I have studied "structure theorem for Artin rings" which states  "An Artin ring $A$ is unique a finite direct product of Artin local rings".  Let $A$ be Artin ring.  By Chinese remainder theorem, $A \cong \prod (A/m_i^k)$, in which $m_i$ is a maximal ideal in $A$ and $R^k= (m_1 \cap m_2 \cap... \cap m_n)^k =0 $.  Since $A$ is Artin then $A/m_i^k$ is also Artin.      However, I do not understand why $A/m_i^k$ is a local ring.   I am beginner then could you help me explain in detail about it?  I guess the maximal ideal of the ring $A/m_i^k$ is $m_i/m_i^k$.  Thank you in advance.
12858
Say I have a commutative ring $R$ with a maximal ideal $m$. Then $m/m^k$ is a maximal ideal in $R/m^k$ for any $k$. Is it the only maximal ideal, i.e. is $R/m^k$ a local ring?    This is a well known result for $k = 1$, as $R/m$ is a field. It seems to be true in other cases, e.g. $p\mathbb{Z}\subset \mathbb{Z}$ for prime $p$ and for $(x,y) \subset \mathbb{F}[x,y]$, the maximal ideal generated by $x$ and $y$ in a polynomial ring over the field $\mathbb{F}$.   Equivalently, if an element $x\notin m$, is $x + m^k$ an invertible element in $R/m^k$?
76942
We're all aware of the theorem that says for a commutative ring $R$ with $1$, $I$ is maximal iff $R/I$ is a field. I'm currently working on a problem where I was asked to show commutative rings have IBN, and I used a lemma saying if $\psi:R\to T$ was a ring homomorphism and $R$ did not have IBN, then $T$ does not have IBN. I used this by sending $R$ to $R/I$, where $I$ is maximal, so $R/I$ is a field and must have IBN, so $R$ has IBN.   This made me wonder why this wouldn't work for any ring $R$. If I could just choose any maximal left ideal $I$ then send it to $R/I$, if $R/I$ is a division ring then I must have $R$ having IBN for the same reason. Therefore there must be a case where $I$ is a maximal ideal but $R/I$ is not a division ring. Does anybody know any examples? Is a "maximal ideal" even a well defined notion if $R$ is not commutative?
10216
Sorry. This might probably be a really easy question, but I am only a beginner in algebraic number theory. So, please bear with me.  Let $K$ be an algebraic number field and $\mathcal{O}_K$ its ring of integers.  Preludium   In my lecture we have defined the norm $N_{K/\mathbb{Q}}(x)$ for an element $x \in K$ in the usual way. In a following lecture we "extended" this to the norm of an ideal $\mathfrak{a}$ of $\mathcal{O}_K$ by setting $$ \mathfrak{N}(\mathfrak{a}) := [\mathcal{O}_K : \mathfrak{a}] := \left| \mathcal{O}_K / \mathfrak{a} \right|$$ and showed that for the principal ideal $(a)$ with $a \in \mathcal{O}_K$ we always have $\mathfrak{N}((a)) = \left|N_{K/\mathbb{Q}}(a)\right|$. Later, we introduced the notion of a fractional ideal in $K$ being a non-zero, finitely generated $\mathcal{O}_K$-submodule of $K$. And we showed that a non-zero $\mathcal{O}_K$-submodule $\mathfrak{a}$ of $K$ is a fractional ideal iff there is a $c \in \mathcal{O}_K \setminus \{0\} $ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ (thus $c\mathfrak{a}$ is an ideal of $\mathcal{O}_K$). We now extended the norm $ \mathfrak{N}$ by defining for each fractional ideal $\mathfrak{a}$ of $K$ the norm $$ \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|}, $$ $c$ being an element of $\mathcal{O}_K \setminus \{0\}$ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ (exists according to 2.).   Question  How can I show that this latter definition is independent of the chosen $c$?  Edit  I changed the definition in (3) from $ \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{N_{K/\mathbb{Q}}(c)} $ to $\mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|} $ and added the absolute value everywhere else, too. I think that was a typo in my lecture notes.  Thoughts (Is it better to add them as comments? I was not sure.)   Let $c, c' \in \mathcal{O}_K \setminus \{0\}$ such that $c \mathfrak{a} \subseteq \mathcal{O}_K $ and $c' \mathfrak{a} \subseteq \mathcal{O}_K$. At the moment I am contemplating if and in which way $c$ and $c'$ are related. I was wondering whether one always has $c'=bc$ (or vice versa) for an $b \in \mathcal{O}_K$. Then we had $$ \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c')\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(bc)\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(b)\right| · \left|N_{K/\mathbb{Q}}(c)\right|}$$ and it remained to show that $[\mathcal{O}_K : bc\mathfrak{a}] = \left|N_{K/\mathbb{Q}}(b)\right| · [\mathcal{O}_K : c\mathfrak{a}]$ (maybe because $[\mathcal{O}_K : bc\mathfrak{a}] = [\mathcal{O}_K : (b)] · [\mathcal{O}_K : c\mathfrak{a}]$ ???).
121070
I want to understand why the following is true:     Let $S \subseteq R$ be commutative rings with $1$ and assume that $R$ is finitely generated as an $S$-module by at most $k$ elements. For every maximal ideal $M$ of $S$ there are at most $k$ maximal ideals of $R$ lying over $M$.   So let $T$ be a maximal ideal of $R$ lying over $M$ then $T \cap S = M$. Now it can be shown that $R/T$ is a finitely generated $S/M$-module. Hence $R/T$ is Artinian and thus it has finitely many maximal ideals. From here I have two questions:  1) Why from this follows that there can be only finitely many $T$ lying over $M$? I suspect they are using some correspondence which I don't see.  2) Why exactly at most $k$ ?  Can you please explain?
36400
I'm making some exercises to prepare for my ring theory exam:     Prove that $I= \{a+bi \in ℤ[i] : a≡b \pmod{2}\}$ is an maximal ideal   of $ℤ[i]$.   I know that $a+2l=b$ with $l\in ℤ$ (or should I say $l \in ℤ[i]$)? Therefore I can write $a+(a+2l)i=a+ai+2li$. So I don't have $0+i$ in my ideal. For even real parts I have even complex parts, and for odd real parts I have odd complex parts. But I don't see what I can conclude now.
81108
Given a Noetherian ring $R$ and a proper ideal $I$ of it.  Is it true that $$\bigcap_{n\ge 1} I^n=0$$ as $n$ varies over all natural numbers? If not, is it true if $I$ is a maximal ideal? If not, is it true if $I$ is the maximal ideal of a local ring $R$? If not, is it true under additional assumptions on $R$ (like $R$ is regular)?
67143
I find difficulty to understand the proof of this theorem :  Theorem : Let $Y\subseteq\mathbb A^n$ be an affine variety. Let $Ρ\in Y$ be a point. Then $Y$ is nonsingular at $Ρ$ if and only if the local ring $\mathcal O_P$ is a regular local ring.  Let $P$ be the point $(a_1,\dots,a_n)$ in $\mathbb A^n,$ and let $\frak {a}$$=(x_1-a_1,\dots,x_n-a_n)$  be the corresponding maximal ideal in $A = k[x_1,\dots,x_n].$ We define a  $\theta : A\longrightarrow k^n$ by $$\theta(f)=\left(\frac{\partial f}{\partial x_1},\dots,\frac{\partial f}{\partial x_n}\right).$$ So $\theta$ induce an isomorphism $\theta':\frak{a}/\frak{a}^2\longrightarrow$$k.$   Now let $\frak b$ be the ideal of $Y$ in $A,$ and let $f_1,\dots,f_n,$ be a set of generators of $\frak b.$ Then the rank of the Jacobian matrix $\left(\frac{\partial f_i}{\partial x_j}\right)_{i,j}$ is just the dimension of $\theta(\frak b)$ as a subspace of $k^n.$ Using the isomorphism $\theta'$ this is the same as the  dimension of the subspace $\frak a^2+b/a^2$ of $\frak a/a^2,$ if $\frak m$ is the maximal ideal of $\mathcal{O}_Р,$ we have $$\frak m/m^2\cong a/(b+a^2).$$   I don't understand how I can use $\theta'$ to get $\dim_k \theta(\frak b)$$=\dim_k(\frak a^2+b/a^2)$ and why  $$\frak m/m^2\cong a/(b+a^2).$$   Thank you for any help.
108333
$\def\Z{{\mathbb{Z}}\,} \def\Spec{{\rm Spec}\,}$ Suppose $R$ a ring and consider $\Spec(\prod_{i \in \mathbb{Z}} R)$. Now for the finite case, I know that holds $\Spec(R \times R) = \Spec(R) \coprod \Spec(R)$. My intutition says that this does not extend to the infinite case. Maybe $\Spec(\oplus_{i \in \Z} R) = \coprod_{i \in \Z}\Spec(R)$ holds, but I am not sure. Can anybody give a proof or counter example for both the infinite direct sum and direct product?
8709
How can I prove this:     Let $A$ be a local regular ring with maximal ideal $\mathfrak m$ and $x \in \mathfrak m-\mathfrak m^2$. Then $A/(x)$ is a regular ring. Prove also that if $x\in\mathfrak m^2$, $x\ne 0$, the result does not hold anymore.   I don't know how to begin! Thanks to everyone who helps!  But if $d=\infty$?
23592
I am learning abstract algebra by myself using Dummit &amp; Foote, and sometimes by working very hard I find very complicated non-proofs of things and I have no idea if they are the best or the worst ideas for those kind of things out there (obviously if they're not working they're probably wrong, but maybe not!). So I feel like talking about it and see what the audience thinks about it. Hence I am asking for a full solution only if you don't feel like giving a hint. If a hint has been given please don't try to come up with the answer and wait for me to accept.  So here's the exercise (page 259, Dummit &amp; Foote's Abstract Algebra) :  Let $R = C[0,1]$ with addition and multiplication as the ring operations. For each $c \in [0,1]$ define $M_c = \{ f \in R \, | \, f(c) = 0 \}$.   It can be easily seen that $M_c$ is a maximal ideal of $R$ because $f(x) = (x-c) \in M_c$ and if $g \notin M_c$ then $g(c) \neq 0$, thus $f^2 &gt; 0$ everywhere except at $c$ and $g^2(c)&gt; 0$, hence $f^2 + g^2$ is a strictly positive continuous function, and since any ideal $I$ such that $M_c \subsetneq I \triangleleft R$ is such that $(M_c,g) \subseteq I$, $f^2 + g^2 \in (M_c,g)$, thus $f^2 + g^2$ is a unit in $(M_c,g)$ and that means that for every $g \notin M_c$, $R = (M_c, g) = I$.   Using this trick with $f^2 + g^2$ I wanted to show the following :  Prove that if $M$ is any maximal ideal of $R$ then there is a real number $c \in [0,1]$ such that $M = M_c$.  I tried considering $K(g) = \{ c \in [0,1] \, | \, g(c) = 0 \}$ and then defining decreasing sequences $K(g) \supsetneq K(g_1) \supsetneq K(g_2) \supseteq \dots$ (using the $f^2 + g^2$ trick) and then looking at the intersection of all those $K$'s but that got me to the weird need of using Zorn's Lemma and got me depressed after multiple tries. Also tried to use the fact that $R/M_c$ is a field but even if I had shown that it was isomorphic to $\mathbb R$ it wasn't clear what I could do in that direction so I gave up pretty fast on that one.   Any hints are welcome!
119971
Prove that $\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}$ implies $\gcd(m,n)=1$.  This is the converse of the Chinese remainder theorem in abstract algebra. Any help would be appreciated.  Thanks!
65205
Is it true that any prime ideal of a finite direct product ring $R=\prod_{i=1}^nR_i$ is of the form $P=\prod_{i=1}^nI_i$, where $I_j$ is a prime ideal of $R_j$ for some $j$ and $I_i=R_i$, for $i\neq j$?   Any ideal of the form above is prime in $R$. Indeed, if $X=\prod_{i=1}^nA_i$ and $Y=\prod_{i=1}^nB_i$ are ideals of $R$ (where $A_i$'s and $B_i$'s are ideals of $R_i$) such that $XY\subseteq P$, then either $A_j$ or $B_j$ is a subset of a  prime ideal $I_j$, for some $j$. Hence either $X$ or $Y$ is a subset of $P$.  Any help/suggestion would be appreciated!
12382
Let $A$ be a commutative ring. Suppose $P \subset A$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors.  This can be proved using localization, when $A$ is noetherian: $A_P$ is local artinian, so every element of $PA_P$ is nilpotent.  Hence every element of $P$ is a zero-divisor. (As Matt E has observed, when $A$ is nonnoetherian, one can still use a similar argument: $PA_P$ is the only prime in $A_P$, hence is the radical of $A_P$ by elementary commutative algebra.)  Can this be proved without using localization?
24262
Prove that $(x^2 + x+2)$ ideal is a maximal ideal in $F_3[x]$ ring.  I know the theorem about factor rings and fields     B ideal from A ring is maximal iff A/B factor-ring is a field   How can I use the theorem here?
9858
Well, the problem I'm trying to solve is this:      Let $A$ be the ring of all continuous functions from $\mathbb{R} \to \mathbb{R}$. Show that $M = \{f \in A: f(0)=0\}$ is a maximal ideal of $A$.   I tried to show that if $J$ is an ideal of A that properly contains M, so $J = A$: we know that there is a function $g \in J$ with $g(0) \neq 0$. So, we have that $$J \supseteq \langle g, M\rangle=\{ag+f;a \in A, f\in M\}$$   I tried to show that 1 $\in$ $\langle g,M\rangle$. However I'd have to find some function $h(x)\in A$ such that $h(x)(ag+f) = 1$, but this function $h$ has to be different from 0 for all x $\in \mathbb{R}$, so I don't know how to find this function.  My second idea to solve this problem was prove that the quocient $A/M$ is a field. We have that $$A/M = \{h(x) + M; h(x) \in A\},$$ but again I have the problem to show that all elements of $A/M$ are units.  If someone could help, I would be really grateful (:
123855
I am studying Artin-Wedderburn structural decomposition theorem for semisimple rings. I understand that it says that any semisimple ring, $R$ is isomorphic (as rings) to $M_{n_1}(D_1) \times M_{n_2}(D_2)\times\cdots\times M_{n_k}(D_k)$ for some $n_i$ and division rings $D_i$. Is this decomposition unique, i.e., suppose $R$ is also isomorphic to $M_{n'_1}(D'_1) \times M_{n'_2}(D'_2)\times\cdots\times M_{n'_{k'}}(D'_{k'})$, then is $k=k'$ and $n_i$ equal to $n'_i$ and $D_i \approx D'_i$ up to some permutation? If so, how?
30362
Denote by $M_{n \times n}(k)$ the ring of $n$ by $n$ matrices with coefficients in the field $k$.  Then why does this ring not contain any two-sided ideal?    Thanks for any clarification, and this is an exercise from the notes of Commutative Algebra by Pete L Clark, of which I thought as simple but I cannot figure it out now.