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6

Note that $(a+1)\oplus (b+1)=(a+b)+1$: This makes us suspicious and we compute $$\begin{align}(a+1)\odot(b+1)&=(a+1)(b+1)-(a+1)-(b+1)+2\\ &=ab+1.\end{align}$$ We conclude that $x\mapsto x+1$ maps the commutattive ring with unity $(\mathbb Z,+,\cdot)$ to the structure $(\mathbb Z,\oplus,\odot)$, which is therefore also a commutative ring with unity. ...


0

It suffices to know the comultiplication induced by $\mu$ on cohomology. For degree reasons the unique possibility is $$\Delta(\alpha) = \alpha \otimes 1 + 1 \otimes \alpha$$ and from here you are in the realm of a purely Hopf-algebraic calculation: you want to apply the "$n^{th}$ comultiplication" $\Delta_n : H \to H^{\otimes n}$ (where $H$ is the Hopf ...


1

I'm assuming the domain is $\mathbb Z$, but the same argument would work for $\mathbb Q$, $\mathbb R$, or really any ring. In general, if $f:X\to\mathbb Z$ is $1-1$ and onto, we can define operators on $X$: $$x\oplus y = f^{-1}(f(x)+f(y))$$ $$x\odot y = f^{-1}(f(x)\cdot f(y))$$ In these cases, the additive identity is $f^{-1}(0)$. In your case above,this ...


5

We want to show that $\odot$ distributes over $\oplus$. However, we can NOT assume that $\odot$ distributes over $+$. For the first part, we have: \begin{align*} a \odot(b \oplus c) &= a \odot (b + c - 1) \\ &= (a)(b + c - 1) - ((a) + (b + c - 1)) + 2 \\ &= (ab + ac - a) - (a + b + c - 1) + 2 \\ &= ab + ac - 2a - b - c + 3 \\ \end{align*} ...


2

Consider the polynomial $X^2-aX+b \in \mathbb{C}[X]$. It has two roots ...


4

If $R$ is a ring, then the two sided ideals in the ring $M_n(R)$ of $n\times n$ matrices are in bijection with the two sided ideals of $R$. The ring $\mathbb{Z}/2^k\mathbb{Z}$ has exactly $k$ proper ideals. If $F$ is a field, then $F\times F$ has exactly two (isomorphism classes of) simple modules (irreducible and simple are synonyms).


1

Intuitively, $rI$ might be very different from $I$ in terms of its members, so intersecting its subset with $I$ might not be useful. If $J$ is a right ideal in $rI$, then, being a subset of $rI$, it has to arise as $rX$ for some $X\subset I$. Can you prove that $X$ is a right ideal? You might need to tighten the specification of $X$ a little to make it ...


2

No. Unfortunately, determining if two multivariate polynomials have a common factor is quite a bit trickier than doing the same with single variable polynomials. Finding out whether $F$ and $G$ have a non-trivial common factor is generally hard to do by hand. However, there are techniques to handle this problem. The greatest common divisor (GCD) of $F$ and ...


1

This has been asked on mathoverflow and received a couple of answers: MO/32875: Lifting units from modulus n to modulus mn. MO/31495: When does a ring surjection imply a surjection of the group of units? Your proof has already the right idea: Using the chinese remainder theorem, one easily reduces to the case of powers of a prime $p$. Now if $n \leq m$, ...


2

The generator $x^2 + x + 2$ of the ideal is quadratic, so every element admits an (in fact unique) linear representative, $$ ax + b . $$ We want to find the element $ax + b + I$ such that $$[(2x + 3) + I][(ax + b) + I] = 1 + I.$$ Simplifying gives $$ 2a x^2 + (3a + 2b) x + (3b - 1) \in I.$$


2

You need $$(2x+3)p(x)\equiv1\pmod{x^2+x+2}$$ in ${\Bbb Z}_5[x]$. That is, $$(2x+3)p(x)+(x^2+x+2)q(x)=1\ ,$$ and you can solve this by using the Euclidean algorithm (not forgetting that all the arithmetic of coefficients is modulo $5$). If you have seen how to use the Euclidean algorithm in $\Bbb Z$ to solve equations such as $23x+112y=1$, this problem is ...


1

Consider a non-noetherian valuation ring of rank two. For such an example you can take a look at Examples of Non-Noetherian Valuation Rings.


1

Suppose $[f] = [\phi], [g] = [\psi].$ Then we need to say that $[f] + [g] = [\phi] + [\psi].$ By definiton, $\exists \,U \ni x_0\,, f|_U = \phi|_U$ and $\exists \,V \ni x_0\,, g|_V = \psi|_V.$ So $(f + g)|_{U \cap V} = (\phi + \psi)|_{U \cap V} \Rightarrow [f + g] = [\phi + \psi].$ This proves that the addition is well defined. Similarly, you can prove that ...


2

What structure has R/I when I is a maximal left ideal? The best we can say is that it is a left $R$ module and not necessarily anything more. moreover if $I$ is maximal then it is a field, No, in general you will only get a simple ring. However, can we put the word bilateral instead of left or right in (⋆⋆)? It will not yield the Jacobson ...


2

A valuation ring of rank $n$ has the property you are looking for.


1

No in the Noetherian case. For the proof it suffices to show there is no Noetherian local domain $(R,{\mathfrak m})$ with precisely one non-zero, non-maximal prime ideal $\{0\}\subsetneqq {\mathfrak p}\subsetneqq{\mathfrak m}$. Namely, if we had such a domain and if $x\in {\mathfrak m}\setminus{\mathfrak p}$, then ${\mathfrak m}$ would be minimal among the ...


1

If char$\,K\mid n\;$, say $\;n=pr\;,\;\;p\;$ a prime, then $\;x^n-1=x^{pr}-1=(x^r-1)^p\;$ and this polynomial has at most $\;r<n\;$ different roots in $\;K^*\;$ and we're done. Otherwise, define $$W_n:=\left\{\;a\in K\;;\;w^n=1\;,\;\;n\in\Bbb N\;\right\}$$ It's easy to check $\;W_n\;$ is a subgroup of $\;K^*\;$ , and since it is finite it is then ...


2

There are commutative domains (for example, any non-Noetherian valuation domain) which are not PIDs but for which every finitely generated torsion-free module is free.


2

By construction, $KG$ has $K$ dimension $|G|$, so yes, it is finite dimensional. The group algebra $KG$ is defined by forming the free $K$-vector space using the elements of $G$ as a basis, and then enforcing the "obvious" multiplication rules to make it into an algebra. Thus if $G$ is finite, $KG$ is finite dimensional.


0

$c_{m+n} = \sum_{i=0}^{m+n} a_i b_{m+n-i} = a_n b_m$ because for $i<n$ we have $m+n-i > m$, hence $b_{m+n-i}=0$, and for $i>n$ we have $a_i=0$.


1

It should be clear that free always implies torsion-free. So it suffices you assume $R$ is a domain where every torsion-free module is free. As noted in the comments, one should assume $R$ is noetherian. Now every ideal $\mathfrak a$ of $R$ is certainly torsion free, so it must be free. Assume this is generated by more than one element. Let $a_1,a_2$ be ...


1

Consider $D=\Bbb R$, $n=2$ and the set of matrices $$\begin{pmatrix}0&a\\0&a'\end{pmatrix}$$ This is a left ideal, and it is proper. You can probably generalize this yourself.


1

There is no need to construct the field of fractions of $\mathbb{Z}[i]$ specifically, since the construction works "uniformly" for every integral domain. The field of fractions of $\mathbb{Z}[i]$ is $\mathbb{Q}[i] \subseteq \mathbb{C}$ because $\mathbb{Q}[i]$ is a field containing $\mathbb{Z}[i]$, and clearly the smallest one.


1

You almost done it. Consider $F=\ker\pi\oplus M$, $x\in\ker\pi$, $x=\sum r_ie_i$ with $r_i\in\mathfrak m$, and write $e_i=u_i+v_i$ with $u_i\in\ker\pi$ and $v_i\in M$. Then $x=\sum r_iu_i+\sum r_iv_i$, so $$x-\sum r_iu_i=\sum r_iv_i\in\ker\pi\cap M=\{0\},$$ hence $x=\sum r_iu_i\in\mathfrak m\ker\pi$.


1

$F \cong \ker(\pi) \oplus M$ implies $F/\mathfrak{m}F \cong \ker(\pi)/\mathfrak{m} \ker(\pi) \oplus M/\mathfrak{m} M$. Count dimensions.


0

Consider $A = k[x, y]$ and $D = \{1, x, x^2, \ldots\}$. The ideals $(y)$ and $(xy)$ are distinct in $A$ and neither intersects $D$, but when we invert $x$ they become the same ideal.


0

It follows as long as you know that an Artinian ring is Noetherian; I assume they're trying to avoid that fact here.


7

Note that $\varphi(1_R)=a$. But also $\varphi(1_R)=\varphi(1_R\cdot1_R)=a\cdot a=a^2$.


2

There is an isomorphism of rings $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \times \mathbb{C}$ (Hint: Use $\mathbb{C}=\mathbb{R}[x]/(x^2+1)$ and then CRT), but $\mathbb{C} \otimes_{\mathbb{C}} \mathbb{C} = \mathbb{C}$. So these are not isomorphic rings, since $\mathbb{C} \times \mathbb{C}$ has zero divisors for example. But they are ...


0

Since you already have two very good answers, I'd like to (just for fun) note that this also follows from very basic commutative algebra. You have an inclusion $D\hookrightarrow F$ which, by assumption, is finite. Then, by the Lying Over Theorem, we'd have a surjection $\text{Spec}(F)\to\text{Spec}(D)$ which implies that $\text{Spec}(D)$ has only one prime. ...


0

For indecomposability it's enough to prove that any two non-zero $R$-submodules of $F$ (say $N_1, N_2$) have non-zero intersection: if $a_1/b_1\in N_1$, and $a_2/b_2\in N_2$ with $a_i\ne 0$ and $b_i\ne 0$, then $a_1a_2\in N_1\cap N_2$. The second part of your question is already solved here: Commutative integral domain does not finitely generate its field ...


2

Remember that to show injectivity, it sufficies to show that if $\phi(x) = 0$, then $x=0$. So, what equation do you get when you write $\phi([m+ni,r+si]) = 0$?


2

($\Rightarrow$) Suppose $I=R$, then $1\in R$ which is an invertible element (unit). ($\Leftarrow$) We need to show that $1\in I$, as if this happens, then by definition of ideal. $1.a\in R$ i.e. every $a\in R$ implies $a\in I$. Thus $I\subseteq R$ Case I: $1\in R$. By definition of ideals $I=R$ Case II: If $a$ is a unit in R and $a\in I$. Then ...


1

If $I = R$, then $R$ contains unity, which is its own inverse, so $I$ contains an invertible element.


1

Let $K$ be the field you want to construct and let $\Bbb Q(i)$ the field of complex numbers of the form $r+si$ with both $r$ and $s$ in $\Bbb Q$. First thing you need to convince yourseld that $K\subset{\Bbb Q}(i)$. If you take a fraction $\frac{a+bi}{c+di}$ with $a,b,c,d\in\Bbb Z$ you can rewrite it as $$ ...


0

Show that the ring is isomorphic to the ring of complex trigonometric polynomials $\mathbb{C}[e^{i\theta},e^{-i\theta}]$. This is a localization of $\mathbb{C}[t]$ so is a PID.


1

Show $\Bbb Q[\Bbb Z]$ is isomorphic to the Laurent polynomials $\Bbb Q[x,x^{-1}]$ via an obvious map. That ring is a domain. But an Artinian domain is a field. Is our ring a field?


1

In algebra, we usually study objects (groups, rings, modules, etc.) up to isomorphism. In other words, our theory treats two isomorphic objects as the same, since an isomorphism tells us when two objects look the same as objects. For instance, any ring-related theorem about the integers is true for any ring isomorphic to them. Here we just need to know that ...


1

These are all hints, but you should be able to turn them into a complete solution. 1) Use the generalized product rule, which you can prove by induction: $$(fg)^{(k)}(x) = \sum_{n=0}^k f^{(n)}(x)g^{(k-n)}(x).$$ 2) Think about polynomials. 3) Yes. (If you showed that it's a bijective homomorphism, then you shouldn't need to ask us if it's an isomorphism!) ...


1

You seem to have misunderstood the question, as well as the notion of an ideal. If $S$ is a subset of a ring $R$, we say that $S$ is an ideal (specifically a two-sided ideal) if $S$ is closed under addition, i.e. for any $s,s' \in S$ we have $s+s' \in S$ $S$ is absorbing under multiplication, i.e., for any $s \in S$ and $r \in R$, both $rs$ ...


0

An ideal of a ring $R$ is a nonempty subset $I$ such that $I$ is closed under addition, and for all $r \in R$ and $s \in I$, we have $rs \in I$.


0

Here's a simple proof for finitely generated algebras. Lemma 1: Let $S$ be an algebra over $R$. Then $\alpha\in S$ is integral over a $R$ iff $R[\alpha]$ is finitely generated as a module over R. (Note that this lemma works even for $S$ not finitely generated over $R$). Proof: (Forward). If $\alpha$ is integral over $R$, then there is some monic ...


2

Suppose $I$ is an ideal in $R$. Clearly $I[x]$ is a subring of $R[x]$ since $I$ must be a subring of $R$. By definition, $\forall r\in R\;\forall x\in I\;rx,xr\in I$. Then since the coefficients of the product of two polynomials result from products of the coefficients, it is clear that $\forall p\in R[x]\;\forall q\in I[x]\;pq,qp\in I[x]$.


2

The easiest way to show that $p$ is not well-ordered, is to show that $p$ has no minimal element. This is trivial once you note that $$\lim_{n \to \infty}(\sqrt{2}-1)^n = 0$$


4

Alternatively, you can recognize that $(x,y)$ is kind of like $x + yi \in \mathbb{C}$ with the given operations. From your definition, $$(u, v) + (x,y) = (u + x, v + y)$$ and $$(u, v)\cdot (x, y) = (ux - vy, uy + vx).$$ But this is the same behavior as $$(u + vi) + (x + yi) = (u + x) + (v + y)i$$ and $$(u + vi)(x + yi) = ux + uyi + vxi + vyi^{2} = ux + uyi ...


6

Solve for $m$ and $n$ to get $n=\frac {-y}{x^2+y^2}$ and similarly for $m$ .Since $x,y$ are not both zero inverse exists


1

You could do it indirectly by proving that the complex numbers form a field, but you could also do it directly as follows. If $(x,y) \cdot (u,v) = (xu - yv, xv+yu) = (0,0)$, then $xu = yv$ and $xv = -yu$. Now assume both $(x,y)$ and $(u,v)$ are not zero. If $x \neq 0$, multiply the first equation by $x$ and get $$ x^2 u = yxv = -y^2u \implies (x^2+y^2)u = 0 ...


1

I don't think so. NOTE: There are no zero divisors in $(\mathbb{C},+,\times)$ Note that when you consider the multiplication in $\mathbb{C}$ $(x+iy)\cdot (u+iv)=(ux-vy)+i(uy+xv)$ If you consider the map $(x+iy)\to(x,y)$ you can see that the operation here and multiplication in $\mathbb{C}$ are equivalent and note that there is no zero divisor in ...


0

No you can't. Your operation is equivalent to the multiplication of 2 complex numbers $x+iy$ and $u+iv$. Multiplying any two Complex numbers is equivalent to multiplying their norm and adding the arguments. The product of two non-zero norms is a product of two non-zero Real numbers, and cannot be $0$, since the Reals are an integral domain ( a field, ...


2

Yes in general, if $R \to S$ is an isomorphism of rings, one easily checks that it restricts to an isomorphism of the unit groups $R^\times \to S^\times$. And for two rings $R_1,R_2$ we clearly have $(R_1 \times R_2)^\times = R_1^\times \times R_2^\times$.



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