Let $P=K[x_1,\dots,x_n]$ be a polynomial ring over a field $K$ and $I = (f)$ be a principal ideal in $P$ generated by $f \in P - \{0 \}$. Moreover let $L \subset \{x_1, \dots, x_n \}$ and $\hat{P} = K\{x_i : x_i \notin L \}$. Show $I \cap \hat{P} = (0) \iff f \notin \hat{P}$. (Exercise 4, Section 3.4 from Kreuzer and Robbiano, Computational Commutative Algebra, vol. 1.)   So I know that in one direction, if $f \not\in \hat{P}$, then $f$ contains a variable in $L$, but then $\hat{P}$ cannot contain any variable that is not in $L$. So the intersection must be $0$. (this need some justification...?) I am not sure about the other way. If someone could show me both, that would be good.
123329
I have a polynomial ring $R = K[x_1,\dots,x_n]$ and $N \subset \{x_1,\dots,x_n\}$. Define $R_N = K[x_i:x_i\notin N]$. If $f\not= 0$, show $(f) \cap R_N = (0) \iff f \notin R_N$.  So my entire argument hinges that $f \in R_{N^c}$ and clearly $f\in (f)$, but $f \not= 0$, so the intersection must be $(0)$ since $0 \in R_N$. I m not sure how to do the other direction. They look similar.
119724
Let $k$ be a field of characteristic $\neq2$ and consider $R=k[x,y]/(y^2-x^3+x)$. Then  (a) Show that $R$ is normal.  (b) Let $P=(x,y)$ be a prime ideal of $R$. Show that $P^2$ is a principal ideal.  (c) Show that $P$ is not a principal ideal.   I came across this in some commutative algebra reading recently. Here are my thoughts so far (really only for part (c)). Assume that $P$ is generated by $\alpha=f(x)+g(x)y$. Then if $\sigma$ is the nontrivial element in the Galois group, $\sigma(\alpha)$ will generate $\sigma(P)=P$, so the norm of $\alpha$ will generate $P^2$. I thought to compare this supposed generator with the one found in part (b) and then arrive at a contradiction.     Any help with this problem would be appreciated!
24506
Consider the polynomial ring $F\left[x\right]$ over a field $F$. Let $d$ and $n$ be two nonnegative integers. Prove：$x^d-1 \mid x^n-1$ iff $d \mid n$.  my tries:  necessity, Let $n=d t+r$, $0\le r&lt;d$ since $x^d-1 \mid x^n-1$, so, $x^n-1=\left(x^d-1\right)\left(x^{\text{dt}+r-d}+\dots+1\right)$... so,,, to prove $r=0$? I don't know, and I can't go on. How to do it.
53797
I want to show that if $K$ is a field then $K[x]$ is principal  Here is what I did:  I took 2 polynomial p(x) and q(x). Since we're on a field we can do a euclidean division of polynomials:  $q(x) = p(x)a(x) + r(x)$ with $deg(r) &lt; deg(p)$  Now if $r= 0$ we obviously have $p \in (q)$   But I don't know how to show that it is principal if $r \neq 0$  Also I would like to show that the assertion is false if we replace "$K$ is a field" by "$K$ is a principal domain and an integral domain". I can't figure out how to do that
83953
Let $p$ be a prime number. Let $M$ denote the ring of $2 \times 2$ matrices over the field $F$ of $p$ elements. For $A \in M$, let $C(A)$ denote the set of those matrices $B \in M$ such that $AB = BA$. What are the possible values of the dimension of $C(A)$ over $F$ for $A \in M$?  What I know is this: Given a square $n \times n$ matrix $A$ over a field $F$, it is always true that $A$ commutes with any $B=p(A)$, where $p(x)$ is a polynomial with coefficients in $F$. Now, because of the theorem of Cayley-Hamilton, the matrix $A$ is root of its  characteristic polynomial (which has degree $n$), so that means that every power of $A$ larger than $n-1$ can be written as a linear combination of powers up to $n-1$. For, our case, it means that: $C(A)=B=p_0I+p_1A$, where $p_0,p_1 \in F$.  So, that means $ 2\le \dim(B)\le 4$ (cases where $p_0=0$ and $p_1=0$). Is this correct? But what about $p$ prime - what it has to do with anything?
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.
83774
Just got a question about ideals of $F[x]$ being principal with $F$ being a field.  The proof basically starts assuming $g(x) \in N \subset F[x]$ where $N$ is an ideal and $g(x)$ is a minimal degree element. The rest is just deducing the what the division algorithm says.  But my question is that shouldn't we let $g(x)$ be arbitrary? Why are we choosing the minimal element?
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!
15896
Here is the theorem and proof in my textbook.     Thm.  Let $P$ be an ideal of a commutative ring $R$ with identity $1$.  Then $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.   Proof.   Suppose that $P$ is a prime ideal of a commutative ring $R$ with $1$.  Then $P$≠$R$ implies $1+P≠0+P$. Hence $R/P$ is a commutative ring $R$ with identity.  Assume that $(a+P)(b+P)=0+P$.  Then $ab+P=0+P$ and $ab∈P$.  By the definition of a prime ideal P we get $a∈P$ or $b∈P$.  That is, $a+P=0+P$ or $b+P=0+P$.  Thus $R/P$ is an integral domain.   Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$ and $R/P$ is a commutative ring $R$ which has no zero divisors.  Hence $P≠R$.  Assume $ab∈P$.  Then $ab+P=0+P$ and $(a+P)(b+P)=0+P$.  Since $R/P$ is an integral domain, we get $a+P=0+P$ or $b+P=0+P$.  So $a∈P$ or $b∈P$.  Thus $P$ is a prime ideal.    I am having problems understanding proof with "Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$.".  I know that $R/P$ has an identity. (cause it is ID)  But, why doesn't "$P$" have an identity?  Isn't there a case $P=R$?  Please help me understanding. I think $P≠R$ should be with condition of Thm..  Thank you in advance.
98738
I am trying to solve the following exercise.      Assume that $m$ and $n$ are co-prime positive integers, and that $k$ is a field. Prove that the ideal generated by $x^{m} - {y^n}$ is a prime ideal in $k[x,y]$.   I tried the following approach : Consider the ring homomorphism ${\varphi}:k[x,y]\rightarrow{k[t]}$ defined by $\varphi(x) = t^n$, and $\varphi(y) = t^m$, and extend $\varphi$ to $k[x,y]$ to ensure that it is a ring homomorphism. Then the ideal $(x^m - y^n)$ is clearly contained in the kernel of $\varphi$. It remains to prove the reverse set inclusion. i.e. to prove that this ideal is exactly equal to the kernel of $\varphi$. I don't know how to go about this!
20206
Let $A$ be a commutative ring with 1, set $V(\frak a)=$ the set of prime ideals of $A$ that contains $\mathfrak a$, and write $\operatorname{Supp}(M)$ for the support of the $A$-module $M$. We always have $\operatorname{Supp}(M)\subseteq V(\operatorname{Ann} M)$; if $M$ is finitely generated then the other inclusion also holds. Assume now $M$ is finitely generated. It is ok to show that if $M_{\frak p}=0$ then $\operatorname{Ann} M\cap A\setminus\frak p\ne\varnothing$, I was just wondering if the following procedure for showing this inclusion is flawed:  Suppose $\frak p$ contains $\operatorname{Ann} M$. Since $M$ is finitely generated we have the equality $(\operatorname{Ann} M)_{\frak p}=\operatorname{ Ann} (M_p)$, hence the annihilator of $M_{\frak p}$ is contained in maximal ideal of $A_{\frak p}$ and is therefore different from (1), so $M_{\frak p}\ne0$.
28416
Alternately, let $M$ be an $n \times n$ matrix with entries in a commutative ring $R$. If $M$ has trivial kernel, is it true that $\det(M) \neq 0$?     deals with the case that $R$ is a polynomial ring over a field. There it was observed that there is a straightforward proof when $R$ is an integral domain by passing to the fraction field.  In the general case I have neither a proof nor a counterexample. Here are three general observations about properties that a counterexample $M$ (trivial kernel but zero determinant) must satisfy. First, recall that the  $\text{adj}(M)$ of a matrix $M$ is a matrix whose entries are integer polynomials in those of $M$ and which satisfies  $$M \text{adj}(M) = \det(M).$$  If $\det(M) = 0$ and $\text{adj}(M) \neq 0$, then some column of $\text{adj}(M)$ lies in the kernel of $M$. Thus:     If $M$ is a counterexample, then $\text{adj}(M) = 0$.   When $n = 2$, we have $\text{adj}(M) = 0 \Rightarrow M = 0$, so this settles the $2 \times 2$ case.   Second observation: recall that by Cayley-Hamilton $p(M) = 0$ where $p$ is the characteristic polynomial of $M$. Write this as $$M^k q(M) = 0$$  where $q$ has nonzero constant term. If $q(M) \neq 0$, then there exists some $v \in R^n$ such that $w = q(M) v \neq 0$, hence $M^k w = 0$ and one of the vectors $w, Mw, M^2 w,\dots, M^{k-1} w$ necessarily lies in the kernel of $M$. Thus if $M$ is a counterexample we must have $q(M) = 0$ where $q$ has nonzero constant term.  Now for every prime ideal $P$ of $R$, consider the induced action of $M$ on $F^n$, where $F = \overline{ \text{Frac}(R/P) }$. Then $q(\lambda) = 0$ for every eigenvalue $\lambda$ of $M$. Since $\det(M) = 0$, one of these eigenvalues over $F$ is $0$, hence it follows that $q(0) \in P$. Since this is true for all prime ideals, $q(0)$ lies in the intersection of all the prime ideals of $R$, hence     If $M$ is a counterexample and $q$ is defined as above, then $q(0)$ is nilpotent.    This settles the question for reduced rings. Now, $\text{det}(M) = 0$ implies that the constant term of $p$ is equal to zero, and $\text{adj}(M) = 0$ implies that the linear term of $p$ is equal to zero. It follows that if $M$ is a counterexample, then $M^2 \mid p(M)$. When $n = 3$, this implies that $$q(M) = M - \lambda$$  where $\lambda$ is nilpotent, so $M$ is nilpotent and thus must have nontrivial kernel. So this settles the $3 \times 3$ case.   Third observation: if $M$ is a counterexample, then it is a counterexample over the subring of $R$ generated by the entries of $M$, so     We may assume WLOG that $R$ is finitely-generated over $\mathbb{Z}$.
91928
I am trying to mimic the interesting proof from  to give a generating set of the ideal correspond to the $d$-uple embedding: We fix the notation $\Delta = \big\{ (\nu_0, \dots, \nu_n) \in {\mathbb{N}}^{n+1}_0 ~:~ \nu_0 + \dots + \nu_n = d \big\}$. The polynomial map is $\theta : k[y_\nu : \nu \in \Delta] \rightarrow k[x_0, \dotsc, x_n]$ given by $$ \theta ( y_\nu ) := x^{\nu_0}_0 \cdots x^{\nu_n}_n $$ and we know that ${\mathrm{ker}}~\theta$ is the ideal of the $d$-uple embedding, i.e., $\rho_d({\mathbb{P}}^n) = Z({\mathrm{ker}}~\theta)$ (cf. Exercise I.2.12 Hartshorne). Now let $W \leq k[y_\nu : \nu \in \Delta]$ denote the ideal $$ W := \Big\langle y_{\tau_1} y_{\tau_2} - y_{\tau^{\prime}_1} y_{\tau^{\prime}_2} ~:~ \tau_1, \tau_2, \tau^{\prime}_1, \tau^{\prime}_2 \in \Delta, \tau_1 + \tau_2 = \tau^{\prime}_1 + \tau^{\prime}_2 \Big\rangle $$ Clearly, $W \subseteq {\mathrm{ker}}~\theta$. Using the lemma in the above link, we need to produce a $k$-subspace $T \subseteq k[y_\nu : \nu \in \Delta]$ with the conditions $$ T + W = k[y_\nu : \nu \in \Delta], \hspace{.2in} \theta \lvert_{T} = {\mathrm{injective}} $$ After a few thoughts it seems reasonable to think of the following subspace as a candidate for $T$: First define the equivalence relation on $\Delta \times \Delta$ given by $(\tau_1, \tau_2) \sim (\tau^{\prime}_1, \tau^{\prime}_2)$ if $\tau_1 + \tau_2 = \tau^{\prime}_1 + \tau^{\prime}_2$. Let $\Omega \subseteq \Delta \times \Delta$ be a subset with a unique representative from each class of $\sim$. Now let $T$ be the $k$-span of the monomials $y^{\alpha_1}_{\tau_1} \dotsc y^{\alpha_r}_{\tau_r}$ (with the distinct elements $\tau_1, \dotsc, \tau_r \in \Delta$ and $\alpha_1, \dotsc, \alpha_r \geq 1$) such that $(\tau_i, \tau_j) \in \Omega$ for each $i \neq j$. Clearly, $\theta$ is injective if restricted to $T$. I got lost in the step : Each monomial $M$ of $k[y_\nu : \nu \in \Delta]$ satisfy $M \equiv b$ mod $W$ for some $b \in T$. Did I do anything wrong?
55731
I've been trying to show that   $$ K[X,Y]/(Y-X^2)\cong K[X] $$  where $K$ is a field, $K[X]$ and $K[X,Y]$ are the obvious polynomial rings over the indeterminates $X$ and $Y$ and $(Y-X^2)$ is the ideal generated by the polynomial $Y-X^2$.  Though I'm sure there's a fairly easy way to find an explicit isomorphism between the two rings, the following argument jumped out at me:   If we substitute in a value for $X$ - $x$, say, then the ideal $(Y-x^2)$ is a maximal ideal of $K[Y]$.  So the quotient $K[Y]/(Y-x^2)$ is a field; in fact, the homomorphism $K[Y]\to K:P(Y)\mapsto P(x^2)$ and is clearly surjective, so the quotient is isomorphic to $K$.    I'd like to be able to deduce from this that $K[X,Y]/(Y-X^2)\cong F[X]$, but I can't see a nice way to do it.  I know that the 'substitution' maps $P(X,Y)\mapsto P(x,Y)$ are homomorphisms, but I can't see a nice way of pulling all these homomorphisms back to the polynomial ring in two variables.    Or maybe I'm completely wrong and there is no way to turn this into a valid argument.  Can anyone help me?
116563
Let $R$ be a finite commutative ring with identity. Let $a\in R$ is a root of $p(x)\in R[x]$ . Prove that $p(x)=(x-a)q(x)$ for some $q(x)\in R[x]$ of degree= (degree of $p)-1$.   If $R$ were a Euclidean domain then it is obvious. But $R$ is not necessarily a ED. So what I was trying to do is showing $p(x)\in (x-a)$; the ideal generated by $x-a$.   I tried defining a homomorphism $f:R[x]\rightarrow R$ by $f(x)=a$ and tried to show ker$f=(x-a)$. But when I was going to show it, I felt like showing the same thing that asks in the question. So no success. Can anybody please give me at least a hint? I feel like this is a standard question. So I searched through MSE for similar questions but could not find. If you find a duplicate please copy a link and then I will delete this question.
24054
Trying to solve a problem, I got stuck on the following question.     If $P\in\mathbb R[X]$ is an irreducible polynomial and $\operatorname {deg} P=2$, then is it true that $$\mathbb{R}[X]/(P)\cong\mathbb C?$$   I believe it must be true from the bits I've been hearing throughout my studies, but of which I haven't seen proofs. I've been unable to find anything on the internet other than . It tells me that there is something called Weierstrass' theorem which should be relevant, but can't find even the precise statement of the theorem, let alone the proof.  Here's what I have tried.   First of all, I can assume that $P$ is monic, because for any polynomial of degree $2$ there's a monic polynomial generating the same ideal in $\mathbb R[X].$ So I have  $$P(x)=x^2+bx+c$$  for some real numbers $b,c$ such that $b^2&lt;4c$.   I want to assume there is an isomorphism $\phi:\mathbb R[X]/(P)\to\mathbb C$ and work out what it should look like. I will denote by $[S]$ the coset of $S\in \mathbb R[X]$ in $\mathbb R[X]/(P).$ I'd like to find $S$ such that $\phi([S])=i.$  I have   $$ [x^2+bx+c-1]\mapsto -1, $$  So the coset that is mapped to $-1$ has to be a square root of $[x^2+bx+c-1]$ in $\mathbb R[X]/(P).$ Such a coset must be represented by a linear polynomial $kx+l$. Thus $$(kx+l)^2-(x^2+bx+c-1)\in (P),$$  that is $$ (kx+l)^2-(x^2+bx+c-1)\equiv a(x^2+bx+c) $$  for some real number $a$, and further  $$ (k^2-1-a)x^2+(2kl-b-ab)x+l^2-c-ac+1\equiv 0. $$  This gives   $$ \begin{cases} k^2=1+a\\ l^2=(1+a)c+1\\ 2kl=(1+a)b \end{cases} $$  The problem is that solving this gives formulas that are very difficult to handle. And I need to handle them if I want to prove that what I'm constructing is indeed an isomorphism. Is there a better way of doing it?
40621
Let $K$ be a field and let $R$ be a $K$-algebra with unity which is finite dimensional as a $K$-vector space. Prove that $R$ has only finitely many prime ideals all of which are maximal. (Hint: Chinese remainder theorem)   My attempt: We have $R \simeq K^n$, for some $n$, as vector spaces. I claim that there are finitely many ideals of $K^n$. Let $I$ be an ideal of $K^n$ and $(a_1, a_2,...,a_n) \in I$. For every entry $a_i$, we have two choices, either $a_i=0$ or not.  So we get $2^n$ many cases, and any case give a particular ideal of $K^n$ which are distinct from others.  Is my proof correct or am I missing some point? If it is true, how can we give a proof by using the Chinese remainder theorem? Thanks!