$R$ is a commutative ring. We need to prove that $R[x]$ has a subring isomorphic to $R$.   Let $S$ be that subring of $R[x]$ which has polynomials of even degree. Now I consider a mapping from $S$ to $R$ as : $f(x) \to f(0)$ where $f(x)$ is a polynomial of even degree and $f(0)$ is clearly in $R$.  The mapping defined as above "seems" one-one and onto. But I couldn't prove it.  I took the kernel of the mapping as $\ker \phi = ( f(x) \in S | \phi ( f(x)) = 0 )$ ,  Which only gives $a_0 = 0$ ,  $ f(x) = a_0 + a_1 x + a_2 x^{2} $ . . . ... $ a_n x^{n} $   We know if kernel only contains the " 0 " element then the mapping is one-one . But I'm unable to prove that here. Can anyone help me out here ?
122539
Say $R$ is a commutative ring. Does there exists a subring of $R[x]$ that is isomorphic to $R$?  My approach would be to define the subring of $R[x]$ that generates $R$. Any thoughts?
120588
In general, I struggle working with quotients of polynomial rings, so I was hoping someone might be able to help with the following exercise:  Show that $\mathbb{Q}[x]/(x^2-2x)$ is isomorphic to $\mathbb{Q}[x]/(x^2-1)$, but the latter is not isomorphic to $\mathbb{Q}[x]/(x^2)$.  Mostly I want to understand what these quotients all &quot;look like.&quot; We did few examples in class and few homework problems with such quotients, and the ones we do work with are much simpler (typically involving $\mathbb{Z}_2$ or $\mathbb{Z}_3$ or $\mathbb{R}$ instead of $\mathbb{Q}$). Any help is appreciated, thank you!
31515
Let $R$ be a commutative ring. Prove that $R[x]/(x)$ is isomorphic to $R$.  Should I use the first isomorphism theorem for rings?  Can someone give me a hint on how to approach this problem? Also how would I use the first isomorphism theorem to prove this? I know that the theorem says $R /\ker f$ is isomorphic to image of $f$.
32824
I'm having troubles with one of the problems in the book Introduction to Commutative Algebra by Atiyah and MacDonald. It's on page 11, and is the last part of the second question.  Given $R$ a commutative ring with unit. Let $R[x]$ be the ring of polynomials in an indeterminate $x$ with coefficients in $R$. We say that a polynomial $f = \sum\limits_{i=0}^n r_ix^i \in R[x]$ (with coefficients $r_0, r_1, \ldots, r_n$) is primitive if $\langle r_0,r_1,...,r_n\rangle = R$, i.e., the ideal generated by the coefficients of $f$ is $R$. Let $f$ and $g$ be two polynomials in $R[x]$. Prove that $fg$ is primitive iff $f$ and $g$ are both primitive.  The $\Rightarrow$ part is easy. Say, $f = \sum\limits_{i=0}^n r_ix^i$, $g = \sum\limits_{i=0}^m s_ix^i$; then $fg = \sum\limits_{i=0}^{m+n} c_ix^i$, where $c_k = \sum\limits_{i + j = k}r_is_j$. Since $fg$ is primitive, there exists a set of $\{\alpha_i\} \subset R$, such that $\sum\limits_{i=0}^{m+n} \alpha_ic_i = 1$, to prove $f$ is primitive, I just need to write all $c_i$'s in terms of $r_i$'s, and $s_j$'s, then group all $r_i$ accordingly, rearranging  it a little bit, and everything is perfectly done. And the proof of the primitivity of $g$ is basically the same. The $\Leftarrow$ part is just so difficult. Say $f = \sum\limits_{i=0}^n r_ix^i$, $g = \sum\limits_{i=0}^m s_ix^i$ are both primitive, then there exists $\{\alpha_i\}; \{\beta_i\} \subset R$, such that $\sum\limits_{i=0}^{n} \alpha_ir_i = 1$, and $\sum\limits_{i=0}^{m} \beta_is_i = 1$. At first, I thought of multiplying the two together, but it just didn't work. So, I'm stuck since I cannot see any way other than multiplying the two sums together. I hope you guys can give me a small push on this. Thanks very much in advance, And have a good day.
73961
Let $R$ be a commutative ring, $a_1, ..., a_n$ its elements and  $\phi: R[x_1, ..., x_n] \to R$  defined by  $ \phi(f(x_1, ..., x_n)) = f(a_1, ... ,a_n)$  a ring homomorphism.  Prove: $\ker \phi = (x_1-a_1, ..., x_n-a_n)$  It is obvious that $(x_1 - a_1, ..., x_n -a_n) \subseteq \ker \phi$. I'm not sure how to prove the converse.  At this point I don't know any division algorithms for multivariable polynomials, only for the ones in $R[x]$(and the book from where I taken the exercise doesn't assume the reader to know something beyond basics of rings and ideals, and the division algorithm for $R[x]$). Though I know this could be solved for $i = 1$ by dividing by $x-a$:   Let $f(x) \in \ker \phi$, divide by $x-a: f(x) = q(x)(x-a) + r, f(a) = r = 0$, so $f(x) = q(x)(x-a) \in (x-a)$.
30081
I am studying from Patrick Morandi's Field and Galois Theory, and I am stuck on Problem 6 from Section 1 (on page 13).     Verify the following universal mapping property for polynomial rings:         Let $A$ be a ring containing a field $F$. If $a_1,\dotsc,a_n \in A$, show that there is a unique ring homomorphism $\varphi \colon F[x_1,\dotsc,x_n] \to A$ with $\varphi(x_i) = a_i$ for each $i$.       Moreover, suppose that $B$ is a ring containing $F$, together with a function $f \colon \{ x_1,\dotsc,x_n \} \to B$, satisfying the following property: For any ring $A$ containing $F$ and elements $a_1,\dotsc,a_n \in A$, there is a unique ring homomorphism $\varphi \colon B \to A$ with $\varphi(f(x_i)) = a_i$. Show that $B$ is isomorphic to $F[x_1,\dotsc,x_n]$.      The  for the book says that we must assume that all the rings are commutative and the ring homomorphisms are $F$-homomorphisms (that is, they are ring homomorphisms that fix $F$).    I have proved the first part as follows: let $\mathbf{a} = (a_1,\dotsc,a_n)$. The evaluation map $\operatorname{ev}_\mathbf{a} \colon F[x_1,\dotsc,x_n] \to A$ given by $\operatorname{ev}_\mathbf{a}(f) = f(a_1,\dots,a_n)$ is an $F$-homomorphism with $\operatorname{ev}_\mathbf{a}(x_i) = a_i$ for each $i$. Moreover, if $\varphi$ is any $F$-homomorphism from $F[x_1,\dotsc,x_n]$ to $A$ with $\varphi(x_i) = a_i$, then we must necessarily have that $$ \begin{align} \varphi\left( \sum_{k_1,\dotsc,k_n} c_{k_1,\dotsc,k_n} x_1^{k_1} \dotsm x_n^{k_n} \right) &amp;= \sum_{k_1,\dotsc,k_n} c_{k_1,\dotsc,k_n} \varphi(x_1)^{k_1} \dotsm \varphi(x_n)^{k_n}\\ &amp;= \sum_{k_1,\dotsc,k_n} c_{k_1,\dotsc,k_n} a_1^{k_1} \dotsm a_n^{k_n}\\ &amp;= \operatorname{ev}_\mathbf{a}\left( \sum_{k_1,\dotsc,k_n} c_{k_1,\dotsc,k_n} x_1^{k_1} \dotsm x_n^{k_n} \right). \end{align} $$ Thus, the existence and uniqueness of $\varphi$ has been established.    I am stuck in proving the second part. From the previous result, we have a unique $F$-homomorphism $\Phi \colon F[x_1,\dotsc,x_n] \to B$ given by evaluation at $(f(x_1),\dotsc,f(x_n))$, and my intuition is that this is the required isomorphism. But I am not able to make much progress.  To show that $\Phi$ is injective, the idea I have in mind is as follows: given $0 \neq g \in \ker(\Phi)$, choose the ring $A$ and elements $a_1,\dotsc,a_n$ in such a way that $g(a_1,\dotsc,a_n) \neq 0$, thereby arriving at a contradiction. To show that $\Phi$ is surjective, my intuition is that the uniqueness of the map $\varphi$ should play a role, since we will have used only the existence in showing injectivity.  Am I on the right track? I haven't been able to convert these thoughts into a proof yet, and I would really appreciate some hints. Thanks in advance!
112818
$R$ is a domain with characteristic $p$ ($p$ is prime). There is a homomorphism $f : R \to R$, $f(a)=a^p$. $f$ is called the Frobenius endomorphism. And I have known this.   When $R$ which is mentioned above is also a field, it is said that $f$ is an isomorphism. I think I just need to ensure $f$ is bijective. But I don't know how to prove it's surjective.    Actually I have got your ideas. In some of your opinions, if it is finite, then it can be surjective. Then what if it is infinite?  In my question, $R$ is a field as well as a domain. Then I wonder whether the "domain" is helpful for the proof of surjection or not.
7149
This is a very simple question but I believe it's nontrivial.  I would like to know if the following is true:      If $R$ and $S$ are rings and $R[x]$ and $S[x]$ are isomorphic as rings, then $R$ and $S$ are isomorphic.    Thanks!  If there isn't a proof (or disproof) of the general result, I would be interested to know if there are particular cases when this claim is true.
42788
How can we show that if $R$ is an infinite commutative ring and $R/I$ is finite for every nonzero $I \unlhd R$, then $R$ is an integral domain?   I tried proceeding by contradiction: assume $a$,$b$ $\in R \backslash \{0\}$ and $ab=0$; then $R/(a)$ and $R/(b)$ must be finite, say $R/(a)=\{k_i + (a) : 1 \leq i \leq m\}$ and $R/(b)=\{l_j + (b) : 1 \leq j \leq n\}$. Does this mean $R$ must be finite? Or what about using the fact that $R/(a,b)$ finite?  Thanks for any help with this!
29348
I don't see how to solve the following problem:     Let $R$ be a commutative and unitary ring. If there exists a monic polynomial $f(x) \in R[x]$ so that $R[x]/(f(x))$ is a field, show that $R$ is a field. What about if $f(x)$ isn't monic?   I don't have a clue about how proving it. I see that if $R$ is a field and $f(x)$ a monic and irreducible polynomial in $R[X]$ then $\frac{R[x]}{(f(x))}$ is a field, but that is not the question!  Thank you in advance for your help
86973
I believe that for a subring to be isomorphic it needs to be a ring homomorphism but i'm not sure how to go about finding isomorphic rings, I'm not looking for a straight up answer I'm trying to understand how to go about the question.
72795
I have the following problem:  Let $R:=\mathbb{R}[X,Y]/(Y-X^2,Y+X)$ i) Show that $R\cong_\text{Ring} \mathbb{R}\times \mathbb{R}$ ii) Conclude that $|\operatorname{Spec}R|=2$. How do these two ideals look like?  So for the first one I got the hint to eliminate one variable at a time and see what happens. Doing that I got: $1.) \mathbb{R}[X]/(-X^2,X)\cong \mathbb{R}[X]/(-X^2)\times\mathbb{R}[X]/(X)\cong\mathbb{R}[X](-X^2)\times \mathbb{R}$ $2.) \mathbb{R}[Y]/(Y,Y)=\mathbb{R}[Y]/(Y)\cong\mathbb{R}$ using the chinese remainder theorem in $1.)$. But since $\mathbb{R}[X]/(-X^2,X)\times\mathbb{R}[Y]/(Y)\ncong\mathbb{R}[X,Y]/(Y-X^2,Y+X)$ and the fact that there is still $\mathbb{R}[X]/(-X^2)$ left in the first equation, I don't see how this leads me to the desired isomorphism. For the second one I don't see how an isomorphism can lead me to the number of ideals, because I'm pretty sure $A\cong B$ doesn't imply $\operatorname{Spec}A \cong \operatorname{Spec}B$ for rings $A,B$ (I've seen it in a book). Can someone help me with this problem? Thanks in advance.
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.
111066
The problem is stated as follows:  "Let $R$ be a Noetherian ring and $\theta$ be a ring homomorphism from $R$ to $R$. Show that if $\theta$ is surjective then it is also injective."  Regardless of the right solution, I don't understand why is the following wrong:     We have $\theta: R\to R$. By the isomorphism theorem $R/\ker\theta\cong\operatorname{Im}\theta$. Since $\operatorname{Im}\theta = R$, it follows $\ker\theta =0$, so it's injective.   Harsh criticism will be appreciated. Thanks.
95572
Let $A$ be a subring of ring $B$, with $B$ integral over $A$. If $x\in A$ is a unit in $B$, then it is a unit in $A$.   I know that $f(t) = t - x$ is in $A[t]$ with $f(x) = 0$, and that there exists some monic polynomial $g(t) = t^n + a_{n-1}t^{n-1}... + a_0$ also in $A[t]$ so that $g(x^{-1}) = 0$, and as I understand it, this proof comes down to showing that $x^{-1}$ must be an element of A. But I'm not sure what kind of manipulation of these polynomials will do that for me. I tried examining $x^n g(t)$ and saw that $0 = x^ng(x^{-1}) = 1 + xa_{n-1} + ... + x^na_0$, but I can't seem to get anything useful out of this. Any hints toward the right direction would be appreciated.
73093
Let $f:R\to S$ be a surjective homomorphism, where $R$ is a commutative ring and $S$ is a field. Prove that $\ker(f)$ is a maximal ideal.  I already know that $\ker(f)$ is an ideal of $R$. I tried to consider some ideal $J$ of $R$ such that $\ker(f) \subset J$. If we can show that for arbitrary $y\in J$, $f(y)=0$ then we are good. But I don't know how to show that. Specifically, how does being surjective come into play?  Another theorem I know is that if $f$ is a surjective homomorphism, then quotient ring $R/\ker(f)$ is isomorphic to $S$. Don't know if that's gonna help.
106259
Dummit and Foote, 7.4.33(a): Let $R$ be the ring of all continuous functions $[0,1] \to \mathbb{R}$ and let $M_c$ be the kernel of evaluation at $c \in [0,1]$, i.e. all $f$ such that $f(c) = 0$. Show that if $M$ is a maximal ideal in $R$ then $M = M_c$ for some $c$.    I am aware of proofs that use the compactness of $[0,1]$; however I have been told that a simpler proof exists using only the isomorphism theorems for rings along with basic facts about ideals. I have tried on my own to prove it this way but still have no success; is it possible?  (My attempts go something like: since $M$ is maximal, the quotient ring $R/M$ is a field. Then it would be nice to use the fact that $R/M$ has no zero divisors, but I can't quite make it work. Alternatively, consider $M \cap M_c$.)