I think that these two sets cannot be turned into fields by re-defining addition or multiplication (or both) but I am not sure how to prove this only from axioms of the field and (if needed) some elementary properties of fields that follow directly from axioms of the field.  My intuition is that $\mathbb N_0$ and $\mathbb Z$ are not "dense enough", as is, for example, $\mathbb Q$, to be able to become fields.  How would you prove that we can(or cannot) turn them into fields?
116185
Is it possible to define addition and/or multiplication on the set of  a) natural numbers (including $0$: $0,1,2,3,...$)  b) integers $(..., -2, -1, 0, 1, 2, ...)$  in such way that they will become fields?  Thanks in advance.
85655
Let $\mathbb{Z}[i]=\left\{a+bi:a,b \in \mathbb{Z}\right\}$  be the ring of Gaussian integers $(i^{2}=-1)$ and let $\mathbb{Q}[i]=\left\{a+bi:a,b \in \mathbb{Q}\right\}$ be its fraction field.      1) Show that the tensor product $\mathbb{Q}[i] \otimes_{\mathbb{Q}} \mathbb{Q}[i]$ is isomorphic to the product $\mathbb{Q}[i] \times \mathbb{Q}[i]$ of two copies of the field.       2) Let $S=\left\{1,2,2^2,2^3,...\right\} \subseteq \mathbb{Z}$ be the multiplicatively closed subset generated by 2. Find two distinct idempotents in the algebra $S^{-1}(\mathbb{Z}[i] \otimes_{\mathbb{Z}} \mathbb{Z}[i])$ other than $0$ and $1$.    For the first part, I thought I would try to use the presentation $\mathbb{Q}[i] \simeq \mathbb{Q}[T]/(T^2+1)$ and see what comes from that. Any help would be much appreciated!
50392
I am very new to proofs so please excuse any trivial errors.  In lecture we were told that:     A set $\mathbb S$ is called finite if there exists a one-to-one    mapping (bijective mapping) between $\mathbb S$ and the set $\mathbb  N_n=\{0,1,2,3,...,n-1\}$ for some natural number $n$. A set is called    countable infinite if there exists a bijective mapping between    $\mathbb S$ and the set $\mathbb N=\{0,1,2,...\} $ of all natural    numbers   I want to answer the following questions:  1. Show that the set $\mathbb S=\{a,b,c\}$ is countable  Does countable mean countable infinite? An attempt from another post on this site: For all triples $(a,b,c)$ there exists an according sum $S=a+b+c \implies ...$ But wouldn't the sum be a surjection between $\mathbb S$ and $\mathbb N$ and not a bijection?  2. Show that the set of integer numbers $\mathbb Z=\{0,\pm1,\pm 2,...\}$ is countable infinite  Before I continue with this proof I have a question about mapping. The definition says that "if there exists a bijective mapping...". In this context, can a function $f(x)$ be considered a mapping? If yes, then can't I just define myself some map $$f:\mathbb Z\to \mathbb N$$  and let $\space f$ be: $$x \mapsto \lvert x \rvert$$  which would then automatically imply that $\mathbb Z$ is countable infinite? (or wouldn't it)?  3. Show that $\mathbb S=\mathbb N \times \mathbb N=\{x |x=(p,q),p,q \in \mathbb N\}$ is countable infinite.   I am not sure about this one but if my "intuition" from 2) turns out to be correct I can maybe define some sort of norm and show that it is a bijection between $\mathbb S$ and $\mathbb N$
117076
How to show that $\mathbb Q(\sqrt 2)$ is not field isomorphic to $\mathbb Q(\sqrt 3)?$   My text provides the hint as: Any isomorphism from $\mathbb Q(\sqrt 2)\to\mathbb Q(\sqrt 3)$ is identity when restricted to $\mathbb Q.$  Ya! I can see that: $\mathbb Q(\sqrt 2)\simeq\mathbb Q[x]/\langle x^2-2\rangle$ and $\mathbb Q(\sqrt 3)\simeq\mathbb Q[x]/\langle x^2-3\rangle.$ Now if $\phi$ be such an isomorphism then $\phi:1+\langle x^2-2\rangle\mapsto1+\langle x^2-3\rangle.$ Hence the restriction of $\phi$ over $\{q+\langle x^2-2\rangle:q\in\mathbb Q\}$ remains the identity.  Now what?
116277
Let $\mathbb{F}, \mathbb{F'}$ be fields, $\mathbb{F} \subset \mathbb{F'}$, and $\alpha \in \mathbb{F'}$. Assume $\mathbb{F(\alpha)}$ is finitely generated algebra over $\mathbb{F}$.   Is this true: $(\mathbb{F}(\alpha):\mathbb{F}) &lt; \infty$ ?   First i thought it's a consequence of this theorem:   But we don't know if $\alpha$ is algebraic or not.
52815
$\mathbb Z[1/2]$ is not finitely generated ?   Maybe I misunderstood, what finitely generated means.  it says, we need finitely many elements and I think $1$ and $1/2$ suffices as generators.  But , at the end of Proposition $5.1.4$, from the fact that $\mathbb Z[1/2]$ is not finitely generated it is concluded that $1/2$ is not integral.
54528
Show that the set $\mathbb{Q}(\sqrt{p})=\{a+b\sqrt{p}; a,b,p\in\mathbb{Q},\sqrt{p}\notin \mathbb{Q}\}$ is a field  I am having a little trouble proving that for each non-zero $b\in\mathbb{Q}(\sqrt{p})$, there is $d\in\mathbb{Q}$ such that $b\cdot d=1$.  My attempt:  Let $b=a_1+b_1\sqrt{p}$  Let $d=a_2+b_2\sqrt{p}$  We want $$b\cdot d=(a_1+b_1\sqrt{p})(a_2+b_2\sqrt{p})$$ $$=a_1a_2+a_1b_2\sqrt{p}+a_2b_1\sqrt{p}+b_1b_2p$$  $$=(a_1a_2+b_1b_2p)+(a_1b_2+a_2b_1)\sqrt{p}$$ $$=1$$  Am I following the right approach? What should be my next step
78267
So i'm trying to understand the proof of:  $\mathbb{Q}$ is not cyclic. So this is the proof:  We proceed by contradiction.  Suppose $\mathbb Q$ is cyclic then it would be generated by a rational number in the form $\frac{a}{b}$ where $a,b \in \mathbb{Z} $ and $a$, $b$ have no common factors. Also, $a,b \neq 0$.   The set $\langle\frac{a}{b}\rangle$ consists of all integer multiples of $\frac{a}{b}$.  Therefore, if $\mathbb{Q}=\langle\frac{a}{b}\rangle$, then $\frac{a}{2b}$ is an integer multiple of $\frac {a}{b}$  PROBLEM: Why is $\frac{a}{2b}$ an integer multiple? or how is it an integer multiple. I'm not seeing it because isn't $\frac{a}{b} \times \frac{a}{b}=\left(\frac{a}{b}\right)^2$  Anyways, here is the rest of the proof:  but if   $c \times \frac{a}{b}=\frac{a}{2b}$ then $c=\frac{1}{2}$ is not an integer.  Thus, $\mathbb{Q}$ cannot be generated by a single rational number and is not cyclic.   If anyone can clarify that would be great. Also, another problem I have is doesn't this show that $\mathbb{Q}-\{0\}$ is not cyclic because I thought $\mathbb{Q}$ under the operation multplication is not a group unless zero is removed.
81649
I found an argument online that the two fields are not isomorphic but I can't make sense of the argument. You find it .  These are the things I'm confused about when it comes to the argument.     It says that the field $\mathbb Q[x]/(x^2+1)$ contains a root of the polynomial $x^2+1$. I suppose this is the congruence class $[x]$? But this is not the same as the zero element of the field $\mathbb Q[x]/(x^2+1)$, right?   It says that every isomorfism takes $0$ to $0$. I interpret this as the zero-element in $\mathbb Q[x]/(x^2+1)$ (which I interpret to be $[0]$) is taken to the zero-element in $\mathbb Q[x]/(x^2+2)$ denoted $[0]_2$. But in the argument it says that this means that if $\mathbb Q[x]/(x^2+1)$ has a root of the polynomial, then also $\mathbb Q[x]/(x^2+2)$ must have a root of this same polynomial. Why is that?     Greatful for any help!
24913
Suppose one doesn't know any Galois theory, but they are familiar with the basic theory of field extensions. Then how would one justify the following picture?    I'm pretty sure they're just saying these are the known subfields, and there could be more but they just write down those ones (which seems weird, which is why I'm asking). Or have they deduced that this is a comprehensive list of subfields just by using basic theory? This occurs on page 573 of Dummit and Foote's Abstract Algebra.
102629
Let $\alpha = 2^{1/3}$.          Prove that the set $\mathbb{Q}[\alpha] = \{a+b\alpha+c\alpha^2 \mid a, b, c \in\mathbb{Q}\}$ is closed under addition and multiplication.    Prove that if $z \in \mathbb{Q}[\alpha]$ is nonzero then there exists $z^{-1} $such that $z*z^{-1} = 1$      I have proved that $\mathbb{Q}[\alpha]$ is closed under addition, but I am having difficulty proving that it is closed under multiplication.   With all coefficients in $\mathbb{Q}$, let $a_1 + b_1\alpha + c_1\alpha^2 $ and $a_2 + b_2\alpha + c_2\alpha^2 \in \mathbb{Q}[\alpha]$. When multiplied, these two elements yield a fourth degree polynomial, and I do not know how to prove that it is an element of $\mathbb{Q}[\alpha]$
77990
The proof I'm familiar with that the algebraic numbers $\mathbb A$ form a field uses the fact that the resultant of two polynomials $p,q\in\mathbb Q[x]$ satisfies the following properties:   It is $0$ iff $p$ and $q$ have a common factor. It is a polynomial in the coefficients of $p$ and $q$.   We then introduce a new variable and cleverly manipulate $p$ and $q$ to get polynomials which vanish at the sums and products of their roots. This is in some ways a nice proof, e.g. it is constructive and so can be converted into an algorithm to find such polynomials (which I in fact just finished doing in C). But I don't find it very enlightening; it seems like the fact that $\mathbb A$ is a field is simply an accident. Is there a more enlightening proof of this fact?
124060
I'm going through the first chapter in a text on real analysis, which contains preliminaries on ordered fields, the real numbers, etc. Supposedly I had learned about such things already, in calculus, but I thought it wouldn't hurt to go over it again.  Up until the paragraph which is the subject of my question, everything is thoroughly proven and examples are provided. However the following excerpt just goes through facts, whose proof I cannot conceive:      Although $\mathbb Q$ is an archimedean field, these properties cannot be used to define $\mathbb Q$ since $\\$ there are many Archimedean ordered fields. What distinguishes $\mathbb Q$ from the other $\\$Archimedean ordered fields is that $(\mathbb Q,&lt;)$ is the $smallest$ ordered field in the following sense: $\\$ if $(X,\prec)$ is an ordered field, then $X$ contains a sub-field which is (field) isomorphic to $\mathbb Q$.$\\$ Furthermore such an isomorphism preserves the order relation.   I've concatenated the paragraph a bit, so the text isn't idem from the book (if anyone wishes to know it's Phillips, An Introduction to Analysis and Integration Theory, Dover Publications). Also $\mathbb Q$ contains no proper subfields, and this can be verified by the apparently easy-to-prove fact that, if a field has characteristic zero, then it contains a subfield isomorphic to $\mathbb Q$. Then it remains to prove the preservation of order.  I'm guessing that some concepts from abstract algebra or field theory would easily suffice, but at the moment such topics are a bit over my head, so all I can think of doing is actually constructing the isomorphism $\phi:\mathbb Q\to\hat{\mathbb Q}$, where $\hat{\mathbb Q}$ is a certain subfield of  $X$ in such a way that field operations are preserved, but I really can't come up with anything. So the question is: how do I construct this isomorphism, or, if there's a better way of proving this, how is it done? Thanks for any help.  Edit: Assume the existence of $(\mathbb Q,&lt;)$ as an archimedean ordered field.
53896
I need help proving  this stunning result.  Let me state it this way.  Let  $\mathbb L$ be a field and $\mathbb K$ be a field with $\mathbb K \subset \mathbb L$  Let $A$ be the set of algebraic numbers ie $A=\{a \in \mathbb L, \exists P \in \mathbb K [X], P(a)=0 \}$  I managed to prove that $A$ is a ring with the following reasoning:  If $a,b \in A$ then $ \mathbb K[a]$ is a ring.  And $ \mathbb K \subset \mathbb K[a] \subset \mathbb K[a][b] $  Some finite-dimension argument(ask details if needed) proves that $\mathbb K[a][b]$ has finite dimension as a vector space over the field $\mathbb K$  Reminding that $\mathbb K[a+b] \subset \mathbb K[a][b]$ and $\mathbb K[ab] \subset \mathbb K[a][b]$ proves that they both have finite dimension, hence $a$ and $b$ algebraic.  How should I proceed with inverses ?
108046
Letting $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ be fields, and letting $n \geq 3$ and $m$ be natural numbers, it is known that $\text{GL}_{m}(\mathbb{F}_{1})$ and $\text{GL}_{n}(\mathbb{F}_{2})$ are elementarily equivalent if and only if $m=n$ and $\mathbb{F}_{1} \equiv \mathbb{F}_{2}$ (as proven in "Elementary Properties of Linear Groups" in the collection "The Metamathematics of Algebraic Systems — Collected Papers: 1936–1967").  So, given a field $\mathbb{F}$, if $n \neq m$, then $\text{GL}_{m}(\mathbb{F}) \not\equiv \text{GL}_{n}(\mathbb{F})$, and thus $\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})$.  Letting $R$ be a commutative ring (with unity), and letting $n, m \in \mathbb{N}$ be such that $n \neq m$, is there a simple "algebraic" way of proving that $\text{GL}_{m}(R)$ and $\text{GL}_{n}(R)$ are not isomorphic (as groups)? Is there a simple group-theoretic way of showing that $\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})$ for a field $\mathbb{F}$?  Certain special cases of this problem trivially hold, for example in the case whereby $\mathbb{F}$ is finite, in which case $|\text{GL}_{m}(\mathbb{F})| \neq |\text{GL}_{n}(\mathbb{F})|$.
96324
Let $K \subset M$, $L\subset K'$, and let $ML$ denote the subfield of $K'$ generated by $M$ and $L$.  Is the following true? $M/K$ and $L/K$ algebraic $\implies ML/K$ algebraic?  Any hints proof or counter-example? Cheers
93710
Statement: There are uncountably many surjections from $\mathbb N$ to $\mathbb N$.  Is the following proof valid:  The real numbers are uncountable.   There exists a bijection between the set of real numbers and the set of all functions from $\mathbb N$ to $\{0,1\}$.   Given a function, $f$, from $\mathbb N$ to $\{0,1\}$, construct a surjection from $\mathbb N$ to $\mathbb N$, denoted $g$, as follows:  We map the $n$th composite number to $n$, and the nth prime, to $f(n)$.   Clearly this is a surjection since the composite numbers are countable. Clearly different functions $f$ lead to different surjections $g$. Since there are uncountably many $f$s, there are uncountably many surjections $g$.