorithm yields integers $r,s$ with $ra+sb=d$. Hence $2\mathbb{Z}+3\mathbb{Z}=\mathbb{Z}$.


2

We have $3+-2=1 \in (3,2)$, hence the generated ideal contains $1$, hence it contains $\mathbb Z$


1

The following is true: $$R^\times=(R[X])^\times$$ Proof: Let $a\in R^\times\subseteq R$. Then the inverse of $a\in R$ is just the inverse of $a$ in $R[X]$. Conversely if $f\in (R[X])^\times$ with the inverse $f^{-1}$ then $$0=deg(1)=deg(f\cdot f^{-1})=deg(f)+deg(f^{-1})\tag{1}$$ Hence $deg(f)=deg(f^{-1})=0$. Thus $f$ and $f^{-1}$ are invertible ...


1

For the first, I think there is no direct proof other than contradiction, since this result is independent of ZF and to prove it one must at least assume axiom of dependent choice. For the second, let $F$ denote the fraction field of $R$. For any ascending chain $(f_1)\subseteq(f_2)\subseteq(f_3)\subseteq\cdots$ of principal ideal in $R[X]$, note that since ...


2

If $\alpha\neq\beta$, then $x-\alpha$ and $x-\beta$ are comaximal in $\mathbb F_p[x]$ (that is $(x-\alpha)+(x-\beta)=(1)$. So Chinese Remainder Theorem implies $\frac{\mathbb F_p[x]}{(x-\alpha)(x-\beta)}\cong\frac{\mathbb F_p[x]}{(x-\alpha)}\times\frac{\mathbb F_p[x]}{(x-\beta)}\cong\mathbb F_p\times\mathbb F_p$.


2

You actually have two claims written, both of which are true. The first is: Is it always true that if $1+i$ divides a gaussian integer then the norm of $1+i$ divides norm of the gaussian integer. Here's the proof: Suppose $1+i \mid a+bi$, meaning $(1+i)z=a+bi$ for some Gaussian integer $z$. Taking absolute values gives $2|z|^2 = a^2+b^2$, as needed. ...


2

As alluded to in the comments, since integral domains can be realized as subrings of fields, one could characterize integral domains as "the class of subrings of fields." The integral domains that aren't fields are just such subrings that don't happen to be fields. This is the most straightforward classification you can hope for. If you are still looking ...


1

For any integral domain $D$, the polynomial ring $D[x_1, \ldots, x_k]$, $k > 0$. For any connected, open subset $U \subseteq \mathbb{C}$, the ring of holomorphic functions on $U$. (Note that the space of merely smooth functions on $U$ is not an integral domain.) (Since all finite domains are fields, all examples of such rings are infinite.)


1

I think computing the quotient (as sugggested in the other answer) is the easiest way to solve this. Nevertheless here is an alternative: If you are fine about $\mathbb Z[i]/(q)$ being a field, you can find the number of elements by looking only at the abelian group structure. As an $\mathbb Z[i]$ is generated by $1,i$, hence so is $G := \mathbb Z[i]/(q)$. ...


0

Just use the definition: $K(u_1,\dots ,u_n)$ is an extension field of $K(u_1,\dots ,u_{n-1})$ which contains $u_n$, so $K(u_1,\dots ,u_{n-1})(u_n) \subset K(u_1,\dots ,u_n)$. On the other hand, $K(u_1,\dots ,u_{n-1})(u_n)$ is an extension of $K$ containing $u_1, \dots , u_n$, so $K(u_1,\dots ,u_n) \subset K(u_1,\dots ,u_{n-1})(u_n)$. Obviously, switching ...


2

You are correct in noting that the case $r= 1$ is impossible because $m$ was assumed to not be a unit. The point is that the condition $r= 0$ or $d(r) < d(m)$ restrict $r$ to be $0$, $1$, or $-1$, and only two of these cases are possible by our assumption on $m$.


1

You can get this field from $\mathbb{Z}[x]$ by first killing $x^2+1$ and then killing $q$, as you have shown above. But you could also get there by killing $q$ first, and then $x^2+1$. In this way, we get that the field is isomorphic to $F_p[x]$ mod $x^2+1$. How many elements does this field have?


1

There is a theorem of Issac which says that if every prime ideal is principal then so is every ideal.


1


3

"Always" is wrong. For example $f(x)=x-a$ is irreducible.


1

Your second guess is a correct description of $\Bbb Q(\pi)$ (as long as you specify that the denominator is not the zero polynomial). The first guess describes the ring $\Bbb Q[\pi]$, which does not have inverses. In particular, $\pi$ has no inverse since $\pi f(\pi) \ne 0$ for any polynomial $f\in \Bbb Q[x]$. It's not hard to see that these rational ...


0

It is your second option. The first is called $\Bbb Q[\pi]$. In the example they make $F=\Bbb Q$ and $\alpha=\pi$.


2

and I still can't seem to prove that if they do [correspond], $P$ can't be some non-principal ideal properly larger than $(x)$. Assuming you convince yourself of the correspondence of prime ideals (which is just fine) here's an elementary way to see that there can only be one prime ideal in this ring. Now $(x)$ is a maximal ideal of $\Bbb F[x]$ for any ...


4

Prime ideals do correspond under the correspondence theorem, so your argument suffices. To see this, Let $I\subset P\subset R$ be any prime in $R$ containing an ideal $I$, then $R/P \cong \frac{R/I}{P/I}$ by the 3rd isomorphism theorem. Since $P$ is prime, $R/P$ is an integral domain, hence so is $\frac{R/I}{P/I}$. Thus, $P/I$ is a prime ideal in $R/I$. ...


1

I'm very far from being an expert on these things, but I suspect that the answer might be the trivial (and disappointing?) one: the property which distinguishes GCD domains is ... the existence of GCDs. Compare the chain of (strict) class inclusions $\qquad$ UFDs $\subset$ bounded factorization domains $\subset$ ACCP domains $\subset$ atomic domains to ...


0

Let us consider the sets$$R=\left\{ \begin{pmatrix} a & b\\ 0 & c\end{pmatrix}:a,b,c\in\mathbb{Z}\right\}\\J=\left\{\begin{pmatrix} 0&a\\0&b\end{pmatrix}:a,b\in \mathbb{Z}\right\}\\I=\left\{\begin{pmatrix}0 &a\\0&0\end{pmatrix}:a\in\mathbb{Z}\right\}$$Now we can easily check that $R$ is a ring with respect to usual matrix addition and ...


1

Without additional conditions on (say) $u$ and $m$, this is certainly false. For example, if $\mathfrak{p}$ is prime in $\mathcal{O}_K$, let $u$ be any nonzero element of $\mathfrak{p}$. Then $(u)\subseteq \mathfrak{p}$, so that $|u|_{\mathfrak{p}}>0$. For $m$, if $\mathfrak{p}$ lies over the rational prime $p$, let $m=p$. Then $|m|_{\mathfrak{p}}>0$.


1

There is a more general way to go about this. Consider the augmentation map $\varepsilon:FG\to F$ defined by $\sum_{g\in G} a_gg\mapsto \sum_{g\in G} a_g$ for $a_g\in F$, which is a ring homomorphism. The kernel $\mathcal{A}_G\subset FG$ of this map is called the augmentation ideal. One can show that $\mathcal{A}_G$ is nilpotent when $G$ is a finite ...


4

I think you refer to a theorem of Kummer: Fermat's Last Theorem is true for an odd prime $p$ if and only if $p$ doe not divide the class number of the cyclotomic extension $\mathbf Q(\zeta_p)$, i. e. the order the the group of fractionary ideals of this field modulo principal ideals. Such a prime number is called a regular prime. Kummer criterion: An odd ...


2

You must keep in mind Proposition 10.12: If $$0 \to M' \to M \to M'' \to 0$$ is an exact sequence of $A$ modules and $\hat{M}', \hat{M}, \hat{M}''$ are their ${\mathfrak a}$-adic completions then $$0 \to \hat{M}' \to \hat{M} \to \hat{M}'' \to 0$$ is exact too. This gives you first $\widehat{A^n} = \hat{A}^n$ by considering the sequence $$0 \to A^{n-1} ...


2

The proof is this: Let $U$ be the set of unit of your ring $R$. Let's suppose that $1 \in R$. Then: 1) It exists the neutral element; 2) If $a,b \in U$, then $a*b \in U$, because $(a*b)^{-1}=b^{-1}*a^{-1}$; in fact: $$(a*b)*(b^{-1}*a^{-1})=a*(b*b^{-1})*a^{-1}=a*a^{-1}=1$$ 3) The associativity follows from the definition of $R$ as a ring.


0

It might be good to write the definition of convex ideal in $C(\mathbb{R})$: An ideal $I\subset C(\mathbb{R})$ is said to be a convex ideal if for every $f,g\in C(\mathbb{R})$ such that $0\leq f\leq g$ and $g\in I$, we must have that $f\in I$. Assume that $0\leq f(x)\leq xg(x)$ with $f,g$ continuous. In particular $g(x)\leq0$ for $x<0$ and ...


1

Let $R=K[X,XY,XY^2,\dots, XY^n,\dots]=K+XK[X,Y]$, and $I=(X)$. Let $P$ be a prime ideal containing $X$. Then we have $(XY^m)^2=X(XY^{2m})\in P$, thus $(X,XY,XY^2,\dots, XY^n,\dots)\subseteq P$. But $(X,XY,XY^2,\dots, XY^n,\dots)$ is maximal, so $(X,XY,XY^2,\dots, XY^n,\dots)=P$. We have $(0)\subsetneq P'\subsetneq P$, where $P'=(XY,XY^2,\dots, ...


2

Suppose 1 is false, so there is a submodule M' of M that is not finitely generated." That is not the negation of "$M$ is finitely generated" unless you are talking about $M=M'$. Using the notation $M'$ basically serves no purpose. So, you are not really having any problems in your proof of 3 implying 1. @MooS isn't a noetherian ring a ring that ...


3

The zero ideal is in fact maximal in $M_2(\mathbb{R})$, as it is for $M_n(R)$ for any $n>1$ and ring $R$ (see Why is the ring of matrices over a field simple?). I will give the proof for $M_n(F)$, $F$ a field and $n>1$ since vector space theory makes it a bit more elegant. To prove the zero ideal is maximal, we take any non-zero ideal $\mathfrak{A}$ ...


1

I'll assume the isomorphism is as $R$-modules (or the result is false, see Can $R \times R$ be isomorphic to $R$ as rings?). Suppose $R/A\oplus R/B\cong R/AB$, in particular is cyclic, and that $(x+A,y+B)$ is a generator. Then, for some $r\in R$, $(1+A,0+B)=r(x+A,y+B)=(rx+A,ry+B)$, so $$ 1=rx+a,\qquad ry\in B $$ for some $a\in A$. Then $y=y1=ay+rxy\in A+B$. ...


3

The proof for part (a) is good. For part (b) you just have to show that any nonzero element in $R$ has a multiplicative inverse and $0\ne1$ if and only if $(R\setminus\{1\},\circ)$ is an abelian group. The other field properties are already guaranteed by the fact that $R$ is a commutative ring. If $(R\setminus\{1\},\circ)$ is a group, then it is nonempty, ...


0

The ring hom $\rm\:g\mapsto \bar g = g + \left<f\right> \:$ preserves $\rm\:\color{#a0f}{sums}\,\ \&\,\ \color{#0a0}{products}\:$ so it preserves $\rm\color{#c00}{polynomials}$ by induction, since polynomials are compositions of sums and products. $ $ More explicitly $$ \begin{eqnarray} \rm 0\, =\ \overline{\color{#c00}{f(x)}}\: &=&\rm\ ...


2

Hint $\ $ Any set bijection $\,h\,:\,R'\to R\,$ serves to transport the multiplicative structure of $\,(R,*,1),$ to $\,(R',\circ,1') \,$ by defining the operations in $\,R'\,$ to make $\,h\,$ be an isomorphism $$\begin{align} &a \circ b\, =\, h^{-1}(h(a)\, * h(b)),\quad\, 1' = h^{-1}(1)\end{align}\qquad \qquad$$ Yours is the special case $\ h(x) = ...


3

As stated in the comments, your proof is correct. However, It would make the proof more readable if you could more thoroughly explain steps following the equality $-yx = xy$. Alternatively, it might be easier to explicitly state that $x = -x$ for all $x \in R$, as MooS explains.


1

$\newcommand{\Ass}{{\mathrm{Ass}}}$ I assume $R$ noetherian. We write $I=Q_1 \cap \cdots \cap Q_r$ as the representation of $I$ as an intersection of primary ideals. Then $(I:x) = (Q_1:x) \cap \cdots \cap (Q_r:x)$ as this is a general property of intersection of ideals. Now $(Q_i:x)$ is equal to $R$ if $x \in Q_i$ and $(Q_i:x) \subseteq \sqrt{Q_i} = P_i$ ...


2

If $a \neq 1$ is a unit, and $b \neq 0$ is not, then $ab$ is not a unit either; further, we cannot have $ab = 0$, thus $ab = b$. But then, since $a \neq 1$, $a - 1$ is a zero divisor, so not a unit, so $a - 1 = b$. So there can be at most one unit which is not equal to $1$. Thus if $R$ is a ring with precisely two nonunits then $3 \leq |R| \leq 4$. When ...


2

No need to use Zorn's Lemma. The proof is quite elementary. Lemma 1. If $A,B,C$ are $R$-modules such that $A \oplus B$ is cyclic, then every bilinear map $\beta : A \times B \to C$ is trivial. Proof. Let $(u,v)$ be a generator of $A \oplus B$. Then $u$ is a generator of $A$ and $v$ is a generator of $B$. It suffices to prove $\beta(u,v)=0$. Choose some $r ...



Top 50 recent answers are included