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7

A good example to add. "There is only one group of order 2 up to isomorphism". The group is $G=\{e,a\}$ with $ea=ae=a$ and $a^2=e$. Now you could argue "I have another group, $H=\{e,b\}$ with $eb=be=b$ and $b^2=e$". Your group IS different than mine, but your group is just the renamed version of mine. "Renaming" is isomorphism.


4

This answer uses the information, given in OP, that $2, 1 + \sqrt{-5}$ are irreducible elements that are not associated. Suppose $I = (a)$. If $\{2, 1 + \sqrt{-5}\} \subset I $ then $a \mid 2$ and $a \mid 1 + \sqrt{-5}$. Thus as $2, 1 + \sqrt{-5}$ are irreducible: either $a$ is a unit, or $a$ is associated to $2$ and $a$ is associated to $1 + ...


4

The field $Q[i\sqrt2]$ has degree two over $Q$ and so does $Q[\sqrt2]$. If the first contained $\sqrt2$ then the two would be equal. Yet the second one is contained in $R$ and the other is not. On the other hand, the minimal polynomial $\sqrt[4]2$ over $Q$ has degree four, so that number cannot be in $Q[\sqrt2]$.


4

I guess that's fine. It would however have been enough to simply compute $A^TA=I$ explicitly and conclude $A^{-1}=A^T$ from that. As a sidenote observe that $A^TA=I$ simply states that the colmn vectors of $A$ are pairwise orthogonal and of length $1$.


3

You have already shown that it is necessary that a unit have norm of (plus or minus) one. To show sufficiency, assume that $a + b\sqrt{2}$ is such that $N(a+b\sqrt{2})=a^2 -2 b^2=\pm 1$. Then notice that $(a+b\sqrt{2})(a-b\sqrt{2})=a^2-2b^2=\pm 1 $. Hence $a-b\sqrt 2$ is the desired multiplicative inverse, and so $a+b\sqrt 2 $ is a unit.


2

A maximal ideal of $\mathbb{Z}\times\mathbb{Z}$ must be of the form $I\times\mathbb{Z}$ or $\mathbb{Z}\times J$ where $I$ and $J$ are maximal ideals of $\mathbb{Z}$. For example $p\mathbb{Z}\times \mathbb{Z}$ is a maximal ideal if $p$ is a prime. $(\mathbb{Z}\times\mathbb{Z})/(\{0\}\times p\mathbb{Z})$ is isomorphic to $\mathbb{Z}\times \mathbb{Z}_p $ ...


2

Note that units can easily have norm $-1$. For instance, if you look at $1 + \sqrt{2}$, the norm is $1^2-2(1^2) = -1$, but $(1+\sqrt{2})(-1+\sqrt{2}) = 1$. You should instead prove that the units are precisely the elements with norm $\pm 1$.


2

I just thought that I would recast ajotatxe’s argument in a slightly different manner. Observe that we have a multiplicative function $ N: \Bbb{Z}[\sqrt{-5}] \to \Bbb{N}_{0} $ defined by $$ \forall (a,b) \in \Bbb{Z}^{2}: \quad N(a + b \sqrt{-5}) \stackrel{\text{df}}{=} a^{2} + 5 b^{2}. $$ Suppose that $ \{ 2,1 + \sqrt{-5} \} \subseteq \langle x \rangle $ ...


2

Since the submodule is torsion free, the mapping from $R$ to $\langle h \rangle$ by $r \mapsto rh$ is an isomorphism.


2

Consider the $A$-module (or $R$-module) epimorphism $$I\oplus J \rightarrow I+J=A, \;\; (i, j)\mapsto i+j \in A,$$ compute its kernel and use the first isomorphism theorem. Now, if you know that $A$ is a projective $A$-module and know what properties projective modules have, this should give you the isomorphism you are looking for. If not, the isomorphism ...


2

Hint: consider $$f(x) = \begin{cases} 1, & x \le 0 \\ 0, & x>0\end{cases}$$ and $$g(x) = \begin{cases} 0, & x \le 0 \\ 1, & x>0\end{cases}.$$ What do you know about $fg$? (You can adapt this to even work for continuous (or even smooth) functions.)


1

The point is exactly that $A$ itself is a projective $A$-module, but you can understand the proof as long as you know that $A$ is a free $A$-module with basis $\{1\}$. First the inclusion maps $I\subset A$ and $J\subset A$ give a homomorphism $\pi:I\oplus J\to A$. Since $I+J=A$, $\pi$ is a epimorphism and there exist $i\in I$ and $j\in J$ such that ...


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 ...


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

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

I think that the automorphism is $(a_{ij}) \mapsto (b_{ij})$ where $$b_{ij} = a_{(n-j+1),(n-i+1)}$$ corresponding to switching entries symmetrically with respect to the 'other diagonal' of the matrix (not the usual diagonal, but the other one). Then rows become column and viceversa, and being upper triangular is preserved, and products row-by column are ...


1

The degree of each commutator is either 0 (for constants) or at least two. The degree of a product of commutators is the product of the degrees. It follows that you cannot have a product of commutators of degree 1, and the only possibility to get a product of commutators of degree 2, is the commutator $[x_i,x_j]$.


1

It's an easy exercise to show that Rank(Def2) is equal to $\dim_K(K\otimes_RM)$, where $K$ is the field of fractions of $R$. This shows that Rank(Def2) is well-defined, and this is by definition the rank of $M$ in such context. "Is every cardinality of a $R$-linearly independent subset of a finitely generated $R$-module finite?" Yes, it is. In fact its ...


1

This is because, as $Rp$ is a free $R$-module, the short exact sequence: $$0\to N'_a\to N_a\to Rp\to 0$$ splits.


1

Clearly, as $n>0$, $$ \lim_{x\to-\infty}-a_0(x)= \lim_{x\to-\infty} \bigl(a_n(x)e^{nx} + a_{n-1}(x)e^{(n - 1)x} + ... + a_1(x)e^x\bigr)= 0 $$ So… (Note: the only needed property is $\lim_{x\to-\infty}x^he^{kx}=0$ for $h\ge0$, $k>0$.)


1

You can do it, well, by using calculus. Suppose $$\sum_{i=0}^{n}a_i(x)e^{ix}=0$$ with $n$ being smallest possible, i.e. $a_n(x)\neq 0$. In particular, $$\lim_{n\rightarrow \infty}a_n(x) \neq 0.$$ Then it follows that for any $\alpha \in \mathbb{R},$ $$\lim_{x \rightarrow \infty} \frac{\sum_{i=0}^{n}a_i(x)e^{ix}}{e^{\alpha x}}=0.$$ Also, for positive ...


1

According to a paper by Dummit, "SOLVING SOLVABLE QUINTICS", a quintic is solvable by radicals iff its Galois group is is contained in the Frobenius group $F_{20}$ of order $20$ in the symmetric group $S_5$. An algorithm and formula is provided as well. Abstract: It is well–known that an irreducible quintic with coefficients in the rational numbers ...


1

Assume that $2$ and $1+\sqrt{-5}$ are in an ideal generated by $a+b\sqrt{-5}$. Let's start with the equation $$2=(a+b\sqrt{-5})(c+d\sqrt{-5})$$ take modules and square: $$4=(a^2+5b^2)(c^2+5d^2)$$ Now it is clear that $b=d=0$, and $a^2c^2=4$. But, on the other hand, $$1+\sqrt{-5}=(a+b\sqrt{-5})(c'+d'\sqrt{-5})=ac'+ad'\sqrt{-5}$$ This yields: $$ac'=1$$ ...


1

On Isomorphism and Equality: Intuitively, it means that there are two group structures of order $4$. But there's a problem: there are many non-equal groups of order $4$ (see below). The proper way to formulate this idea is to say there are only two groups of order $4$ up to isomorphism (again, see below). To illustrate, here are several groups of order ...


1

The following is a very general discussion of the idea of an isomorphism without going into detail about what types of structures we are talking about (e.g., groups, rings, vector spaces, etc.) since the general idea should be the same across each of these. What does it mean for two structures to be isomorphic to each other? It means that there is a way to ...


1

The difficulty may be that $A/P$ is set-theoretically not a subring of $A/Q$. You rather have an inclusion ring homomorphism $A/P → A/Q$ coming from the composition $A → B → B/Q$, factoring through its kernel which is $A ∩ Q = P$. From your point of view, you should say that the equation $$b^n + a_{n-1}b^{n-1} + … + a_0 = 0$$ holds in $B$, and therefore ...


1

No need for $P,Q$ to be prime. For $b\in B $$$b^n+a_{n-1}b^{n-1}+\dots+a_0=0$$ so $$(b^n+Q)+(a_{n-1}b^{n-1}+Q)+\dots+(a_0+Q)=Q$$ so $$(b+Q)^n+(a_{n-1}+Q)(b+Q)^{n-1}+\dots+(a_0+Q)=Q$$ so $$(b+Q)^n+(a_{n-1}+P)(b+Q)^{n-1}+\dots+(a_0+P)=Q$$ Therefore $b+Q$ is integral over $A/P$


1

Hint: $\alpha^{24}=(\alpha^2)^{12}$ Alternative hint: Don't even rewrite $\alpha$ as product of disjoint cycle. Just note that it is a permutation of $\{1,3,5,6\}$ only and that $|S_4|=24$.


1

"Up to isomorphism" just means that two groups that are isomorphic are considered the same group. For example on the set $\{a,b\}$ you can define two groups, one whose neutral element is $a$, and one whose neutral element is $b$. However those two groups are isomorphic, thus up to isomorphism there's only one group.


1

If $\mathbb{Q}[\sqrt{2}i]$ contains $\sqrt[4]{2}$, then $\mathbb{Q}[\sqrt{2}i]=\mathbb{Q}[\sqrt{2}i][\sqrt[4]{2}]$ and therefore $$2=[\mathbb{Q}[\sqrt{2}i]:\mathbb{Q}]=[\mathbb{Q}[\sqrt{2}i][\sqrt[4]{2}]:\mathbb{Q}[\sqrt[4]{2}]]\cdot [\mathbb{Q}[\sqrt[4]{2}]:\mathbb{Q}]= k\cdot 4$$ where $k=[\mathbb{Q}[\sqrt{2}i][\sqrt[4]{2}]:\mathbb{Q}[\sqrt[4]{2}]]$, ...



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