Say $R$ is a ring, and $I_1,...,I_n,J\subseteq R$ ideals, s.t $J\subseteq\bigcup_{i=1}^nI_i$. If there exists $\phi:K\rightarrow R$ homomorphism where $K$ is an infinite field, then there exists an $i$ s.t $J\subseteq I_i$. I split it into two cases: Since $\phi$ is an homomorphism and $K$ is a field, then $\phi^{-1}(J)=\langle0 \rangle $ or $\phi^{-1}(J)=K$. If it's the first case, then since $\phi$ is injective (since $K$ is a field) we then get that $J=\langle 0 \rangle$ and obviously for every $i$, $J\subseteq I_i$. Now, if $\phi^{-1}(J)=K$, then there must be at least one index $i$ s.t $\phi^{-1}(I_i)=K$ - otherwise the union is $\bigcup_{i=1}^n I_i=\langle 0 \rangle$ - contradiction since $J\neq\langle 0 \rangle$. So I can see that $\phi(\phi^{-1}(J))\subseteq \phi(\phi^{-1}(I_i))$, but this doesn't immediately mean that $J\subseteq I_i$ - even with $\phi$ being an injection. What am I missing here? Any help would be appreciated.
45456
Let $R$ be a ring and let $I_1,\dots,I_n,J\subseteq R$ be ideals such that $J\subseteq I_1\cup\dots\cup I_n$. Suppose that there exists an infinite field $K$ and a homomorphism $\phi:K\to R$. Prove that $\exists k$ s.t. $J\subseteq I_k$.   Attempt: (Edited)  If $J\subseteq I_1$ then we're done. Otherwise, let $x\in I_1$ and let $y\in J\setminus I_1$. $\forall r\in K, x+ry\notin I_1$ (because otherwise we'll get $y\in I_1$).  $K$ is infinite so we have infinitely many elements of the form $x+ry\in J$. So there's some $k$ for which $I_k$ contains infintely many elements of the former form.  If $n=2$ so $x+y\notin I_1$ implies $$x+y\in I_2\Rightarrow x\in I_1\Rightarrow I_1\subset I_2\Rightarrow J\subseteq I_2$$  Suppose the theorem holds for some $n$. $x+y\notin I_1\Rightarrow \exists k,I_k\ni x+ry$ for infintely many elements as above.
89299
Let $f \colon A \to B$ be a ring homomorphism, and let $N$ be a $B$-module. Regarding $N$ as an $A$-module by restriction of scalars, form the $B$-module $N_B = B \otimes_A N$. Show that the homomorphism $g \colon N \to N_B$ which maps $y$ to $1 \otimes y$ is injective and that $g (N)$ is a direct summand of $N_B$.   [Define $p \colon N_B \to N$ by $p (b \otimes y) = by$, and show that $N_B = \operatorname{Im} (g) \oplus \operatorname{Ker} (p)$.]   My idea was to define an isomorphism \begin{align*} \phi \colon N_B &amp; \to \operatorname{Im} (g) \oplus \operatorname{Ker} (p) \\ b \otimes y &amp; \mapsto (p (b \otimes y), b \otimes y - 1 \otimes (g \circ p) (b \otimes y)) \end{align*}  It's easy to check that $\phi$ is injective and surjective. If I take the $B$-module structure of $N_B$ to be $$ b' . (b \otimes y) = (b'b) \otimes y $$ I only manage to show that $\phi$ is a homomorphism of $A$-modules, because $g$ seems to be only a homomorphism of $A$-modules: $$ b.g(y) = b.(1 \otimes y) = b \otimes y \neq 1 \otimes (b.y) = g (b.y) $$ (where $\neq$ means "not equal in general").  However, the way the question is phrased seems to suggest that the isomorphism should be one of $B$-modules.  Should I take the $B$-module structure of $N_B$ to be something else, for example $$ b'.(b \otimes y) = b \otimes (b'.y) $$ or should I show the result only for $A$-modules?  I've been confusing myself about this issue of "as $A$-module" or "as $B$-module" and would appreciate if someone could clarify. (I also wonder if, since homomorphisms are only fancy functions of sets, it somehow doesn't matter.)
91403
I have been having a lot of trouble teaching myself rings, so much so that even "simple" proofs are really difficult for me. I think I am finally starting to get it, but just to be sure could some one please check this proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field. Thank you.  Proof: Notice that $$\langle 1 - i \rangle\\  \Rightarrow 1 = i\\  \Rightarrow 2 = 0.$$ Thus all elements of the form $a+ bi + \langle 1 - i \rangle$ can be rewritten as $a+ b + \langle 1 - i \rangle$. But since $2=0$ this implies that the elements that are left can be written as $1 + \langle 1 - i \rangle$ or $0 + \langle 1 - i \rangle$. Thus  $$ \mathbb Z[i]/ \langle 1 - i \rangle = \{ 0+ \langle 1 - i \rangle , 1 + \langle 1 - i \rangle\}. $$  This is obviously a commutative ring with unity and no zero-divisors, thus it is a finite integral domain, and hence is a field. $\square$
44835
If $R$ is a ring, J is an ideal in $R$, and $I$ is an ideal of $J$ (with $J$ considered as a ring), does it follow that $I$ is an ideal of $R$?  That is, is $I$ necessarily closed under multiplication by elements of $R$?  Surely $I$ is closed under multiplication by elements of $J$ (since $I$ is an ideal of $J$).  The "obvious" approach to proof fails so quickly that it must be false, that "an ideal of an ideal is not necessarily an ideal of the big ring".  Please provide a counterexample.  (Or a proof?)
38817
If $A$ is a semilocal ring and $f:A\rightarrow B$ is a surjective homomorphism, then $f(\operatorname{rad}A)=\operatorname{rad}B$.   I know that if $A$ is a semilocal ring and if $I_{1},\dots, I_{n}$ are all of its maximal ideals, then $\operatorname{rad}A=I_{1}\cdots I_{n}=I_{1}\cap\cdots\cap I_{n}$. Let $y\in\operatorname{rad}B$, then there is $x\in A$ such that $f(x)=y$. I would like to show that $x\in\operatorname{rad}A$.  Any hint?
76964
Prove that if an integral domain $R$ is an injective $R$-module then $R$ is a field.  "Proof"  Choose a non-zero element $x$ of $R$ and consider the map $f: R \rightarrow R$ given by $f(t)=tx$.  This yields an exact sequence $0 \rightarrow R \rightarrow R \rightarrow R/\langle x \rangle \rightarrow 0$ where the first map is $f$.  By assumption $R$ is injective and thus $R \cong R \oplus R/\langle x \rangle$.   Now the claim is that $1 \in \langle x \rangle$. Otherwise $(1,0 + \langle x \rangle)$, $(0,1+ \langle x \rangle)$ are zero divisors of $R \oplus R/\langle x \rangle$ which is impossible since $R$ is an integral domain.  Thus $1 \in \langle x \rangle$ so $x$ is invertible and we're done. Is this OK?
67788
Let $S$ be a multiplicatively closed subset of a ring $R$, and let $I$ be an ideal of $R$ which is maximal among ideals disjoint from $S$. Show that $I$ is prime. If $R$ is an integral domain, explain briefly how one may construct a field $F$ together with a ring homomorphism $R\to F$. Deduce that if $R$ is an arbitrary ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset disjoint from $I$, then there exists a ring homomorphism $f\colon R\to F$, where $F$ is a field, such that $f(x)=0$ for all $x\in I$ and $f(y)\neq 0$ for all $y\in S$.   [You may assume that if $T$ is a multiplicatively closed subset of a ring, and $0\notin T$, then there exists an ideal which is maximal among ideals disjoint from $T$.]  Here is a question I need to answer. For the first part I can show if $x, y$ are not in $I$ then nor does $xy$. The second part is just the field of fractions.   For the third part, I think I need to find an ideal $J$ containing $I$ such $J$ is prime, so that $R/J$ is an integral domain, and use the second part. To find $J$ prime, I think as suggested by the hint, I should go for a maximal ideal disjoint from $S$ containing $I$, but how can I do that?
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.
34488
I've been doing some old exam problems and I've come across a problem that I've answered, but my gut is telling me that there's something I'm glossing over.     Let $R$ be a commutative ring with identity and let $U$ be an ideal that is maximal among non-finitely generated ideals of $R$. I wish to show that $U$ is a prime ideal.   Assume that $U$ is not prime. Let $x, y\not\in U$ be such that $xy\in U$. $U$ is contained in a maximal ideal $M$ and $xy\in M$, so either $x$ or $y$ is in $M$; assume $x\in M$. The condition $U\subset M$ then implies that there is a ring homomorphism $$\varphi: R/M\to R/U$$  Since $R/M$ is a field, $\varphi$ is injective. Hence, $\varphi(x)\in U$. This is a contradiction, so $U$ must be prime.  The thing that worries me is that I never explicitly used the hypothesis that $U$ was not finitely generated or the result that $M$ must be finitely generated.
79380
I'm stuck on this exercise and any help would be well appreciated:     Let $R$ be a commutative ring with ideals $I,J$. Show that $R=I+J$ if and only if $\phi(x)= (x + I, x + J)$ is surjective from $R$ to $R/I \times R/J$.   Assuming surjectivity I got as far as to realize that I need to show $R\subset I+J$ since $I+J\subset R$ always holds (as they are both subrings of R). Now I'm not sure how to proceed.  Also, I don't even know where to begin in the direction $R=I+J$ implies $\phi$ is onto.  Again, any help, pointers or references are much appreciated.  Thanks in advance.
49308
If $ \phi: R \to R'$ is a homomorphism of $R$ onto $R'$ and $R$ has a unit element,  $1$, show that $\phi(1)$ is the unit element of $R'$.  I am having trouble proving this. I think I just need a hint on how to start and I will be able to solve. But basically I have tried a few things like $\phi(1) = \phi(a)\phi(b)$ but I cannot assume it is a division ring. Also is it okay to assume from the wording of the question that $R'$ has a unit element.  Thanks in advance.
89578
I'm trying to learn some basic abstract algebra from Pinter's A Book of Abstract Algebra and I find myself puzzled by the following simple question about ring homomorphisms:     Let $A$ and $B$ be rings.  If $f : A \to B$ is a homomorphism from $A$ onto $B$ with kernel $K$, and $J$ is an ideal of $A$ such that $K \subseteq J$, then $f(J)$ is an ideal of $B$.   I'm clearly missing the obvious, but I don't see where the requirement $K \subseteq J$ comes into the proof.  Since $f$ is also a homomorphism of additive groups, the image $f(J)$ must be closed under addition and negatives (correct?). Then for $f(J)$ to be an ideal we have to show that it is closed under multiplication by an arbitrary element $b \in B$. Since $f$ is onto, there is some $a \in A$ such that $b = f(a)$. Let $j'$ be any element of $f(J)$, so $j' = f(j)$ for some $j \in J$. Then $bj' = f(a)f(j) = f(aj)$, and $aj \in J$ since $J$ is an ideal.  Then it seems that $f(J)$ is closed under multiplication by $B$.  What mistaken assumption am I making?
67264
Quoting " Let $\phi : \Bbb Z \rightarrow \Bbb Z$ be given by $\phi(n) = 7n$. Prove that $\phi$ is a group homomorphism. Find the kernel, and the image of $\phi$."   My understanding:  Part 1:   Given $n,m \in \Bbb Z$, let's check that $\phi(n+m) =\phi(n)+\phi(m)$  $$\phi(n+m) = 7(n+m) = 7n + 7m = \phi(n)+\phi(m)$$  Therefore as the group operation in $\phi$ is preserved, $\phi$ is a group homomorphism.  Part 2 (revised):   $\phi$ is one-to-one as:  $\space \space \phi (n) = \phi(m) \Rightarrow 7n =7m \Rightarrow n=m  $. We also know that $\phi(e)=e$ by the property of homomorphism. Therefore, no other object than $0$ in the domain can map to $0$ in the codomain. It follows that: $$\ker(\phi)=\{ x \in \Bbb Z : \phi(x)=e\}=\{e\}$$   Part 3 (revised):  I claim that the only possible image-elements of $\phi$ are the multiples of $7$ in $\mathbb Z$ denoted $7\mathbb Z$.   To prove this I have to show that Im$(\phi) \subset 7 \Bbb Z$ and that $ 7 \Bbb Z \subset $ Im$(\phi)$.  Proving Im$(\phi) \subset 7 \Bbb Z$ :      $\forall y \in $ Im$(\phi)$ such that $  y =\phi(x)$ where $x \in 7 \Bbb Z$. It follows that $y \in 7 \Bbb Z$, therefore Im$(\phi) \subset 7 \Bbb Z$   Proving $ 7 \Bbb Z \subset $ Im$(\phi)$:   $\forall y \in  7\Bbb Z$ such that $  y =7x$ where $x \in \phi(x)$. It follows that $y \in \phi(x)$, therefore $ 7 \Bbb Z \subset $ Im$(\phi)$.   Any input to my understanding is much appreciated.
117791
Let the ring of fractions be denoted as $S^{-1}R$.       Proposition (from Robert Ash's textbook "Basic Abstract Algebra")      Define $f: R \rightarrow S^{-1}R$ by $f(a) = a/1$. Then f is a ring homomorphism. If S has no zero divisors then f is a monomorphism, and we say that R can be embedded in $S^{-1}R$. In particular:      (i) A commutative ring R cna be embedded in its complete (or *full) ring of fractions ($S^{-1}R$, where S consists of all non-divisors of zero in R).       (ii) An integral domain can be embedded in its quotient field.    Why does the proposition say "If S has no zero divisors..."? If S really had zero divisors, then this wouldn't even be properly defined, right? Because we would have a/0 for some a in R.
87998
It is well-known that if $S$ is a graded ring, and $f$ is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localization $S_f$ and the prime ideals of $S_{(f)}$, the subring of $S_f$ comprising the homogeneous elements of degree $0$.    This is proposition II.2.5b in Hartshorne, exercise 5.5B in Ravi Vakil's notes $[1]$ (p. $130$ of the February $24$, $2012$ version), and proposition 8.1.21 of Akhil Mathew's notes $[2]$ (p. $136$). Unfortunately I cannot follow any of those proofs to my own satisfaction, perhaps because I'm not well-versed in commutative algebra.  The crux of the proof appears to be to show that, given a homogeneous prime ideal $\mathfrak{p}$ of $S$ not containing $f$, the construction used to obtain a prime ideal $\Psi (\mathfrak{q})$ of $S$ from a prime ideal $\mathfrak{q}$ of $S_{(f)}$ will recover $\mathfrak{p}$ when $\mathfrak{q} = S_f  \mathfrak{p} \cap S_{(f)}$. To be precise, let $\Psi (\mathfrak{q})$ be the homogeneous ideal of $S$ generated by $$\bigcup_{d \in \mathbb{N}} \{ s \in S_d : s / f^d \in \mathfrak{q} \}$$ and let $\Phi (\mathfrak{p}) = S_f \mathfrak{p} \cap S_{(f)}$. It's easy to see that $\Phi \circ \Psi$ acts as the identity on $\operatorname{Spec} A$ (or, for that matter, the set of all ideals of $A$), but I cannot see any obvious reason why $\Psi \circ \Phi$ should act as the identity on the set of prime ideals of $S$ not containing $f$. A detailed proof of this point would be much appreciated.  References  $[1]$ .  $[2]$ Algebraic geometry notes (covers material at the level of the first and second volume of EGA): , and .
118281
[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]  Below, $R$ is a commutative ring with $1$.  In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.  The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.  I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.  So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.  My questions:   If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution :-) I'll post my solution once I have one. In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great.
116926
Assume R is noetherian and $f:R \to S$ is a ring homomorphism. Is $f(R)$ noetherian?     Reading Dylan Moreland commentary: "If $\mathfrak{b}$ is an ideal of $f(R)$, then $f^{-1}(\mathfrak{b})$ is an ideal of R and $f(f^{-1}(\mathfrak{b}))=\mathfrak{b}$ ." Then he writes: "if you have an  increasing chain of ideals in $f(R)$" . So because there are only finitely many ideals in $f^{-1}(\mathfrak{b})$, which is noetherian, there can also only be finitely many ideals in $f(R)$. So we can say from this that $f(R)$ is noetherian?  Reading Arturo Magidins commentary: I don't know the lattice isomorphism theorem. I can't find it on google either. Do you maybe have a link for me so I can read it? For the second definition of a noetherian ring: every ideal is finitely generated because: $f(a_{0},a_{1},....,a_{n}) = f(a_{0}),f(a_{1}),f(a_{2}),...,f(a_{n})= f(R)$ follows directly because f is a ring homomorphism. Is this a correct proof already?    Thank you.