I'm trying to show that the kernel of the homomorphism $$\varphi: \mathbb{K}[X,Y]\rightarrow \mathbb{K}[t^2,t^3],$$ $$X\mapsto t^2,$$ $$Y\mapsto t^3,$$ is the ideal $I=\langle Y^2-X^3\rangle$ in $\mathbb{K}[X,Y]$, where $\mathbb{K}$ is a field. I can show that $I$ is contained in $\ker \varphi$, but not the other way around. For a given polynomial $p(X,Y)$ such that $\varphi(p(X,Y))=0$, I wrote it as $$ p(X,Y)= \sum_{i=1}^n a_iX^i + \sum_{j=1}^mb_jY^j + \sum_{r,s&gt;0}c_{rs}X^rY^S +d, $$ but that didn't help me as much as I hoped. At least I can show that $d=0$, but that was it. The change of variables made by $\varphi$ suggested some integer relationships between $i$, $j$ and $r,s$ but I could not conclude anything. This was a part of an example seen in a Commutative Algebra online course under the section of Integral Extensions with a spoiler from what owuld be seen in an Algebraic Geometry course (I can provide link for the Youtube video, if needed, though I believe it's not necessary)
53261
How to prove that the ideal $I=(x^3-y^2)$ in $k[x,y]$ is prime?    I have constructed a map from $k[x,y]$ to $k[t]$, which maps $x$ to $t^2$, and $y$ to $t^3$. Then, I want to show that the kernel is exactly $I$. But the difficulty is that how to prove the kernel is contained in $I$.  Could you help me? Thanks a lot!
24914
My professor proved that the kernel is a normal subgroup under a homomorphism f by saying. Let $h\in H$  by applying f to it  $f(ghg^{-1})=f(g)f(h)f(g)^{-1}=e $  My question is how is this a proof that the kernel is normal. What is the reason we apply f to it ? By applying f dont we just prove that $f(ghg^{-1})$ is normal ?
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$$(\mathbb{Z}/(n))[x] \simeq \mathbb{Z}[x] / (n)$$  I've shown this by showing that the map that sends $\overline{1} \mapsto [1+(n)]$ (where the bar denotes the congruence class mod $n$) and $x \mapsto [x+(n)]$ is a homomorphism that is injective and bijective, which is a bit cumbersome. Is there an easier way to show this? I can't tell if this is trivial or not.
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I am trying to figure out the normalisation of $k[x,y]/(y^2-x^2(x-1))$, for an algebraically closed field $k$.    I can show that it is not normal and I have the information that the normalisation is $k[t]$, where $t=\frac{y}{x}$. I can't figure out how to prove that this is the normalisation i.e. that the fraction fields of $k[x,y]/(y^2-x^2(x-1))$ and $k[t]$ are the same, that $k[t]$ is integrally closed and that $k[t]$ is an integral extension of $k[x,y]/(y^2-x^2(x-1))$.   Also, how does one get to the idea of $t=\frac{y}{x}$? Algebraically if possible, since this is from a commutative algebra class, not algebraic geometry.
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For each $m \ge 1$, how do I construct a homeomorphism $(\mathbb{CP}^1)^m/S_m \overset{\sim}{\to} \mathbb{CP}^m$?  My thoughts so far: I probably want to identify $\mathbb{C}^{m+1}$ with the vector space $\mathbb{C}^m[z, w]$ of degree $m$ homogeneous polynomials in two variables and use that a polynomial in one variable is determined by its roots up to a nonzero constant factor somehow... but from here, I'm kinda of stuck. Could someone help?  EDIT: Progress so far. After reading Eric Wofsey's response, I have the following. Since a projective variety is determined by its homogeneous coordinate ring, it suffices to determine the homogeneous coordinate ring of $(\mathbb{CP}^1)^m/S_m$.  For $(\mathbb{CP}^1)^m$, the coordinate ring is the graded ring $\mathcal{S}$ whose $d$th degree part is the $m$-fold tensor product (over $\mathbb{C}$) of $\mathbb{C}[s, t]_{d}$. That is, $S_d$ consists of polynomials in $s^1, t^1, s^2, t^2, \dots, s^m, t^m$ of multidegree ($d, d, \dots, d$). And $S_m$ acts in the obvious way, i.e. if $\sigma(i) = j$, then $\sigma$ sends $s^i$ to $s^j$ and $t^i$ to $t^j$. The homogeneous coordinate ring of $(\mathbb{CP}^1)^m/S_m$ is just the ring of invariants of the $S_m$ action on $\mathcal{S}$. But I still have two questions.   Which polynomials are invariant under this action? I still don't see how to conclude that Eric Wofsey's map is a homeomorphism. How do I see that Eric Wofsey's map is a homeomorphism?
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I understand that uni-variate polynomial rings with coefficients in a field only have principal ideals. For example, $\mathbb{C}[x]$. But how can I tell if an ideal of integer polynomial ring is principal, please? For example, a textbook claims that "the kernel of the map $\mathbb{Z[x]} \rightarrow \mathbb{Z[i]}$ sending $x \mapsto i$ is the principal ideal of $\mathbb{Z}[x]$ generated by $f=x^2+1$" without any justification. How to show this is true, please?
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Let $A$, $B$ be two reduced $k$-algebras. Then if an element of the form $$\sum a_{i}\otimes b_{j}$$ is nilpotent, we can compose it with any $k$-homomorphism $f$ from $A$ to $k$ to get a homomorphism from $A\otimes B$ to $B$. This map's image is $$\sum f(a_{i})\otimes b_{j}$$must be nilpotent as well. So if we assume $b_{j}$ is linearly independent in the first place, we may reason that $f(a_{i})=0$. This showed that $a_{i}$ are in the Jacobson radical of $A$. However, we only know $\mathrm{Nil}(A)=0$, and it is not clear to me how to pass from the Jacobson radical to the nilradical without using some commutative algebra machinery.   Here is what I think that might work. If $A$ is of finite type, then by Nullstellensatz we can assert the above is true. But if $A$ is not of finite type, then I can write $A=k[x_{1},...,x_{n}][X]$, where $n$ is the transcendence degree of $k(A)$ and $k[X]$ is finite over $k$. Then I should be able to pass from $k[X]$ to $A$ by inductive argument on the transcendental degree of $k(A)$, since prime ideals are maximal in a UFD (I am working with $k[x]$ by introducing $[x]$).   The proof I have for the case of finite type is not satisfying. Given $I\subset k[X]$, using Nullstellensatz one can show that $$I_{X}V_{X}[I]=\sqrt{I}$$ where $V_{X}[I]$ is the set of all points in $X$ on which $I$ vanishes, and $I_{X}(V_{X}[I])$ be the ideal in $k[X]$ which vanishes on $V_{X}[I]$. Granting this, let $I=(a_{i})$, then we want to show $I=0$, which is $V_{X}[I]=X$. This follows since if $U\subset W$, then $V_{X}[U]\supset V_{X}[W]$. Since $V_{X}[I]\supset V_{X}[m_{x}]$ for all maximal ideal corresponding to points, it must be $X$. This proof is unnecessarily winding by using the ideal-variety correspondence. Ideally I should be able to construct a pure algebraic proof.   The thing not clear to me is whether the above strategy really works. Namely I am not sure if I can really induct on the transcendental degree of $k(A)$. To make matters simple assume we showed the statement holds for $A$ with transdental degree $&lt;n$. For $A$ with transdental degree $n$ we can thus write it as $k[A'][x]$. Then any maximal ideal $m'$ of $k[A'][x]$ must correspond to a maximal ideal $m=m'\cap k[A']$ in $k[A']$. But it seems not possible to show that any prime ideal is maximal. Indeed considering the simple example of $k[x,y]$, where $(x)$ is a prime ideal but not maximal since $k[x,y]/(x)=k[y]$ is not a field. The problem is in the inducting process the base ring is not neccessarily a field, and so this strategy cannot work.   I am thinking maybe I can show $m'=m[f(x)]$ with $f(x)$ some irreducible polynomial in $k[A'][x]$. Because we know $k[A'][x]$ is noetherian by Hilbert's basis theorem, so in particular $m'$ must have a finite generating set. And we can reduce the number of irreducible polynomial to 1 since $m'$ is a maximal ideal. But I still do not know how to use this to finish the proof.
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Let $T$ be a linear transformation of an $n$-dimensional vector space $V$ over a field $k$. It's pretty easy to define the minimum polynomial of $T$ and make sure its degree is between $1$ and $n^2$, inclusive.     Observe $ I = \{ p(x) \in k[x] : p(T) =0\}$ is an ideal in $k[x]$. Indeed, $I$ is the kernel of the evaluation homomorphism $\mathrm{eval}_T: k[x] \to \mathrm{End}(V)$.  Notice also that:         $\mathrm{eval}_T$ is unital homomorphism, so $I$ is a proper ideal.    The $n^2 + 1$ transformations $I, T,T^2,T^3,\ldots, T^{n^2}$ must be linearly dependent, since $\mathrm{dim}(\mathrm{End}(V)) = n^2$, so there exist scalars $a_0,\ldots,a_{n^2}$, not all zero, such that $a_0I + a_1 T + \ldots + a_{n^2}T^{n^2} = 0$, whence the nonzero polynomial $p(x) = a_0 + a_1 x + \ldots + a_{n^2}x^{n^2}$ belongs to $I$.           Since $k[x]$ is a p.i.d., we may define the minimum polynomial $m(x)$ of $T$ to be the monic generator of the ideal $I$. By the preceding two observations, we have $1 \leq \mathrm{deg}(m(x)) \leq n^2$.    Now, of course, we know that the degree of $m(x)$ actually satisfies $1 \leq \mathrm{deg}(m(x)) \leq n$. One way to see this is to use the Cayley-Hamilton theorem which shows that the characteristic polynomial $c(x)=\det(xI - T)$, whose degree is $n$, annihilates $T$, whence $m(x)$ divides $c(x)$.     Question: Is there another way to see that $T$ is annihilated by a polynomial of degree $\leq n$ which does not require use of the characteristic polynomial?
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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!
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Let $\phi: R \to S$ be a surjective homomorphism. Prove that $\frac{R/ \ker \phi}{(\ker \phi + J)/ \ker \phi} \cong \frac{S}{\phi(J)}$ for an ideal $J$ of $R.$  Obviously, $S \cong R/ \ker \phi$(first isomorphism theorem) and $\phi(J)$ is an ideal of $S$, since $\phi$ is surjective( $\forall s \in \phi(J), s = \phi(j)$ for some $j \in J, \forall s' \in S, s' = \phi(r)$ for some $r \in R$, so $ss' = \phi(j)\phi(r) = \phi(jr) \in \phi(J)$, and $ss' = \phi(r)\phi(j) = \phi(rj) \in \phi(J)$ ).  However, I'm having trouble with going on. If it's important, I mean a unitary ring by a ring.
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If $\phi$ is a homomorphism from the group $G$ with identity $e$ to the group $G'$ with identity $e'$, then $\phi$ is injective if and only if $\ker\phi=$?  I think the $Ker\phi = \{x\in G:\phi(x)=1\}$  because you are looking at an identity then it is all reals $0$ does not work a because if $x=0, e'x=0$  but then i realize it is injective how does that affect the kernel?
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How many ring homomorphisms there is between $\mathbb{Z}[x,y]/(x^3+y^2-1)$ and $\mathbb{Z_7}$? Here $\mathbb{Z_7}$ denote ring of integers mod 7.  I don't know how to approach this problem,so far I've only worked with one variable ring so I can't tell the properties of $\mathbb{Z}[x,y]/(x^3+y^2-1)$. Thanks in advance
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Let $G = (\mathbb{C} - \{0\}, \cdot)$, a subgroup $U = \{x+yi \mid x^2 + y^2 = 1\}$. Use the Fundamental Theorem to show that $G/(\mathbb{R}_{&gt; 0},\cdot)$ is isomorphic to $U$ where $\mathbb{R}_{&gt; 0}$ denotes the positive reals.  I am struggling to find the correct mapping from $G$ to $U$ that results in the the kernel being the nonnegative reals under multiplication. I'm hoping to get help with this mapping then I can show it is a homomorphism and onto and can use the Fundamental Theorem.
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I am trying to prove that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic. I was thinking of arguing the following:  Suppose there exists an isomorphism $\varphi: \mathbb{Z}[x]\rightarrow\mathbb{Q}[x]$. Because isomorphisms are by definition surjective, there exist $x, y \in\mathbb{Z}[x]$ such that $\varphi(x) = c \in \mathbb{Q}[x]$ and $\varphi(y) = d \in \mathbb{Q}[x]$ for any $c, d\in\mathbb{Q}[x]$. Because $\varphi$ is an isomorphism we must have $\varphi(x+y) = \varphi(x) + \varphi(y)$ for all $x, y \in \mathbb{Z}[x]$. Namely, because polynomial addition is defined componentwise, we must have that the constant term of $\varphi(a + b) = c_{0} + d_{0}$ (where $c_{0}, d_{0}$ are the constant terms of $c$ and $d$ respectively. I would then argue that because $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic as additive groups, no such isomorphism $\varphi$ exists. Is this a valid proof?  I've seen proofs that argue that because $\varphi(1) = 1$ for any homomorphism we have $1 = \varphi(2(1/2)) = 2(\varphi(1/2))$ so $\varphi(1/2)$ must be contained in $\mathbb{Z}[x]^{\times}$. Then because $\mathbb{Z}[x]^{\times} = \mathbb{Z}^{\times} = \{\pm1\}$ we have $2 \times \pm1 \neq1$, a contradiction. Is this any different than arguing that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ each have a different number of units?
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I've been given a homework problem that requires me to show that the rings $\mathbb{C}[x,y]/(y - x^2)$ and $\mathbb{C}[x,y]/(xy-1)$ are not isomorphic.    This is my attempt at a solution:  For $\mathbb{C}[x,y]/(y - x^2)$, we can parametrize in the following way: $x = t$ and $y = t^2$. Then this ring is isomorphic to $\mathbb{C}[t]$.  For $\mathbb{C}[x,y]/(xy-1)$, we can parametrize $x = t$ and $ y = 1/t$. Then this is isomorphic to $\mathbb{C}[t, 1/t]$.   But $\mathbb{C}[t, 1/t]$ is not isomorphic to $\mathbb{C}[t]$.  Am I on the right track? If not, any helpful hints?
62204
I came across the following problem Let $k$ be a field and $R=k[t^2,t^3]/(t^4)$. The ring $k[t^2,t^3]$ is defined as the set $\{f(t^2,t^3)|f \in k[x,y]\}$ with the obvious ring structure.  prove that the ring R is Artinian find ideals $A_0, \dots , A_n \unlhd R$ such that $$R=A_0 \supseteq \cdots \supseteq A_n$$ with each $A_i/A_{i+1}$ be a simple $R$ module.  for the first part, I know that in a short exact sequence of R-modules, the middle R-module is Artinian iff the two others are. I thing that $R \simeq k[x,y,z]/(x-z^2, y-z^3, z^4)$ but I don't now how is this helpful. for the second part I don't have a single clew how to proceed. Any hints, useful remarks or theorems that may help are appreciated.
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I'm trying an old problem here:   Suppose $n\mid m$, and I have a natural ring homomorphism $\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $\varphi(j)=j\mod{n}$. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups $(\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times$?  My question in particular is why is it surjective? I take $k\in(\mathbb{Z}/n\mathbb{Z})^\times$. Then there exist integers $s,t$ such that $sn+tk=1$. Since $n\mid m$, I can also write $m=na$. So multiplying through I get $sm+tka=a$. I'm lost here. How can I find a unit in $(\mathbb{Z}/m\mathbb{Z})^\times$ which maps to $k$ to show surjectivity?