Let $Y_1,Y_2$ are two (Zariski) closed subsets of $\mathbb{A}_k^n$, for any field $k$. Then how do I show that $\mathcal{I}(Y_1 \cap Y_2)= \text{rad}(\mathcal{I}(Y_1)+\mathcal{I}(Y_2))$.  Clearly $\text{rad}(\mathcal{I}(Y_1)+\mathcal{I}(Y_2)) \subseteq \mathcal{I}(Y_1 \cap Y_2)$. How can I prove the converse ? I need some help. Thanks.
123861
I want to prove that     $I(X_1 \cap X_2) = \sqrt{I(X_1)+I(X_2)}$   for algebraic sets $X_1=Z(G_1)$ and $X_2=Z(G_2)$, with $G_1,G_2 \subseteq \mathbb{K}[X_1,\ldots,X_n]$.  Remark: Unfortunately I named the indeterminates $X_1,X_2,\ldots$ but obviously the first ones $X_1,X_2$ are algebraic sets and not indeterminates.  I've already proved $\supseteq$ but I'm struggling with proving $\subseteq$.  The problem is equivalent to prove that, if $g \in I(X_1 \cap X_2)$, then $g^n \in I(X_1)+I(X_2)$ for some $n \in \mathbb{N}$, but the $+$ is bothering me... some hint can definitely put myself in the right way.  Also, I'd like to:     Interpret geometrically what it means to have $I(X_1 \cap X_2) \neq I(X_1)+I(X_2)$,   thing that I don't see at all.
26346
Let $K$ be a Galois extension of $F$, and let $a\in K$. Let $n=[K:F]$, $r=[F(a):F]$, and $H=\mathrm{Gal}(K/F(a))$. Let $z_1, z_2,\ldots,z_r$ be left coset representatives of $H$ in $G$. Show that    $$\min(F,a)=\prod_{i=1}^r\left( x - z_i(a)\right).$$   In this product $\min(F,a)$ is of degree $r$. That is true from fundamental theorem.  And if one of $z_i$ is the identity, then $a$ satisfies the polynomial.  My question is:  What is the guarantee that $z_i(a)$ belongs to $F$ for all $i$?  What if I choose a representative which is not the identity of $\mathrm{Gal}(K/F)$?
70478
I started learning topology recently, so I'm very new to this.    The question is as follows:     Given the set $A = \{ (x,y)\;;\;x, y\in\mathbb{Q} \}=\mathbb{Q}^2$, show that the complement set $A^c=\mathbb{R}^2\setminus A$ is polygonally connected.   As an attempt on this question, I plan to show that the set $A^c$ is connected, thus using a theorem, I can say $A^c$ is polygonally connected.  But here is the part that I got confused: By definition, the set $A^c$ is connected iff it cannot be written as the union of $2$ nonempty separated sets.    In this case, I believe $A^c$ is the set of all $(x,y)$ where $x,y$ belong to $\mathbb R-  \mathbb Q$.  But how can I express the fact that $A^c$ is not the union of $2$ nonempty separated sets ?   Based on my knowledge, I know both sets $\mathbb Q$  and $\mathbb R-\mathbb Q$ are dense, but how is this helpful to solving this problem?  I think a reason is because for any collection of $(x,y)$ where $x,y $are in $\mathbb Q$, I can always find a point in $\mathbb R-  \mathbb Q$ which is a boundary point of that certain collection.  But is my thought ok ?  Would someone please help me on this?  I'm kinda lost >_&lt; Thanks ^^
80201
I have $2$ similar questions :    Is  the following set $K$ a field:  $$K=\underset{i=even}{\bigcup} \mathbb{F}_{2^{i}}$$ where $\mathbb{F}_{2^{i}}$  values are all finite fields?   Is so, what is the characteristic of this field?   Is $K \cup \mathbb{F}_{2^{3}}$ a field?    My answer is: yes both $1$ and $2$ are infinite fields by subfield criterion of finite fields, but I am not sure!  And what about the characteristic? Is it $2,$ which is the base prime, or is it $0,$ since it is infinite?
85750
Are there two closed sets $A,B\subset\mathbb{R}^2$ with the following properties?   $A\cap B=\emptyset$ $\forall \epsilon&gt;0$ there exist $a \in A$ and $b\in B$ such that $\|a-b\| &lt; \epsilon$
117564
This is an exercise from Miles Reid, Undergraduate Algebraic Geometry. The proof has two parts, one of which I can do and one of which I can't. I could have some misunderstandings about notations here as well so anything people can point out will be helpful.  a) Let $k$ be an infinite field and $f\in k[X_1,...,X_n]$ be nonconstant. Prove  $$V(f)=\{P\in \mathbb{A}^n_k\mid f(P)=0\}\neq \mathbb{A}^n_k.$$  $A^n_k$ here is the point set $k^n$.   My proof (which follows from a hint of his) is write $f=\sum a_i(X_1,...,X_{n-1})X_n^i$. Then $V(f)$ consists of two kinds of points   $P=(p_1,...,p_{n-1},p_n=0)$ $P=(p_1,...,p_{n-1},p_n\neq 0 )$ with $\sum a_i(p_1,...,p_{n-1})p_{n}^i=0$   In the first case we have $V(f)\neq \mathbb{A}^n_k$. Assume we have $P$ like in the second case and now write $f=\sum b_i(X_1,...,\hat{X}_{n-1},X_n)X_{n-1}^i$. Here again, either $P=(p_1,...,p_{n-1}=0,p_n\neq 0)$ and $V(f)\neq \mathbb{A}^n_k$ or $p_{n-1}\neq 0$, and by induction we should get to $V(f)\neq \mathbb{A}^n_k$.  b) Now let $k$ be algebraically closed. Let $a_m(X_1,...,X_{n-1})X_n^m$ be the leading term of $f$. Show that when $a_m\neq 0$, there is a finite set of points of $V(f)$ corresponding to every value $(X_1,...,X_{n-1})$. Thus, $V(f)$ infinite for $n \geq 2$.  So I have kinda no idea what to do. A given value of $(X_1,...,X_{n-1})$ corresponds to points in $V(f)$ $P=(p_1,...,p_{n-1},p_n)$ where $f(P)=0$, but I don't know anything about the value of $p_n$; if it vanishes then clearly there is a correspondence $(X_1,...,X_n)\to (p_1,...,p_{n-1},0)$ but I guess I need to show there are a finite number of $p_n$ where this is true and I don't know how.
114681
$K$ and $L$ are number fields and $K \subset L$ . Now, if $I$ and $J$ are two ideals in number ring of $K$ and  $IS \mid JS$ then we have to show that $I \mid J$.     Here, $S$ is number ring of $L$. The book suggests to factor $I$ and $J$ into primes of number ring of $K$ and then consider what happens in $S$.  This question has been asked here before:     once but it doesn't use the hint the book gives and instead part $(b)$ of the same problem to prove this result.  Can anybody explain what the author means by 'considering what happens in $S$' or give hints but no full solution please
17589
I am trying to prove the following statement.     If a field $K$ is not algebraically closed, then any $K$-variety $V\subset\mathbb{A}^n$ can be written as the zero set of a single polynomial in $K[X_1,X_2,...,X_n].$     Suppose $V=V(f_1,f_2,...,f_n)$, and I would want to find a polynomial $g\in K[X_1,X_2,...,X_n]$ such that $V=V(g)$.  I know that if I can find a polynomial $\phi\in K[X_1,X_2,...,X_n]$ whose only zero is $(0,0,...,0)$, then it is done, since we could take $g=\phi(f_1,f_2,...,f_n)$.  For $n=1$, then I would just choose $\phi=X$, but I do not even know how to proceed to the case $n=2$. Maybe we could choose an irreducible polynomial of degree $&gt;1$.  Any help will be appreciated.
107715
My question is  Let $A: \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation, how can one prove that $\{Ay|y\geq 0\}$ is a closed set?  Here $y\geq 0$ means each component of $y$ $\geq 0$. I can simplify the question in two ways  Since the image of $A$ is a closed linear subspace of $\mathbb{R}^m$, by replacing $\mathbb{R}^m$ with this linear subspace, we may assume A is surjective.  After a change of basis of the codomain and a reordering of the basis of the domain, we may assume $A$ is of the form $[I|B]$   Any help or hints are appreciated, thank you.  Ideas: Let $v_i=Ae_i$, then my original statement is equivalent to saying the set $\{\sum a_iv_i |a_i\geq 0\}$ is a closed set. So I should now prove the following:  For any $v_1,...,v_k \in \mathbb{R}^n$, the set $S=\{\sum a_iv_i |a_i\geq 0\}$ is a closed subset of $\mathbb{R}^n$.   I suspect that the set $S$ is an intersection of half spaces, from this the closedness will be obvious. Equivalently, I suspect that for each $x\notin S$, there exists a half space $H$ (cut by a codimension 1 linear subspace)s.t. $S\subset H$. Is this statement true? I guess this should be true after drawing some examples.  New ideas: I am proving that the set $\{\sum a_iv_i|a_i\geq 0 \}$ is closed, and this is easy if all the $v_i$'s are indeed independent, so I would want to prove the following:  Let $S=\{$linearly independent subsets of $\{v_i\}\}$, then $\{\sum a_iv_i|a_i\geq 0\}=\cup_{T\subset S}\{$non negative linear combinations of $T\}$  And this union is closed since it is a finite union of closed set. Equivalently, let $x=\sum a_iv_i,a_i\geq 0$, I want to write $x=\sum b_ie_i$, where $b_i\geq 0$ and $\{e_i\}\subset \{v_i\}$ is linearly independent. My proof is as follows: I can replace $\{v_i\}$ with a smaller subset of itself s.t. $x$ cannot be expressed by non negative linear combinations of any strictly smaller subset of it. Now I claim that this set is linearly independent. Suppose not, then we can write $c_1v_1+c_2v_2+\cdots + c_nv_n=0$, with some $c_i$'s are positive(since we can multiply the whole expression by $-1$) and I assume that $c_i&gt;0$ for $i=1,2,\dots ,k$ and $c_i \leq 0$ for $i&gt;k$ Now I assume $b=a_1/c_1=\min _{i=1,2,\dots ,k}{a_i/c_i}$, Then I have $$x=\sum a_iv_i-b(\sum c_iv_i)=                      (a_2-bc_2)v_2     +(a_3-bc_3)+\cdots + (a_n-bc_n)v_n$$ Which is a non negative linear combinations of fewer vectors, contradicts to my assumption and hence the minimal set $\{v_i\}$ is linearly independent.
77166
If $R_1$ and $R_2$ are both subrings of $R$ , how to prove that $R_1 \cap  R_2$ is also a subring of $R$.  here is my attempt  (1)Since $R_1$ is a subring of $R$ then it must contain zero (identity with respect to addition) same thing for $R_2$ and so $R_1 \cap R_2$ must have at least zero so it is nonempty.  (2)Now how to prove that if $x,y \in R_1 \cap R_2$ then $x-y \in R_1 \cap R_2$ and $xy \in R_1 \cap R_2$ how can we guarantee that?
53041
$(X, \mathcal{T}_X), (Y,\mathcal{T}_Y)$ be topological space, and let   $A, B$ be subset of $X,Y$ respectively.      Now, $(X \times Y, \mathcal{T}_{X \times Y})$ be product topology. $A  \times B \subset X \times Y$. Consider subspace topology   $\mathcal{T}_1$ on $A \times B$      Also, $(A, \mathcal{T}_A), (B, \mathcal{T}_B)$ where $\mathcal{T}_A$   and $\mathcal{T}_B$ are subspace topologies on $A, B$   respectively,consider product topology $\mathcal{T}_2$ on $A \times B$      Prove $\mathcal{T}_1 = \mathcal{T}_2$   My idea is to show that the bases of $\mathcal{T}_1$ and $\mathcal{T}_2$ are same. where $\mathcal{T}_1$ basis is {$(A \times B) \cap (U \times V)  | (U \times V) \in \mathcal{T}_{X \times Y}$}  and $\mathcal{T}_2$ basis is {$U_0 \times V_0 | U_0 \in \mathcal{T}_A,  V_0 \in \mathcal{T}_B$}. I know how to show these two bases are equal. However, I don't if this is sufficient to prove two topologies are equal.
100257
Let $A, B$ be sets. Show that $\mathcal P(A \cap B) = \mathcal P(A) \cap \mathcal P(B)$.   I understand what this question is asking. The power set of an intersection equals the intersection of two power sets. I just have no idea how to prove it.
106818
Full Question: Show that if $\Delta(\Gamma)$ is the deductive closure of a consistent $\Gamma$ then $\Delta(\Gamma)$ is the intersection of all maximal consistent sets of formulas $\Lambda$ with $\Gamma \subseteq \Lambda$.  I'm really stuck with this question, I honestly don't know how to even start. Can someone please provide some assistance?
75065
Let $k$ be a field and let $A \subset B$ be two finitely generated $k$-algebras. Prove that the contraction of any maximal ideal of $B$ is a maximal ideal of $A$.   thank you very much again!
101189
In Hartshorne section 1.1 he gives a problem (ex 1.11) which says that,   Let $Y \subset \mathbb A^3$ be the curve given parametrically by $x=t^3, y=t^4, z=t^5$. Show that $I(Y)\subset k[x,y,z]=A$ is a prime ideal of height 2 which cannot be generated by two elements.  Firstly, my question is that how can I check whether $Y$ is Zariski closed or not. Is it always the case that, a subset given by $x=f(t), y=g(t), z=h(t); f,g,h \in k[t]$ is always Zariski closed ?  My attempt was to prove that $A/I(Y) \cong k[t^3,t^4,t^5]$ to show $I(Y)$ is prime. And then I wanted to show the map $ \mathbb A^1 \rightarrow \mathbb A^3$ sending $t \rightarrow (t^3,t^4,t^5)$ is a topological embedding. I am stuck in proving that the map is continuous.   After proving this I could have said $Y$ (having proved that it is Zariski closed) is a variety of dimension 1, hence $I(Y)$ has height 2. I don’t know how to determine the least number of generators. Any help from basic algebraic results would be helpful.  I am a beginner in algebraic geometry so I don’t know many things, please keep it as basic as possible.
32100
I am trying to show that $k[x,y,z]/(xz-y^2)\not\cong k[x,y]$.    The latter is a UFD, so I am trying to show the former is not. Clearly $x$ is not prime, since $x\mid xz$ which implies $x\mid y^2$, but $x\mid y$.      So if I can show that $x$ is irreducible then $k[x,y,z]/(xz-y^2)$ is a not a UFD, because irreducible implies prime in UFDs.    Attempt: Assume $x$ is reduced into $ab$ in $k[x,y,z]/(xz-y^2)$, then we know there exists $c\in k[x,y,z]$ such that $x=ab+c(xz-y^2)$ in $k[x,y,z]$, or more usefully as $x-ab=(xz-y^2)c$. Someone suggested that I write $a,b$ such they are degree at most 1 in $y$, that is, replace every $y^n$ using $xz$. This means the LHS is degree at most 2 in $y$ and the RHS is degree at least two in $y$ in  $x-ab=(xz-y^2)c$. I am not sure where to go from here just by knowing that $c$ has no $y$ terms though. Ultimately I want to show that this forces $x=ab$ in $k[x,y,z]$ which, since $x$ is irreducible there, shows that either $a,b$ is a unit, which means $x$ is irreducible in $k[x,y,z]/(xz-y^2)$.
104753
If there is a closed subset $B$ of $\mathbb{R}^d$ and $z\in \mathbb{R}^d$ then show there is a point $b_0\in B$ so that we have $\delta(z,B) = \|z-b_0\|.$  I'm not sure how to begin this exercise and what theorems are available to help me solve it