I have understood that since $\mathbb Z_3$ is a finite field so $R$ has to be a finite dimensional vector space but is there any way of proving this from purely ring theoretic concepts ?
124095
i) Check whether $R$ is a finite ring or not. ii) Check whether $R$ has zero divisors. iii) Check whether $R$ has nilpotent elements.  i) The field $Z_3$={0,1,2} has 3 elements but as repetitions are possible while forming a polynomial for infinite times and power of x can also have infinitely possible positive integers thus $Z_3[x]$ is not a finite ring. Again as $Z_3$[x] is not finite so will not be R. Please help rectify if I am wrong here and also need help for part ii) &amp; iii).
55249
I've seen similar questions asked here, but I've not been able to find a comprehensive answer.  I know that for a ring $R$, $R[X]$ denotes the ring of polynomials over $R$ and $R(X)$ denotes the field of fractions of $R[X]$.  But if $\alpha \in S$, where $S \supseteq R$ are rings, what is the distinction between $R[\alpha]$ and $R(\alpha)$?  It is my understanding that if $R$ is a field then $R[\alpha] \cong R(\alpha)$, but that this is not generally true for any ring.  Is this correct?  Adding to my confusion is the fact that $\mathbb Z[i]$ and $\mathbb Z(i)$ are used interchangeably, despite $\mathbb Z$ not being a field.  However, it's clear to me that they are equivalent ( i.e. $\mathbb Z[i] = \mathbb Z(i) = \{ a + bi | a,b \in \mathbb Z \}$); is this because $i$ is algebraic over $\mathbb Z$?  Or for some other reason?  Thanks for your help!
42912
It's well known that if $K$ is a finitely generated extension of some field $E$, then any intermediate field $F$, $E\subseteq F\subseteq K$, is also finitely generated over $E$.  I'm curious, does the same hold for rings?      Say $S$ is a ring, and $R\supset S$ is a finitely generated extension of $S$. If $T$ is any intermediate ring, is it necessarily true that $T$ is finitely generated over $S$ as a ring?   Is it as simple as saying that for any $t\in T$, $T$ can be generated by the generators of $R$ over $S$? I feel unsure about this statement, since it's not clear to me that the generators of $R$ over $S$ need be in $T$.  If not, what is an example what shows otherwise? Thanks.
13698
I am given that $R$ is a commutative ring, $A$ is an ideal of $R$, and $N(A)=\{x\in R\,|\,x^n\in A$ for some $n\}$.  I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons.  FIRST - We decided to assume $n\in\mathbb{Z}^+$ even though this restriction was not given. We decided $n\ne 0$ because then $x^0=1$ and we are not guaranteed unity. We also decided $n\notin\mathbb{Z}^-$ because $x^{-1}$ has no meaning if there are no multiplicative inverses. Is this a valid argument?  SECOND - We want to assume $x,\,y\in N(A)$ which means $x^m,\,y^n\in A$ and use the binomial theorem to expand $(x-y)^n$ which we have already proved is valid in a commutative ring and show that each term is in A so $x-y$ is in $N(A)$. The biggest problem is how to approach the $-y$ if we are not guaranteed unity. Does anyone have any suggestions?  Thank you in advance for any insight.
42525
Let $R$ be a ring (unital, not necessarily commutative), $M$ a finitely presented left $R$ module. Suppose $m_1,\ldots,m_n\in M$ generate $M$. This determines a surjection $f:R^n\to M$.     Must the kernel of $f$ be finitely generated?   This is true by definition if the generating set $m_1,\ldots,m_n$ is the one coming from the finite presentation of $M$.
1484
Let $R$ be a commutative ring with identity. Show that $R$ is a field if and only if the only ideals of $R$ are $R$ itself and the zero ideal $(0)$.  I can't figure out where to start other that I need to prove some biconditional statement. Any help?
111914
Suppose that $V$ is a vector space, and $W$ is a subspace of $V$. If   $V$ is finite dimensional, then prove $W$ too must be finite dimensional.   It seems intuitively obvious that the dimension of $W$ must be less than or equal to that of $V$, but I am uncertain of how specifically to show this to be true analytically.  This question has been asked before (). However, there is not a great degree of explanation in the given answers. The answer which labels itself as a hint is most similar to my thoughts towards an approach to proving this particular theorem.  I was thinking that if $\exists n\in\mathbb{R}:n=\text{dim}(V) \implies \exists B = \{\vec{v}_1,...,\vec{v}_n\}$, a basis for $V$, with $n$ vectors $\vec{v}_1, ..., \vec{v}_1 \in V$. Since $W$ is a subspace of $V \implies \beta$, a basis for $W$, which has $m$ vectors in it can be constructed by removing $k\in\mathbb{R}$ vectors from $B$ such that $B\setminus\{\vec{v}_i,...,\vec{v}_j\}=\beta$ where $\vec{v}_i,...,\vec{v}_j$ are those $k$ vectors. This means a basis for $W$ has $m=n-k$ vectors in it, and by the definition of bases $\implies \text{dim}(W)=m$, and since $n-k \in \mathbb{R}$ and $n-k=m \implies m\in\mathbb{R}$. Thus $W$ is finite. $\square$  It seems like there may be some fault with this argument, though.
62747
Let $\mathbb Z$ be the ring of integers. The question asks to show  that every ideal of $\mathbb Z$ is principal. I beg someone to help me because it is a new concept to me.
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.
32099
Check my proof please.  Let $\Bbb Z_p:=\Bbb Z/p\Bbb Z$ be the quotient ring modulo $p$. I want to prove that if $p$ is prime then $\Bbb Z_p$ is a field.  Known facts about the quotient rings of the kind $\Bbb Z_n$ (I dont list all the characteristics of a ring, just some ones):   They are commutative rings with unity. $[0]=[n]$ is the identity of addition or zero. Then $[n]+[a]=[a]$ and $[n]\cdot[a]=[n]$ for all $[a]\in\Bbb Z_n$. $[1]$ is the multiplicative identity or unity. Then $[1]\cdot[a]=[a]$ for all $[a]\in\Bbb Z_n$. The addition is defined as $[a]+[b]=[a+b]$, and the multiplication is defined as $[a]\cdot[b]=[ab]$. $[a]=[b]$ means that exists $z\in\Bbb Z$ such that $a=b+nz$. $|\Bbb Z_p|=p$.   A ring $\Bbb Z_p$ dont have zero divisors, i.e. doesnt exist $[a],[b]\in\Bbb Z_p$ distinct of $[p]$ such that $[a]\cdot[b]=[p]$. Proof by contradiction: suppose that exists such $[a],[b]$ distinct of $[p]$ that are zero divisors. Then exist some $z_j\in\Bbb Z$ such that:  $$(a+pz_1)(b+pz_2)=pz_3\iff ab+pz_4=pz_3\iff ab=pz_5$$  but this is a contradiction with the assumption that $a,b\neq p$ and $p$ prime, so dont exist divisors of zero on any $\Bbb Z_p$.$\Box$  For any $[a]\in\Bbb Z_p$ distinct of $[p]$ exist $[b]$ such that $[a][b]=[1]$. Because $\Bbb Z_p$ dont have zero divisors and is finite for some $[a]\neq[p]$ we have that if $[b]\neq[c]$ then $[a][b]\neq[a][c]$. Proof by contradiction: if $[a][b]=[a][c]$ in the above conditions then  $$[a][c]=[ac]=[ab]=[a][b]\iff ac+pz_1=ab+pz_2\iff a(c-b)=pz_3$$  for some $z_j\in\Bbb Z$. But $pz_3\in[p]$, then $[a][c-b]=[p]$ but because there are no zero divisors in $\Bbb Z_p$ then or $[a]=[p]$ or $[c-b]=[p]$ what contradicts the above conditions.$\Box$  Because $[a][b]\neq[a][c]$ for $[a]\neq[p]$ and $[b]\neq[c]$ then for any $[a]\neq[p]$ we have that  $$[a]\cdot\Bbb Z_p=\Bbb Z_p\implies \exists [b]\in\Bbb Z_p:[a][b]=[1]$$  In other words: every element of $\Bbb Z_p$ but $[p]$ have a multiplicative inverse as stated above.$\Box$  Because $\Bbb Z_p$ is a commutative ring with $[1]\neq[0]$ and have multiplicative inverse for all their elements but $[p]=[0]$ then $\Bbb Z_p$ is a field.$\Box$
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
122001
Here we have $F$ a field and $f: \mathbb{Z} \to F$ a ring epimorphism. We are to prove that $F$ is a finite field with non-zero characteristic.   I know that since $f$ is an epimorphism, we have that $\forall x\in F$, $\exists y \in \mathbb{Z}$ such that $f(x) = y$. (As an epimorphism is a surjective ring homomorphism).   I am unsure about two things: Am I to prove that F is finite, or just prove that it has a non-zero characteristic. Either way, I could use some guidance. Thank you :)   PS This is for HW, so I only require a good hint about how to approach the problem.
67015
How are $2\mathbb{Z}\ncong3\mathbb{Z}$ different as rings? What interesting properties does one have that the other doesn't?
34826
Let $V_{n}(F)$ be a vector space over the field $F=\mathbb Z_{p}$ with $\dim V_{n} = n$, i.e., the cardinality of $V_{n}(\mathbb Z_{p}) = p^{n}$. What is a general criterion to find the number of bases in such a vector space? For example, find the number of bases in $ V_{2}(\mathbb Z_{3})$. Further, how can we find the number of subspaces of dimension, say, $r$?   I need a justification with proof. I have a formula, but I am unable to understand the basic idea behind that formula.
67717
Every finite commutative ring $R$ must be of the form $\frac{ \mathbb{Z} }{n_1 \mathbb{Z}} \times \frac{\mathbb{Z}}{n_2 \mathbb{Z}} \times \dots \frac{\mathbb{Z}}{n_k \mathbb{Z}}$. I am able to proceed upto the step that it should be direct product of artinian commutative local rings. I used structure theorem for commutative ring. But I am unable to proceed further.
83432
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.
123394
Let $V$ be an $n$-dimensional vector space over $\mathbb R$. Prove that $V$ has a subspace of dimension $r$ for each $0 \le r \le n$.  Is this as simple as saying that $V$ has a basis of $n$ elements, and then take $r$ elements from this basis and this will generate an $r$-dimensional subspace? Or am I missing something? It seems too simple.  Thanks