Let $R$ be the ring of all real valued  continuous functions on $[0,1]$ and $M$ be a maximal ideal of $R$. Then how to prove that $\exists c \in [0,1] $ such that $M:=\{f \in R:f(c)=0\}$ ?
123461
Let $R$ be the ring of all continuous real valued functions on the unit interval $[0,1]$ (with pointwise operations), and let $I$ be a proper ideal of $R$. Show that there exists $λ\in [0,1]$ such that   $$I\subseteq M_{λ}= \left\{f \in R\mid f(λ)=0\right\}.$$   Can't really get anywhere with this one. Appreciate the help.
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?
119369
Let $R$ be a commutative ring and let $I$ be a prime ideal of $R$. Prove that $S = R \setminus I$ is a multiplicative closed subset containing $1$.  Please can anyone help me out here?
50854
$$Rad(I)=\{a \in R | \exists n \in \mathbb{N} \text{ such that } a^n \in I\}$$   R is a commutative ring, I is an ideal.   To show that $Rad(I)$ is an ideal of $I$, we have to show that for $a,b \in Rad(I)$ it stands that $a-b \in Rad(I)$ and for $r\in R$ it stands that $a\cdot r\in Rad(I)$, right??   When $a,b \in Rad(I)$ we have that $a,b \in R$ and $a^n, b^m \in I$.   $a-b \in R$.   But how can we show that $(a-b)^x \in I, x \in \mathbb{N}$ ??
107696
Let $R$ be a finite commutative ring. Show that an ideal is maximal if and only if it is prime.  My attempt: Let $I$ be an ideal of $R$. Then we have $I$ is maximal $\Leftrightarrow$ $R/I$ is a finite field $\Leftrightarrow$ $R/I$ is a finite integral domain $\Leftrightarrow$ $I$ is a prime ideal.  Is my proof valid ?
49526
For part a) of this exercise we need to show that for a scheme $X$ and an open affine subset $U=$ Spec $B$ that $U \cap X_f = D(\bar{f})$ where $\bar{f}$ is the restriction of $f \in \mathcal O_X(X)$ to $\mathcal O_X(U)$ and $X_f= \{x \in X :$ the stalk of $f$ at $x$ $(f_x)$ is not contained in the maximal ideal $(m_x)$  of $\mathcal O_x \}$.  Now I believe the proof goes something along the lines of using the fact that this maximal ideal in the affine scheme is just $xB_x$ but I can't see how $f_x \not\in m_x$ implies $\bar{f}_x \not\in xB_x$. Like how can we go from the local ring $\mathcal O_x$ to $B_x$?
84555
Let $A, A'$ be commutative rings with $1 \neq 0$.  Let $h : A \to A'$ be such that $h(1) = 1$.  Then $f: {\rm Spec}(A') \to {\rm Spec}(A)$ defined by $f(\mathfrak{p}') = h^{-1}(\mathfrak{p}')$ is continous with respect to the spectral topologies.
119447
Let $K=F[x]/(p(x))$, where $F$ is a field, $p(x)\in F[x]$ is an irreducible polynomial.  What are all the maximal ideals in $K[y]$?  Attempt: I have an attempt, but not entirely sure it is correct.  Since $(p(x))$ is a maximal ideal in $F[x]$, this means $K$ is actually a field.  Thus $K[y]$ is a PID.  So maximal ideals in $K[y]$ are $(g(y))$ for irreducible polynomials $g(y)\in K[y]$?
59335
Suppose $R$ is a ring and every prime ideal of $R$ is also a maximal ideal of $R$. Then what can we say about the ring $R$?
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!
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 $d=\infty$?
88321
Let $R$ be a (not necessarily commutative) ring and suppose $R$ has a maximal left ideal $I$ consisting of nilpotent elements, that is for $x\in R$ there exists some $n\in\mathbb{Z}_{\geq 0}$ such that $x^{m}=1$. I want to show that in this case, $I$ is the unique maximal left ideal.  To attempt to prove this, assume $J$ is some other maximal left ideal of $R$. Since $x^{m}=0$, we have $x^{m}\in J$. This is where I get stuck.  In the commutative case we can note that $I$ is necessarily a prime ideal, so either $x\in J$ or $x^{m-1}\in J$. Iterating this process shows that $x\in J$, then $I=J$ by maximality.  In the non-commutative case I can't seem to proceed. This is probably really easy but I'd appreciate a hint.
112879
If $R$ is a ring and $I$ an ideal of this ring, is that true that $(R/I)[x]=R[x]/I$?   Note that can $I$ be identified as a subset of $R[x]$ and we need that in order to $R[x]/I$ makes sense.  I think it's true because every polynomial $(a_0+I)+(a_1+I)x,\ldots,(a_n+I)x^n$ of $(R/I)[x]$ can be written as a polynomial of $R[x]/I$ and vice versa since we have this equality: $$(a_0+I)+(a_1+I)x,\ldots,(a_n+I)x^n=(a_0+a_1x+\ldots a_nx^n) +I$$  Is my reasoning right?  Thanks
1790
Let $f$ be a real valued and continuous function over $[a,b]$ with $a &lt;b$ such that $f(a)=f(b)$ , My question here is :     Question (claim): How to show that : there  exists  $c \in [a, \frac{a+b}{2} ]$ such that : $f(c)=f(c+\frac{b-a}{2})$.   Note: I have tried intermediate values but I didn't succeed.
29100
Let $R$ be a commutative ring containing a field $k$ as sub-ring. Assume that $R$ is a finite dimensional $k$-vector space. Show that every prime ideal of $R$ is maximal.   My approach:  Let $I$ be an prime ideal of $R$, as $R$ is a finite dimensional vector space, assume $I= \text{span}(\beta_1, \beta_2...\beta_m$). Assume $I$ is not a maximal ideal and $I \subset J$, where $J$ is a proper ideal. Since $J$ is maximal, $R/J$ is a field.  Say, we choose $r_1$, $r_2\in$ $R$, such that $(r_1+J)(r_2+J)=1+J$, since $R/J$ is a field.  Now say $r_1\in I$, so, $r_1=\sum_{i=0}^m a_i\beta_i$, where all $a_i$'s are from field $k$.   As $r_i \in I$, $r_1r_2 \in I$, so $r_1r_2=\sum_{i=0}^m b_i\beta_i$. Where all $b_i$'s are from field $k$.  We have chosen $r_1, r_2$ in such a way $(r_1+J)(r_2+J)= r_1r_2+j_1=1+j_2$.  As $I \subset J$, $J$ has a basis of ($\beta_1, \beta_2...\beta_n$), where ofcourse $n&gt;m$.  Say $j_1 = \sum_{i=0}^n c_i\beta_i$ and $j_2=\sum_{i=0}^n d_i\beta_i$.  Now, $$r_1r_2+j_1=1+j_2\Rightarrow \sum_{i=0}^m b_i\beta_i +\sum_{i=0}^n c_i\beta_i=1+\sum_{i=0}^n d_i\beta_i  $$  $$ \Rightarrow \sum_{i=0}^m b_i\beta_i +\sum_{i=0}^n c_i\beta_i -\sum_{i=0}^n d_i\beta_i=1 $$ $$ \Rightarrow 1 \in span(\beta_1, \beta_2... \beta_n)$$ So the multiplicative unity $1 \in J$. Hence, J is not a proper ideal of R, rather $J=R$. Hence $I$ is maximal.  So here we need not to use the property that $I$ is a prime ideal. Therefore I think either I'm missing something, or this method is completely wrong. Please help.
28057
Let $R$ be a ring of the real functions with "+" and "x" and let $I_1:=\{f(x)\in R\mid f(1)=0\}$, $I_1$ is a maximal ideal?   I think yes, but don't know how to prove it.
84681
Let $A,B$ be commutative rings, and let $\phi: A \to B$ be a ring homomorphism where $B$ has finitely many elements. Prove that if $I \subset B$ is a maximal ideal then $\phi^{-1}(I)$ is also a maximal ideal in $A$  I've tried to construct a proof by contradiction but with no success, im not looking for answers but rather hints.