Explain why $\mathbb{Z \times Z}$ and $\mathbb{R \times R}$ is not a field and that why any external direct sum of two fields cannot be a field. I believe it has much to do with the lack of every non-zero element having an inverse, however I am having difficulty seeing it.
4797
I have a question, and this is it in entirety:     Let R be a non-zero ring(that is, R contains at least one element other than the zero element). Prove that $R \times R$ is NOT an integral domain.   Now I haven't done a problem like this before, and it seems to be(from the Wikipedia definition) that this ring $R$ needs to be commutative to be an integral domain. Which it isn't stated to be here. What exactly am I meant to do? I haven't been given anything else than the area in grey.
59080
Let F be a ring. Show that if $a$ is a zero divisor of F, then $a$ fails to have a multiplicative inverse.  All I understand is that $ab = 1$ doesn't hold and that $ac = 0$ for some $c \in F$ holds. Basically, only the definitions, but I don't know what to do with them.
40688
Is there a (proper) subfield $K$ of $\mathbb{R}$ such that $\mathbb{R}$ is an algebraic extension of $K$?  From this question, , it is clear that for such an $K$, $[\mathbb{R}:K]=\infty$.  This question seems to be non-trivial, and I suspect any existence result will be non-constructive. Any references will be appreciated.
9284
I have to show that the ring $\mathbb{Z}[X]/(X^2-1)$ is not isomorphic with $\mathbb{Z}\times \mathbb{Z}$.    I know that $(\mathbb{Z}\times\mathbb{Z})^*=\{(\pm1,\pm1)\}$, so I thought I should be looking for elements which have inverses in $\mathbb{Z}[X]/(X^2-1)$ and hopefully find more or less than 4. But I didnt succeed, so I need hints. Thanks.
77634
So I took an introductory abstract algebra course a few semesters ago, and we were shown that groups and rings can both be made into products, i.e. if I have some group $G$ (resp. ring $R$) and some indexing set $I$, then I can make a group $G^{I}$ (resp. ring $R^{I}$) with the appropriate cardinality. However, it was also demonstrated that for fields, you cannot keep the multiplicative inverse under a product structure, i.e. if I have fields $F_{1}, F_{2}$, then $F_{1} \times F_{2}$ will still be a ring, but you will not (not just "not in general", but won't) have multiplicative inverses for all non-zero (i.e. not $(0, 0)$) elements, as you could pick $(0, x)$, where $x \neq 0$. According to Wikipedia, an ultraproduct will preserve field structure, but I'm not sure what the cardinality of $\prod_{i \in I} F_{i} / \mathscr{U}$ would be in general.  I saw  that said you could make a field of arbitrarily large cardinality by extending a given field through throwing in a whole bunch of "transcendental" elements. The example given was that if I had $\mathbb{Q}$, I could start by dumping in the complexes, but then I suppose after that I'd just start throwing in dogs and cats or something; I'm really not sure what that responder meant, and ceteris paribus, my dogs tend to stay inside, so I'd rather keep them out of my fields, particularly the large ones the thread was interested in.  Moreover, the thread mentioned earlier seemed more concerned with just making the fields big. My question is a bit more nuanced: Given any cardinality $\kappa &gt; 1$, can I generally construct a field $F$ of a cardinality $\kappa$? Moreover, how vague do I have to be about it (i.e. do I have to use some really choicy methods, or can I make it a bit more straightforward)?  Thanks.  EDIT: Sorry. I am aware that for finite fields, you are limited to powers of primes. I meant for infinite cardinals.
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In $\mathbb{Z}_n$ the elements are fully partitioned between the units and the zero-divisors. I believe this is the case, am I correct?  Now, I take it this does not hold true in general, there may be rings with elements that are neither units nor zero divisors?
115037
So, I was taught that $\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}$.  But since the set of complex numbers is by definition $$\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\},$$ doesn't this mean $\mathbb{R}\subseteq\mathbb{C}$, since for each $x \in \mathbb{R}$ taking $z = x + 0i$ we have a complex number which equals $x$?
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This is my first time posting on Math exchange. I have been self-teaching myself Mathematics and recently started learning some Algebra. I was posed the following questions, in which I would like to answer before I proceed, but having some trouble finding proper solutions. So the format will be question, and my answer.      What name do we give an element $a \in \mathbb{Z}_n$ that has a multiplicative inverse?   A modular multiplicative inverse of an integer $a$ is an integer $x$ such that $ax \cong 1 (\mod m)$. This means that the remainder after dividing $ax$ by integer $m$ is equal to $1$.      Give an example, for some $n \in \mathbb{N}$, of an element of $\mathbb{Z}_n$, that does not have a multiplicative inverse?   This question is throwing me off because - does the $n$ we speak of matter for both $n \in \mathbb{N}$ and $\mathbb{Z}_n$ ot solely $n$? My guess here would be $0$ because we cannot have an inverse of $0$.      Prove that multiplicative inverse are unique; an element $a \in \mathbb{Z}_n$ cannot have two multiplicative inverses.    Suppose that an element $a\in\mathbb{Z}_n$ has an inverse $a'$. Next, suppose that $a^*$ is also an inverse of $a$. Then, $a'$ is a solution to $a \cdot_n x=1$, and similarly $a^*$ has a solution to $a \cdot_n x=1$.   Lemma: Suppose $a'$ is multiplicative inverse of $a$ in $\mathbb{Z}_n$, then for any $b \in \mathbb{Z}_n$ the equation $a \cdot_n x=b$ has a unique solution, such that $x = a' \cdot_n b$.   By Lemma, equation $a \cdot_n x = 1$ has a unique solution, namely $x=a'=a^*$. Therefore, an element $a\in \mathbb{Z}_n$ cannot have two multiplicative inverses.    Thank you for the help. I would love any constructive feedback for these. If you can help me elaborate on some ideas, that would be really helpful also.  Thank you very much!
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!
120060
Again, my instructor's explanation in class is so confusing to me.  Here we cannot take the original approach of finding the order and limiting our choices.  So instead my instructor says that this is isomorphic to $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}_3$.  To justify it he said that $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$ is 3 dimensional and the denominator is one dimensional but this doesn't quite explain where the $\mathbb{Z}_3$ comes from?  Can anyone provide a better way to think about this?  It is making no sense to me at all.
36141
Show that $(\mathbb{Z}[x]/(x^{n+1}))^{\times}\cong \mathbb{Z}/2\mathbb{Z}\times\Pi_{i=1}^n\mathbb{Z}$.   Anyway how and what method is used to prove this. I still have no idea now. really thanks for your help
26386
Let $(\mathbb K,+,\cdot)$ be a field. Is there a group isomorphism between $(\mathbb K,+)$ and $(\mathbb K^\times,\cdot) $ ?  The answer should clearly be negative. I tried to proceed via contradiction, but this has not led me very far. Since it's some kind of &quot;trick problem&quot;, I'm merely looking for hints.
97650
So I am dealing with a problem from Dummit (specifically 2.1.7) and am having some issues. The  part of the problem in question is:      Prove the set of elements of the direct product $\mathbb{Z} \times (\mathbb{Z} / n \mathbb{Z})$ of infinite order together with the identity is not a subgroup of $\mathbb{Z} \times (\mathbb{Z} / n \mathbb{Z})$.    I am assuming this question is referring to addition as the binary operation for both sets. I know the proof basically involves showing that the product is not closed; however, do not all the elements of the integers have infinite order (except the identity)? In that case, the elements of infinite order together with the identity would form all of $\mathbb{Z} \times (\mathbb{Z} / n \mathbb{Z})$ would it not?  Thanks
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.
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Though I've seen several cool axiomizations of $\mathbb{R}$, I've never seen any at all for $\mathbb{Z}$.  My initial guess was that $\mathbb{Z}$ would be a ordered ring which is "weakly" well-ordered in the sense that any subset with a lower bound has a minimal element.  However, after seeing  of a discrete ordered ring, I'm less sure. I made that guess under the impression that the fundamental characteristic of $\mathbb{Z}$ is that every nonzero element has exactly one representation of the form $\pm (1+1+\dots+1)$, but that seems to be shared by every DOR.   Presumably, this definition wouldn't exist if every instance of it was isomorphic to $\mathbb{Z}$, so can someone give me an example of another discrete ordered ring? More to the point, what is a sufficient characterization of $\mathbb{Z}$? (and a proof sketch of uniqueness would be nice)  I'm aware that $\mathbb{Z}$ is pretty easily constructible from $\mathbb{N}$, but I want to use this for a seminar and given the audience I am expecting, I would rather not deal with Peano. (And I guess it feels like cheating to say "$\mathbb{N}$ is a well-ordered rig")
86430
Problem: Consider the subfield $$E = \mathbb{Q} \big(\left\{\sqrt[n]{2} \mid n \in \mathbb{N} \setminus \left\{0\right\} \right\} \big)$$ of $\mathbb{R}$. Prove that $[E: \mathbb{Q}]$ is not finite.  Attempt: I proved that $E$ is algebraic over $\mathbb{Q}$. Because let $u \in E$. Then there is a $n \in \mathbb{N} \setminus \left\{0\right\}$ such that $u = \sqrt[n]{2}$. Then $f(x) = x^n - 2 \in \mathbb{Q}[x]$ is the minimal polynomial of $u$ over $\mathbb{Q}$ such that $f(u) = 0$. Hence $E$ is algebraic over $\mathbb{Q}$.  I now want to prove the degree of extension is not finite. I assume it is finite. Suppose then $[E: \mathbb{Q}] = k \in \mathbb{N}$. I don't know how to derive a contradiction from this.   I know that if $u$ is algebraic over $\mathbb{Q}$ and $f$ is the minimal polynomial of $u$ over $\mathbb{Q}$, then $[\mathbb{Q}(u) : \mathbb{Q}] = \deg(f)$.   Help/suggestions are appreciated.
109391
I am working through Paolo Aluffi's new GSM text on my own (self-study). On page 63, he asks the reader to     Prove that $\Bbb{Q}$ is not the direct product of two nontrivial groups.   For some context, this is an exercise following a section entitled "The category Grp".  I am assuming that he means "is not isomorphic to the direct product of two nontrivial groups", and I can see two possible ways to proceed with this proof, but have been unsuccessful with either approach.    Approach 1:  Show that the additive group of rationals has a property that is preserved by isomorphism that the direct product of two nontrivial groups does not have or vice-versa.  This seems challenging unless I can significantly narrow down the properties that a direct product of two nontrivial groups that was isomorphic to $\Bbb{Q}$ would necessarily have.  Approach 2:  Considering the section in which this question occurs, show that if $G$ and $H$ are nontrivial groups and $\Bbb{Q} \cong G \times H$, then there are homomorphisms $\varphi_{G}:\Bbb{Q} \rightarrow G$ and $\varphi_{H}:\Bbb{Q} \rightarrow H$ which do not factor or do not factor uniquely through the product $G \times H$.  This would be a contradiction, as $G \times H$ is a final object in the category Grp.  I would greatly appreciate suggestions on how to proceed further with either of these approaches or with alternate approaches.