I have to prove that $\mathbb{R}$ is a field.  One of the properties for $\mathbb{R}$ to be a field is that addition in $\mathbb{R}$ is associative. I wanted to use this property as an example for the other ones.  My question is, how can I show something so intuitive? Can I say that it is simply assumed that addition in $\mathbb{R}$ is associative? Or should I use $a+(b+c)=(a+b)+c$ for $a,b,c \in \mathbb{R}$ in some way?  I am sorry for the silly question, it has been a while since I proved something. Thank you.
121495
How does one prove that addition on $\mathbb{R}$ is associative?  We can define addition to be a binary function $+ : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, by that definition $\mathbb{R}$ is closed under addtition as for any $(a, b) \in \mathbb{R}^2$ we have $+(a, b) = a + b = c$ for some $c \in \mathbb{R}$.  But if we take $a, b, c \in \mathbb{R}$, how can we prove rigorously that $(a + b) + c = a + (b + c)$?   We know also that $(\mathbb{R}, +)$ is a group, but most proofs of this fact assume that addition in $\mathbb{R}$ is associative, the proofs do not actually prove that addition is associative.
102484
The Question:  Regard $\mathbb{R}$ as a vector space over $\mathbb{Q}$, where vector addition and scalar multiplication is just addition and multiplication as they would usually be done in the real numbers.  Assuming Zorn's lemma, and hence that $\mathbb{R}$ has a basis, show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is such that $f(x+y)=f(x)+f(y)$, then $f$ is a linear transformation.  From this, show that there is a function $g: \mathbb{R} \rightarrow\mathbb{R}$ where $g(x+y)=g(x)+g(y)$, and where $g$ is not of the form $g(x)=kx$, for any real number $k$.  What I have done so far:  So we need to show that for any rational $q$, that $f(qx)=qf(x)$. Since $\mathbb{R}$ has a basis, $f(qx)=f(q(p_1x_1+...+p_rx_r))=f(qp_1x_1)+...+f(qp_rx_r)$, where the $x_i's$ are basis vectors.   So I tried to show that $f(qy)=qf(y)$ for any basis vector $y$ by defining a set $P$ of linearly independent subsets of $\mathbb{R}$ such that every vector $x$ in the subsets has the property $f(qx)=qf(x)$ for all $q$, and then using Zorn's lemma to get the maximal subset which will be a basis with the basis vectors having the desired property.  But I'm thinking $P$ may be empty because $\{0\}$ is not linearly independent.
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
56262
I have a non-empty set $R$ with a binary operation $*$. If the pair has an identity element $e\in R$ and  $$(a*b)*(c*d)=(a*c)*(b*d)$$  holds for all $a,b,c,d \in R$, how do I then prove that this is associative?   I understand the concept that $x*(y * z) = (x * y)*z$, but I am unsure how to apply it here.
85167
I'm currently reading Marcus' "Number Fields," and I'm having difficulty proving the following result:  Corollary 2.2: Let $m$ be a squarefree integer. The set of algebraic integers in the quadratic field $\mathbb{Q}(\sqrt{m})$ is $$ \left\{ a + b \sqrt{m} : a,b \in \mathbb{Z} \right\} $$ if $m \equiv 2,3 \pmod{4}$; and$$ \left\{ \frac{a+b\sqrt{m}}{2} : a,b \in \mathbb{Z} ~,~ a \equiv b \pmod{2} \right\}$$ when $m \equiv 1 \pmod{4}$. $\square$  It is easy enough to see that the claimed sets do, indeed, consist of algebraic integers. The other half of the proof proceeds something like this: suppose that $\alpha = r + s \sqrt{m}$ is an element of $\mathbb{Q}(\sqrt{m})$. Then $\alpha$ is a root of the polynomial $$ x^2 - 2rx + r^2 - s^2m $$ In particular, this means that $\alpha$ is an algebraic integer if and only if both $2r$ and $r^2 - s^2 m$ are integers. At this point, the proof stops, and says that this implies the result. And therein lies the problem: I don't see how to make the connection. I do know, for example, that if one of $r,s$ is an integer, then the other one must be, too, i.e., either both are integers, or both are not. However, I cannot get any further than that. Can anyone give me a gentle nudge in the right direction? Thank you very much!
16658
I think I've been confusing myself about the language of subspaces and so on.  This is a rather basic question, so please bare with me.  I'm wondering why we do not (or perhaps "we" do, and I just don't know about it) say that $\mathbb{ R } $ is a subspace of $\mathbb{ R }^2 $.  It's elementary to prove that the set $$ S:= \left\{ c \cdot \mathbf{x} \mid c \in \mathbb{ R }, \mathbf{x} \in \mathbb{ R }^2  \right\}$$ is a vector subspace of $\mathbb{ R } ^2$.  What is confusing me is that there seems to be an isomorphism between the set $S$ and $\mathbb{ R } $: \begin{align*} \varphi: S &amp;\rightarrow \mathbb{ R }   \\  c \cdot \mathbf{x} &amp;\mapsto c \\ \end{align*} If this is indeed true, as I believe it is having checked that $\varphi$ gives an isomorphism, wouldn't we say that $\mathbb{ R } $ is a subspace of $\mathbb{ R } ^2$? Any help sorting out this (language) problem will be greatly appreciated!
75081
Let $R$ be a commutative ring with $1$. Suppose that all ideals $I \neq R$ are prime. Prove that $R$ is a field.   Help me some hints.  Thanks a lot.
29348
I don't see how to solve the following problem:     Let $R$ be a commutative and unitary ring. If there exists a monic polynomial $f(x) \in R[x]$ so that $R[x]/(f(x))$ is a field, show that $R$ is a field. What about if $f(x)$ isn't monic?   I don't have a clue about how proving it. I see that if $R$ is a field and $f(x)$ a monic and irreducible polynomial in $R[X]$ then $\frac{R[x]}{(f(x))}$ is a field, but that is not the question!  Thank you in advance for your help
32924
Show that the set $\mathbb{Q}[\sqrt{2}] = \{a + b \sqrt{2} \mid a, b \in \mathbb{Q}\}$ is a field with the usual multiplication and addition.  It is easy enough to show that it is closed under addition and multiplication ( $\forall a,b \space)$ we have $a+b \in \mathbb{Q}\sqrt{2}$ and $a * b = ab \sqrt{2} \in \mathbb{Q}\sqrt{2}$.  However, I had trouble proving the other axioms (associativity, commutativity, unique neutral element, unique inverse, and distributivity of multiplication over addition). I would really appreciate it if someone could help me with those.
98608
Does commutativity imply associativity? I'm asking this because I was trying to think of structures that are commutative but non-associative but couldn't come up with any. Are there any such examples?  NOTE: I wasn't sure how to tag this so feel free to retag it.
49027
This question is from the book "Abstract Algebra" by Dummit and Foote.(Exercise: 7.6.4):  Problem: Prove that if $R$ and $S$ are any two non-zero rings then $R \times S$ is never a field.  My problem with this question is that it doesn't feel right. If $R$ and $S$ are rings then $R \times S$ is also so. (The addition and multiplication on $R \times S$ is defined as $(r_1,s_1)+(r_2,s_2)=(r_1+r_2,s_1+s_2)$ and $(r_1,s_1)(r_2,s_2)=(r_1r_2,s_1s_2)$ respectively where $(r_1,s_1),(r_2,s_2) \in R \times S$.) So proving that $R \times S$ is a field reduces to proving commutativity of multiplication, existence of multiplicative indentity and existence of multiplicative inverse for each element of it. Now, if the two rings are fields themselves then multiplication is commutative, the identy is $(1_R,1_S)$ and inverse of an element $(r,s)$ is $(r^{-1},s^{-1})$. So $R \times S$ is also a field. Am I right? Please help.  Update: The question comes from a section about the Chinese Remainder Theorem. Is it possible to prove this fact using Chinese Remainder Theorem?
71710
I am doing a self-study using Artin's algebra. There is a problem which asks us to prove that adjoining an element to a ring doesn't do anything if that element was already in the ring.  More formally, say that $\alpha=a$ where $a\in R$. We want to show that $R\approx R[\alpha]$. (Where $\approx$ means "isomorphic to".)   My proof: We can consider $R[\alpha]\approx R[x]/(x-\alpha)$, which intuitively means "killing" $x-a$ in the ring of polynomials. Any function $f\in R[x]$ can be written as $(x-\alpha)g(x)+r(x)$ for some g,r by the division algorithm. Since the degree of r must be less than the degree of $x-a$, r must be a constant polynomial. Since $(x-\alpha)=0$, the residue of $f$ in $R[x]/(x-\alpha)$ is just r, which is some element of R. We can see that every $s\in R$ has a corresponding constant polynomial $f(x)=s$ in $R[x]$, which means the simple inclusion map $s\mapsto f(x)=s$ is our desired isomorphism.  So, this proof seems fine to me, except that I seem to have proven that adjoining any element keeps the ring the same. This sounds implausible to me. Where did I go wrong?
45049
a+b and ab are the usual integer addition and multiplication. You can assume that this new operation forms a ring, say R is the set of integers with these operations. Then does R have zero-divisors? And Are R and Z (the integers) isomorphic?  Any help will be appreciated!
117005
Are the groups $\mathbb{C}$ and $\mathbb{R}$ isomorphic under addition?  And how could I prove this ?  What about $\mathbb{Q}$ and $\mathbb{Q}[i]$ ?
86031
Is it correct to say that $\mathbb{R}$ is a subset of $\mathbb{R}^2$? Or, put more generally, given $n,m\in\mathbb{N}$, $n&lt;m$, is $\mathbb{R}^n$ a subset of $\mathbb{R}^m$?  Also, strictly related to that: what is then the "relationship" between the set $\{(x,0)\in\mathbb{R}^2,x\in\mathbb{R}\}\subset\mathbb{R}^2$ and $\mathbb{R}$? Do they coincide (I would say no)? As vector spaces, do they have the same dimension (I would say yes)?  If you could give me a reference book for this kind of stuff, I would really appreciate it.  Thank you very much in advance.  (Please correct the tags if they are not appropriate)
97559
how to prove that if $A$ and $B$ are open in $(\mathbb{R},|.|)$ then  $A+B$ is open ?  Where   $A+B=\{a+b\mid a\in A,b\in B\}$   Thank you
34331
Say $R$ is a commutative ring and $I\in R$ is an ideal. Let us consider the quotient $R/I$. It is created by taking every element $a\in R$, and adding all the elements of $I$ to it. The elements of $R/I$ are of the form $a+I$ and $b+I$, $\forall a,b\in R$.   Now $$(a+I).(b+I)=(a+i_{1}).(b+i_{2}),\forall (i_{1},i_{2})\in I \times I$$  $$(a+i_{1}).(b+i_{2})=ab+a.i_{2}+b.i_{1}+i_{1}i_{2}=ab+I$$  We know that $a.i_{2}+b.i_{1}+i_{1}i_{2}\in I$.   If $(a+I).(b+I)=ab+I$, then taking suitable $(i_{1},i_{2})\in I\times I$, we should be able to prove $(a+i_{1}).(b+i_{2})=ab+i_{3},\forall i_{3}\in I$. However, can every element in $I$ be generated by taking suitable $i_{1},i_{2}\in I$? And if not, is that the reason why $(a+I).(b+I)$ has to be defined as equal to $ab+I$ in violation of the distributive property of ring elements? This has confused me for a long time.   EDIT: I figured addition does not have to be defined as such because $(a+i_{1})+(b+i_{2})=a+b+i_{1}+i_{2}$, where $i_{1}+i_{2}\in I$. In fact, every element $i\in I$ can be constructed by taking $i_{1}=0$ and $i_{2}=i$. Hence, $(a+I) + (b+I)=a+b+I$ naturally. This is the logic I followed to determine that $(a+I).(b+I)$ doesn't quite work as nicey. I'm not sure if this logic is flawed or not as I haven't had the opportunity to ask anybody.   Thanks for your help in advance!