Let $\mathbb{Z}\left [ i \right ]=\left \{ a+bi:a,b \in \mathbb{Z} \right \}.$   Show that the field of quotient of $\mathbb{Z}\left [ i \right ]$ is ring isomorphic to $\mathbb{Q}\left [ i \right ]=\left \{ r+si: r,s \in \mathbb{Q} \right \}$   Let's try to construct the field of quotient:  The field of quotient is the smallest field of fraction contained in an integral domain.   Try: $F=\left \{ \frac{a+bi}{c+di} : a+bi, c+di \in \mathbb{Z}\left [ i \right ]\right \}$  I never liked these sort of questions involving function construction. I wish I could obtain a bit of help here to get me further.  Thanks in advance.
31092
I'm looking to prove that $\Bbb Q[i] = \{ p + qi : p, q \in \Bbb Q \}$ is the field of fractions of $\Bbb Z[i] = \{p + qi : p, q \in Z\}$.  I am familiar with definition of a field of fractions. For example, I understand that if one has an integral domain $D$, it can be embedded in a field of fractions $F_D$, and every element of $F_D$ can be written as the quotient of two elements in $D$. However I have been confused by two questions:  (1) If one has a field $F$, and one can show that any element in $F$ can be expressed as the quotient of two elements in an integral domain $D$, does that imply that $F$ is the field of fractions of $D$?  (i.e. Would it suffice in my case to show that every element in $\Bbb Q[i]$ can be written as a the quotient of 2 elements of $\Bbb Z[i]$?)  (2) How can I show that element in $\Bbb Q[i]$ can be written as a the quotient of 2 elements of $\Bbb Z[i]$? When I write an element of $\Bbb Q[i]$ like  $$q = \frac ab + \frac cdi$$  I get mixed up trying to come up with 2 elements of $\Bbb Z[i]$ that could equal that. I set $z = u+vi$ and $w = x+yi$, equate $q = z/w$ and I end up with these equations like this:  $$a = ux + vy; b = x^2 + y^2; c = vx - uv; d = x^2 + y^2$$  This doesn't strike me as the way to go. Possibly there is a way to take advantage of fact that the elements are equivalence classes, but I'm not seeing it. Thank you very much!
90939
In $\mathbb{Z}\left [ x \right ]$, let $I = \left \{ f\left ( x \right ) \in \mathbb{Z}\left [ x \right ]: f\left ( 0 \right ) \text{ is an even integer} \right \}$   In fact, $I=\left \langle x,2 \right \rangle$  How do I show that the order of $\mathbb{Z}\left [ x \right ]/I$ is 2?
53094
How to show that $\mathbb Z_{mn}$ is isomorphic to $\mathbb Z_m \times\mathbb Z_n$ when $m$ and $n$ are coprime? It is easy to show that the natural map from $\mathbb Z_{mn}$ to $\mathbb Z_m \times\mathbb Z_n$ is a ring homomorphism. How to show that it is bijective?
43828
Let $I$ and $J$ be two ideals of a commutative ring $R$ with $1.$ Give a necessary and sufficient condition so that    $$R/IJ\cong R/I\times R/J.$$   Prove your claim. Then decide whether the following ring isomorphisms are true or not:   $$\mathbb{Q}[x]/\langle x^2-1\rangle\cong \mathbb{Q}[x]/\langle x-1\rangle\times \mathbb{Q}[x]/\langle x+1\rangle,$$   $$\mathbb{Z}[x]/\langle x^2-1\rangle\cong \mathbb{Z}[x]/\langle x-1\rangle\times \mathbb{Z}[x]/\langle x+1\rangle.$$   It has been a while since I took Abstract Algebra and I am preparing for the prelims. I am not sure how to tackle this one. Any help/suggestion/hint will be much obliged. This is what I have so far:     This is a simplified version of the chinese remainder theorem. The map   $R\to R/I\times R/J$ defined by $r\mapsto (r+I,r+J)$ is a ring homomorphism with kernel $I\cap J.$ If the ideals $I$ and $J$ are comaximal, then this map is surjective and $I\cap J=I\cdot J$, so    $$R/(I\cdot J)=R/(I\cap J)\cong R/I\times R/J.$$   My question is about the second part of the question, do I have to check whether: $\langle x-1\rangle$ and $\langle x+1\rangle$ are comaximal and $\mathbb{Q}[x]$ and/or $\mathbb{Z}[x]$ are ring homomorphisms? I believe that these ideals are comaximal in both cases but forget how to prove it.
7288
I still struggle with elementary knowledge about quotient rings.  For example, why can I just write down the second following isomorphism:  $$\frac{\mathbb{C}[x, y, z]}{(x^2 + y^2 - 1, y \pm iz)} \simeq \frac{\frac{\mathbb{C}[x, y, z]}{(y \pm iz)}}{\frac{(x^2 + y^2 - 1,y \pm iz)}{(y \pm iz)}} \simeq \frac{\mathbb{C}[x, y]}{(x^2 + y^2 - 1)} $$  I know the first isomorphism is right by the third isomorphism theorem. But why can I write down the second one?
96113
My questions are:     $1.$ Is there a field morphism $\Bbb R(X) \hookrightarrow \Bbb R$ ?      $2.$ If the answer to $1.$ is "yes", are $\Bbb R$ and $\Bbb R(X)$ isomorphic as fields?   $ $    $ $  For $1.$, here are some comments:   This is true for $\Bbb C$, see . The key point here is the fact that $\Bbb C$ is algebraically closed, of cardinality $2^{\aleph_0}$, and of characteristic $0$. The same argument doesn't work for $\Bbb R$. We can actually find a subfield $F$ of $\Bbb R$ such that $F \cong F(X) \cong F(X,Y) \cong \cdots$, see .  that $\Bbb Q(X)$ does not embed in $\Bbb Q$ (as a field).  This is equivalent to: Is there an injective ring morphism $\Bbb R[X] \hookrightarrow \Bbb R$ ? (By taking the fraction fields).   As for 2. :   They are isomorphic as abelian groups, and actually as $\Bbb Q$-vector spaces. They are not isomorphic as $\Bbb R$-algebras, because they haven't the same transcendence degree over $\Bbb R$. Let's try to see how to construct such a field isomorphism. We could try$^{[1]}$ to find a subset $B$ of $\Bbb R$ such that $\Bbb R = \Bbb Q(B)$ and $C \subsetneq B \implies \Bbb R \neq \Bbb Q(C)$.  The set $B$ must contain transcendental elements, pick such an $x_0 \in B$.  Then $\Bbb R = \Bbb Q(B) = \Bbb Q(B \setminus \{x_0\})(x_0)$ and we could try$^{[2]}$ to find a field isomorphism $\Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)$, so that $\Bbb Q(B \setminus \{x_0\})(x_0) \cong \Bbb Q(B)(x_0) \cong \Bbb R(X)$.  For ${[1]}$, I tried to use Zorn's lemma on $\mathscr B = \{B \subset \Bbb R \mid \Bbb R = \Bbb Q(B)\}$ and the partial order $B_1 ≤ B_2 \iff B_1 \supset B_2$. But this is not clear to me why this should be an inductive set.  For ${[2]}$, my idea was to define $f : \Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)$ by choosing a bijection $b : B \setminus \{x_0\} \to B$ (these sets should be uncountable), and then trying to extend $f\vert_{B \setminus \{x_0\}} = b$ to a field morphism.     Edit for $^{[1]}$: in  article (MINIMAL GENERATING SETS OF GROUPS, RINGS, AND FIELDS, by LORENZ HALBEISEN, MARTIN HAMILTON, AND PAVEL RUZICKA) the theorem 2.4. proves that $\Bbb R$ has no minimal generating set as field (or $\Bbb Q$-algebra, i.e. as ring). The argument uses Artin-Schreier theorem. Minimal generating sets of modules are   .  I think that $B$ is a minimal generating set of a subfield $K \subset \Bbb R$ over $\Bbb Q$ if and only if any element of $B$ cannot be written as a rational fraction of other elements of $B$. Under these conditions and using (transfinite) induction, we can probably prove (and this was my first idea) that any function $f: B \to \Bbb C$ such that $b$ and $f(b)$ are conjugate over $\Bbb Q$ (i.e. have the same minimal polynomial) extends to a (unique) field morphism $\sigma : K = \Bbb Q(B) \to \Bbb C$.  Actually, the collection $\mathscr B$ defined above doesn't have to be an inductive set, at least if we replace $\Bbb R$ by $K =\Bbb Q(\sqrt 2, \sqrt[4]{2},\sqrt[8]{2},\dots)$, because the descending chain $\left(E_n = \{\sqrt[2^m]{2} \mid m \geq n\}\right)_{n \geq 1}$ satisfy $K = \Bbb Q(E_n)$, but $K \neq \Bbb Q\left(\bigcap_{n≥1} E_n\right)=\Bbb Q$, so the chain has no upper bound w.r.t. $≤$ in $\mathscr B$.
12128
Let $A$ be an integral domain, $a$ and $b \in A-\{0\}$, and let $B = A[X]/(aX+b)$. Show that, if $Aa \cap Ab=Aab$, then $B$ is an integral domain.   My attempt at proof (following a hint). Denote by $K$ the field of fractions of $A$. Let $\phi: A[X] \to K$, with $\phi(X)=-b/a$ and $\phi(y)=y, y \in A$.  Then $\phi(aX+b)=-b+b=0$. If $p(X) \in A[X]$ and $p(X) \not \in (aX+b)$, then $p(X)=q(X)(aX+b)+r$, $r \in A$. Therefore, $\phi(p(X))=\phi(r)=r$, so $p(X) \in \ker(\phi) \iff r=0 \iff p(X) \in (aX+b)$.   Hence $\ker(\phi)=(aX+b)$; by the first isomorphism theorem, $A[X]/(aX+b)$ must be isomorphic to Im$(\phi)$. In conclusion $B$ is isomorphic to a subfield of $K$, so it's an integral domain.  The problem is, I don't know where I used the condition $Aa \cap Ab=Aab$. Am I doing something wrong?
81674
How do I prove that a finite ring $R$ of order 10 is isomorphic to the ring $\mathbb{Z}/10 \mathbb{Z}$?    I know that as a group under addition, $(R,+)$ is isomorphic to the group $(\mathbb{Z}/10 \mathbb{Z}, + )$, but the multiplication is rather mysterious to me.
79588
I was recently reading about integral ring extensions. One of the first examples given is that $\mathbb{Z}$ is integrally closed in its quotient field $\mathbb{Q}$. Another is that $\mathbb{Z}[\sqrt{5}]$ is not integrally closed in $\mathbb{Q}(\sqrt{5})$ since for example $(1+\sqrt{5})/2\in\mathbb{Q}[\sqrt{5}]$ is integral over $\mathbb{Z}$ as a root of $X^2-X-1$, but $(1+\sqrt{5})/2\notin\mathbb{Z}[\sqrt{5}]$.     Now I'm  curious, can we find what are all integers $n$ such that $\mathbb{Z}[\sqrt{n}]$ is integrally closed (equal to its integral closure in its quotient field)?   One thing I do know is that unique factorization domains are integrally closed, so I think rings like $\mathbb{Z}[\sqrt{-1}]$, $\mathbb{Z}[\sqrt{-2}]$, $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{3}]$ are integrally closed, as they are Euclidean domains, and thus are UFDs. But can we say what all integers $n$ are such that $\mathbb{Z}[\sqrt{n}]$ is integrally closed? Thanks!
30253
Let $\mathbb{Q}(\sqrt{d}) = \{a + b \sqrt{d}: a,b \in \mathbb{Q} \}.$ Show that $\mathbb{Q}(\sqrt{d})$ is a field.    Everything seems obvious except for existence of inverses in the multiplicative group of $\mathbb{Q}(\sqrt{d}).$  Suppose $\alpha \in \mathbb{Q}(\sqrt{d}) \Longrightarrow \alpha = a + b \sqrt{d}.$ To prove that $\mathbb{Q}(\sqrt{d})$ contains multiplicative inverses, observe that for a non-square $d$ and $a - b\sqrt{d} \ne 0$  one has $$\frac{1}{\alpha} = \frac{1}{a+b\sqrt{d}} \cdot \frac{a - b \sqrt{d}}{a - b \sqrt{d}} = \frac{a}{a^2-b^2d}- \frac{b}{a^2-b^2d} \sqrt{d}.$$  If $a,b \in \mathbb{Q} \Longrightarrow a^2,b^2 \in \mathbb{Q} \Longrightarrow b^2 d \in \mathbb{Q}$ (since $d \in \mathbb{Z}$) $\Longrightarrow a^2-b^2d \in \mathbb{Q} \Longrightarrow \frac{a}{a^2-b^2d}, \frac{b}{a^2-b^2d} \in \mathbb{Q}.$ Hence $\frac{1}{\alpha} \in \mathbb{Q}(\sqrt{d}).$   Is this correct?
67241
I'm interested in the following problem from Artin's Algebra text:     Determine the structure of the ring $\mathbb Z[x]/(x^2 + 3,p)$, where (a) p = 3, (b) p = 5.   I know that by the isomorphism theorems for rings we can take the quotients successively, and so  $$\mathbb{Z}[x]/(p) \cong (\mathbb{Z}/p \mathbb{Z})[x] $$  as the map $\mathbb{Z}[x] \to (\mathbb{Z}/p \mathbb{Z})[x]$ defined by $\sum_{n} a_n x^n \mapsto \sum_{n} \overline{a_n} x^n$ is a surjective ring homomorphism with kernel $(p)$. Thus it remains to study the quotients  $$(\mathbb{Z}/p \mathbb{Z})[x]/(x^2+3) $$ for $p \in \{3,5\}$.    If $p=3$, $(x^2+3)=(x^2)$ in $(\mathbb{Z}/3 \mathbb{Z})[x]$, and by using polynomial division all distinct coset representatives can be reduced to the following list of 9 elements  $$\{0,1,2,x,1+x,2+x,2x,1+2x,2+2x\}. $$  Moreover, it can shown that the list above gives 9 distinct cosets, as no difference of two distinct elements of the list is a multiple of $x^2$. Since $1$ and $x$ generate two distinct additive groups of order $3$, the additive group of our quotient ring is not cyclic. Elementary group theory then shows   $$(\mathbb{Z}/3 \mathbb{Z})[x]/(x^2)^+ \cong (\mathbb{Z}/ 3 \mathbb{Z})^2 $$  as additive groups.   I was then about to conclude that the multiplication on the quotient is then compatible with the usual one in $(\mathbb{Z}/3\mathbb{Z})^2$, but this is wrong!  It can be seen that the quotient is not isomorphic to $(\mathbb{Z}/3\mathbb{Z})^2$ as a ring, because the former contains a nonzero element (represented by $x$) whose square is zero, while the latter contains no such elements.    If $p=5$, a full list of coset representatives is of length 25  $$\{0,1,2,3,4,x,1+x,2+x,3+x,4+x,2x,1+2x,2+2x,3+2x,4+2x,3x,1+3x,2+3x,3+3x,4+3x,4x,1+4x,2+4x,3+4x,4+4x \} .$$  And once again, one can see that these represent 25 distinct cosets. Similarly to the $p=5$ case, I've managed to prove that the additive group of this ring is isomorphic to $(\mathbb{Z}/5 \mathbb{Z})^2$.    My questions:   Have I made any mistakes in my argument? What exactly am I supposed to do in this question? determine the number of elements? Write down the tables for addition and multiplication?   Any further information about these quotients will be appreciated, thanks!
42042
Let $D$ be an integer such that it is not a perfect square in   $\mathbb{Z}$.       If $D \equiv 1$ mod $4$, the set $$T:= \left\{\left( \begin{matrix} a &amp; b  \\ \frac{(D-1)b}{4} &amp; a+b \\ \end{matrix} \right) : a, b \in  \mathbb{Z} \right\}$$ is a subring of $M_2(\mathbb{Z})$ which is isomorphic   to the ring $$O := \left\{a+b\frac{1+\sqrt{D}}{2}: a, b \in \mathbb{Z} \right\}.$$   My problem is to find an isomorphism. However, I find this relation : $$\det \left( \begin{matrix} a &amp; b  \\ \frac{(D-1)b}{4} &amp; a+b \\ \end{matrix} \right)$$ $$= a(a+b)-\frac{b^2(D-1)}{4}$$ $$= \left(a+b\frac{1+\sqrt{D}}{2}\right)\left(a+b\frac{1-\sqrt{D}}{2}\right)$$  and $D \equiv 1$ mod $4$ $\iff$ $\left(\frac{1+\sqrt{D}}{2}\right)\left(\frac{1-\sqrt{D}}{2}\right) = -k$ for some $k \in \mathbb{Z}$  Does someone could give me a little hint to help me complete this problem?
101981
Here I am under the impression that $2\mathbb Z$ and $3\mathbb Z$ are the sets of even numbers and multiples of $3$ respectively and the operations are usual addition and multiplication. This is an exercise in Fraleigh Abstract Algebra.  I believe that $2\mathbb Z$ and $3\mathbb Z$ are isomorphic as groups on addition.  I started a proof by supposing that there was a ring isomorphism, $f$,  in hopes of finding a contradiction with $f(x\cdot y) = f(x)\cdot f(y)$.
106161
I have to prove that $\mathbb Z [X] / (X) \cong \mathbb Z[X] / (X+1) \cong \mathbb Z [X] / (X+2015)$.  I think that one answer could be that $\mathbb Z[X]/(X) \cong \mathbb Z(0)$, $\mathbb Z[X]/(X+1)\cong Z(-1)$ and $\mathbb Z [X] /(X+2015)\cong \mathbb Z(-2015)$, as $0, -1, -2015$ are roots of these polynomials. Also, as $0, -1, -2015\in \mathbb Z$, then all of them are isomorphic to $\mathbb Z $, hence isomorphic between them.  But I don't know very well how to justify this rigorously and my teacher has said to me that there is a simpler way to proof this, without using the roots of these polynomials.
122369
Prove that $\mathbb{Z}$ is not isomorphic to additive group of any vector space over any field.  Proof.  Surpose that: $\phi : (A, +) \rightarrow  \mathbb{Z} $ is an isomorphism. Then there is some $r∈A $ such that $ϕ(r)=1_\mathbb{Z}$. We have $$ 1_\mathbb{Z}=ϕ(r)=ϕ(2(r/2))=2ϕ(r/2)$$  or equivalently that $ϕ(r/2)=\frac{1}{2}$. But this is not in $ \mathbb{Z}$, so there can be no such morphism.  Is my proof above correct? Thanks
65976
I'm trying to prove that every finite Boolean ring, $R$, is isomorphic to a finite number of copies of $\mathbb{Z} / 2 \mathbb{Z}$:  \begin{align} R \cong \mathbb{Z} / 2 \mathbb{Z} \; \times \cdots \; \times \; \mathbb{Z} / 2 \mathbb{Z} \end{align}  I'm aware that there are answers to this question on this website. However, almost all of the answers seem to use facts about modules and vector spaces defined over fields. This question is posed in section 7.6 in Dummit and Foote's Abstract Algebra. This section is part of the chapters which introduces the basics of ring theory (rings, homomorphisms, ideals, Chinese Remainder Theorem etc.).  I'm not even sure how to being answering this question. I know that if $R$ were an integral domain, then we would have that $R \cong \mathbb{Z} / 2 \mathbb{Z} $.   Please only give hints, considering that that the statement has to be proved using the material only included up till chapters 9 in the textbook.
111961
How do I show that $\text{Hom}(\mathbb{Z},G)$ and $G$ are isomorphic?  The method suggested by the book is to define a map $k$: $\text(\mathbb{Z},G) \to G$ by $f \to f(1)$.   I am stuck on how exactly to show that it is an isomorphism. Could anyone shed some light into this? Is $f(1)$ an element of G?  (For those who wanted to know, this is not homework)