Is there any way to show that for $a,b \in F$ , the ideal $\langle x-a , y-b\rangle$ is maximal in  $ F[x,y]$ , by showing that the quotient $F[x,y]/\langle x-a , y-b\rangle$ is a field ? Is the quotient  $F[x,y]/\langle x-a , y-b\rangle$ isomorphic with $F$ ? ( Here $F$ is a field )   I can show that the map $g : F[x,y] \to F$ given by $g( p(x,y))=p(a,b)$ is a surjective ring homomorphism with $\langle x-a , y-b\rangle \subseteq \ker g$ but I cannot show that the kernel is exactly that ; PLease help, Thanks in advance
42805
$K$ is field. $a_1$,$a_2$ elements of $K$. Show that $(x_1-a_1,x_2-a_2)$ is a maximal ideal of $K[x_1,x_2]$.   $K[x_1,x_2]$ is UFD so if $K[x_1,x_2]/(x_1-a_1,x_2-a_2)$ is field then $(x_1-a_1,x_2-a_2)$  is maximal ideal.  If I can show that $K[x_1,x_2]/(x_1-a_1,x_2-a_2)$ isomorphic to $K$, we can verify that $(x_1-a_1,x_2-a_2)$ maximal ideal of $K[x_1,x_2]$.  thanks for helps and comments.
2436
I have been having trouble with an exercise in my abstract algebra course. It is as follows:     Let $f: \mathbb{C}[x,y] \rightarrow \mathbb{C}[t]$ be a homomorphism that is the identity on $\mathbb{C}$ and sends $x$ to $x(t)$ and $y$ to $y(t)$ such that $x(t)$ and $y(t)$ are not both constant. Prove that the kernel of $f$ is a principal ideal.
25049
I studying Brezis' book and I have somes partial solutions of the exercise $1.17.$  Let be $E$ a normed space and $f\in E^*$ be a linear functional nonzero. Consider the set  $M=[f=0]$ given by  $$M=\{x\in E \ : \ \langle f,x\rangle = 0 \}.$$  Here, $\langle f,x\rangle$ denote $f(x)$ and note that $M$ is just $\ker f$.     Dertemine $M^\perp.$   By definition   $$M^\perp=\{g\in E^* \ : \ \langle g,x\rangle=0, \forall x\in M \}.$$  So, the first doubt is: $M^\perp$ is just  $$M^\perp=\{g\in E^* \ : \ \ker f \subset \ker g  \} ?$$     Show that $\forall x\in E$, we have $d(x,M)=\inf_{y\in M}{\|x-y\|}=\frac{|\langle f,x\rangle|}{\|f\|}.$   Since $y\in M$, $|\langle f,x\rangle|=|\langle f,x-y\rangle|\le \|f\|\|x-y\|$. Then  $$\frac{|\langle f,x\rangle|}{\|f\|}\le d(x,M).$$  How can I show the equality? I know that since $\{x\}$ is a compact set and $M$ is a closed subspace of $E$, there is $y_0\in M$, such that $d(x,M)=\inf_{y\in M}{\|x-y\|}=\|x-y_0\|,$ but I don't know if that it is a help.  I tried to show that $\forall \varepsilon &gt;0$, there is $z\in M$ such that  $$\|x-z\|&lt;\frac{|\langle f,x\rangle|}{\|f\|}+\varepsilon,$$  but I could not. The last doubt is:     Let be $E=\{u\in C([0,1],\Bbb{R}) \ : \ u(0)=0\}$ and $f\in E^*$ given for each $u\in E$ by   $$\langle f,u\rangle = \int_0^1 u(t)dt.$$ Show that if $u\in E-M$ then $d(u,M)$ is never achieved.   For that, I tried something by absurd method but I found nothing. Can someone give a little help? Thanks.
82517
I would like to know, why $ \mathfrak{p} A_{\mathfrak{p}} $ is the maximal ideal of the local ring $ A_{\mathfrak{p}} $, where $ \mathfrak{p} $ is a prime ideal of $ A $ and $ A_{\mathfrak{p}} $ is the localization of the ring $ A $ with respect to the multiplicative set $ S = A -\mathfrak{p} $ ? Thanks a lot.  N.B. : I have to tell you that I'm not very good at Algebra, so please, be more kind and generous in your explanation, and give me a lot of details about this subject please. Thank you.
100013
I have a problem when I am proving the following question:  Let $f:A \to A'$ be a surjective homomorphism of rings, and assume that $A$ is local, $A' \neq 0$, show that $A'$ is local.  I have found an answer in . However in my proof, suppose that $\mathfrak{m}$ is the unique maximal ideal in $A$, and I have proved that $f(\mathfrak{m})$ is an ideal and contains any other ideals. To finish this proof, I should prove $f(\mathfrak{m})\neq A'$. But I can't prove $f(1)\notin   f(\mathfrak{m})$.  For example, let  $g:\mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z}$ be the natural projection onto its first coordinate. We know $\mathbb{Z}\oplus 2\mathbb{Z}$ is a maximal ideal in $A$, but $g((1,1)) \in g(\mathbb{Z}\oplus 2\mathbb{Z})$.So it's not true for a general ring.    Is there any direct way to prove $f(1)\notin   f(\mathfrak{m})$ for a local ring?
36939
Let $K$ be a field and $A=K[x_1,x_2,x_3,...]$. Prove that the ideal $I:=\langle x_i: i \in \mathbb N\rangle$ is not finitely generated as $A$-module.   I have no idea what can I do here, I mean, suppose $I$ is finitely generated, then there is a subset $S \subset I$ with $S=\{x_{i_1},...,x_{i_n}\}$ such that $\langle\{x_{i_1},...,x_{i_n}\}\rangle=I$. How can I arrive to a contradiction? Any suggestions would be appreciated.
35987
Let $R$ be a PID, and $r\in R- \{0\}$. Prove that   $\langle r\rangle$  maximal $\iff r$ irreducible.   "$\Leftarrow$"Easy.  "$\Rightarrow$"If $J=\langle r \rangle$ then we will prove that $r$ is irreducible. If $r=ab$, we want to prove that $a\in U(R)$ or $b \in U(R)$.  If we take the ideal which is generated by $\langle a\rangle$ then (because $J$ is maximal)$$\langle a\rangle \subseteq\langle r \rangle \iff r\mid a \iff a=kr, k\in R \Longrightarrow r=krb\iff r(1-kb)=0_R \iff kb=1_R$$ so $b\in U(R)$. Same way if we work with $\langle b \rangle$.  Is this proof right?
15293
In John Fraleigh's book, A First Course In Abstract Algebra Exercises 15.7 and 15.11, one shows that  $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (1,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \  \mbox{ and  } \ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$  One does this with the first isomorphism theorem. With the same idea I proved for example that  $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,3)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \ \mbox{ and }\ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,4)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$ So I conjectured that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_{k}$ where $k=\mathrm{gdc}(m,n)$. For the previous four cases, the homomorphism $\phi: \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z} \times\mathbb{Z}_k$ given by $$ \phi(x,y)=\left(\frac{nx-my}{k}, \ x \ \ (\mathrm{mod} \ k) \right) $$ is surjective with kernel =$\langle (m,n)\rangle$. However, this is not the case when $(m,n)=(4,6)$. So, I cannot use what I did for the four cases to prove the general case.     What I want to know is if my conjecture is true. If so, how can I give a general homomorphism? If it is not true, how can I find $k$, such that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_k$?   Thanks in advance for any help/hint/comment!
69770
I'm asked to prove that the quotient ring $\mathbb{Z}[x]/\langle x, 2\rangle$ is isomorphic to $\mathbb{Z}_2$.  The first thing that came to my mind was to apply the Theorem of Homomorphism of Rings, but to do so I need to construct a homomorphism $\phi: \mathbb{Z} \to \mathbb{Z}_2$, such that $\mathrm{Ker}\phi = \langle x, 2\rangle$ ($\langle x, 2\rangle$  is an ideal of $\mathbb{Z}[x]$). But     What elements are in $\mathbb{Z}[x]/\langle x, 2\rangle$ in the first   place?   I know that $\langle x, 2\rangle$ of all the polynomials with even constant term, but I can't figure out how the elements of $\mathbb{Z}[x]/\langle x, 2\rangle$ look like.  Any help would be appreciated.
50159
This problem goes as follows:     Prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle$ is an integral domain and that its field of fractions is isomorphic to the ring of rational functions $\mathbb{Q}(t)$.    The first part is straightforward, $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$, and  hence a prime ideal, as $\mathbb[Q][x,y]$ is a euclidean and hence a UFD. The quotient by a prime ideal is an integral domain.   How do I do the second part? I guess, I have to set a isomorphism by making a substitution such that only one variable remains. But I don't see how $x^2+y^2-1$ is going to help me make a choice.
34972
I learnd that the maximal ideals in $\mathbb C[x,y]$ have the form $(x-z_1, y-z_2)$ by the Nullstellensatz. But if we set $I=(xy-1)$ then $\mathbb C[x,y]/I$ is isomorphic to $\mathbb C[x,1/x]$ which is in my opinion a field, thus $(xy-1)$ is maximal. Where did I make a mistake?
113362
I need some help to show that:  If  $| \langle u,v \rangle | = \|u\| \|v\|$ then $u=\lambda v$ for some scalar $\lambda$.  We have to consider this over an arbitrary field $\mathbb{F}$.  I appreciate your suggestions.  Thanks.
81501
In Dummit and Foote, Exercise 15.18 (p. 727 3rd edition) one is asked to prove that for a commutative ring with unity $R$, and $f$ a non-nilpotent element of $R$, that the localization $R_f$ is isomorphic to $R[x]/(xf - 1)$.  I have done this, using the strategy outlined in the problem, using the universal property of $R_f$ proved in Theorem 36. The problem I have is this:  If $R_f \cong R[x]/(xf - 1)$ and $\phi:R[x] \to R_f$ is a surjective homomorphism, it stands to reason that $\text{ker}(\phi)$ should be $(xf - 1)$. I cannot show this directly. The surjective homomorphism I have in mind is, for:  $p(x) = a_0 + a_1x + \cdots + a_mx^m$  $\phi(p) = \dfrac{a_0}{1} + \dfrac{a_1}{f} + \cdots + \dfrac{a_m}{f^m}$  It is clear to me that $\phi$ annihilates $(xf - 1)$, so that ideal is contained in the kernel. What I cannot show is the other containment, or equivalently:  If $\phi(g) = 0_{R_f}$ then $g(x) \in (xf - 1)$. The trouble I am running into is that $R$ might be a "bad" ring, containing a lot of zero-divisors, $f$ might even be a zero-divisor (being non-nilpotent is a considerably weaker property), so the usual polynomial tricks of the division algorithm or using the irreducibility of $xf - 1$ aren't much use, here.  I feel I am missing something obvious....what is it?
40062
Let $I = \langle 2\rangle$. Prove $I[x]$ is not a maximal ideal of $\mathbb Z[x]$ even though $I$ is a maximal ideal of $\mathbb Z$.   My professor mentioned that I should try adding something to it to prove that $I[x]$ is not the maximal ideal. I want to assume he meant to add $ \langle 2 \rangle$ to $I[x]$?
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.
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 (:
38567
The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not a principal ideal.   I don't know how to consider it. Any suggestions?