We know the dimension of a affine algebraic variety $X\subseteq$ $\mathbb{A}^{n}_k$ (k a field, and $X$ does not have to be irreducible) is the biggest integer r such that there exists a strictly increasing chain $Z_{0}\subset Z_{1}\subset ... \subset Z_{r}\subset X$ of affine algebraic varieties in X. On the other side,  $\overline{X}$ $\subseteq$ $\mathbb{P}^{n}_k$ is the smallest algebraic subset in $\mathbb{P}^{n}_k$ (k a field) such that $X\subseteq$ $\overline{X}$. I tried proving that the projective closure preserve the inclusions, i.e., if $Z\subseteq Y$, for $Z, Y\subseteq X$, then $\overline{Z}\subseteq \overline{Y}$, for $\overline{Z},\overline{Y}\subseteq\overline{X}$, but I don't know how to proceed or if this will help me with the initial problem.
122559
Let $V$ be an irreducible variety of finite type over a field $k$, $V_{0}\subseteq V$ open and nonempty (or at least dense). Why is \begin{equation*}\operatorname{dim}_{\operatorname{Krull}}(V_{0})=\operatorname{dim}_{\operatorname{Krull}}(V)\text{?}\end{equation*}  My professor uses this result (without mentioning a proof) as if it were somehow self-evident. I have tried to prove it, but I am at a loss. Who can help?  The general statement is NOT true for arbitrary irreducible topological spaces, not even for varieties of finite type over an arbitrary integral domain, see   for instance.
50384
If $K$ and $k$ are fields, $K\supset k$ is a field extension and $\alpha \in K$ is algebraic over $k$, then we denote by $k[\alpha]$ the set of elements of $K$ which can be obtained as polynomial expressions of $\alpha$.   $$k[\alpha] = \left\{P(\alpha): \quad P\in k[X]  \right\} $$  Also, we denote by $k(\alpha)$ the smallest subfield of $K$ containing both $k$ and $\{\alpha\}$. This is easily seen to be equal to the set of fractions of polynomial expressions in $\alpha$ (just thinking about the way the "smallest" subfield must be generated):  $$k(\alpha) = \left\{\dfrac{P(\alpha)}{Q(\alpha)}: \quad P,Q\in k[X] \text{  and } Q(\alpha)\neq 0 \right\} $$  If $M$ is the minimal polynomial of $\alpha$, it is easily shown that $k[\alpha]\simeq k[X]/\langle M\rangle$ is a field  and therefore a subfield of $k(\alpha)$ containing both $k$ and $\{\alpha\}$. Therefore $k(\alpha)=k[\alpha]$, because $k(\alpha)$ is the smallest.   It follows that all quotients $\dfrac{P(\alpha)}{Q(\alpha)}$ are equal to $H(\alpha)$ for some polynomial $H\in k[X]$.   What I want to see is a more direct proof of this fact. Given $P$ and $Q$, how do you produce this $H$?
46247
I saw it's been answered before but I could not understand the answers as it delved into material that I did not study yet.  My question is as follows:  Suppose $p \in F[x]$ is the minimal polynomial of some matrix $A \in M_n(F)$ and $F \subseteq K$  I need to show that $p$ is also the minimal polynomial of $A$ over the field $K$. In other words: Show that the minimal polynomial remains the same when we move to a bigger field.  What I thought of:  Obviously $p$ is irreducible in $F$, suppose it is reducible in $K$, which means there are $q,t \in K, q,t \notin F$ such that $p=qt$, now we just need to show that $q(A) \neq 0$ and $t(A) \neq 0$...But why is that so?  Edit: Another way of presenting this question is "Show that if $p$ is the minimal polynomial of $A$ over the field $K$ while the entries of$~A$ are contained in a subfield $F \subseteq K$, then all the coefficients of $p$ are elements of $F$".
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.
67143
I find difficulty to understand the proof of this theorem :  Theorem : Let $Y\subseteq\mathbb A^n$ be an affine variety. Let $Ρ\in Y$ be a point. Then $Y$ is nonsingular at $Ρ$ if and only if the local ring $\mathcal O_P$ is a regular local ring.  Let $P$ be the point $(a_1,\dots,a_n)$ in $\mathbb A^n,$ and let $\frak {a}$$=(x_1-a_1,\dots,x_n-a_n)$  be the corresponding maximal ideal in $A = k[x_1,\dots,x_n].$ We define a  $\theta : A\longrightarrow k^n$ by $$\theta(f)=\left(\frac{\partial f}{\partial x_1},\dots,\frac{\partial f}{\partial x_n}\right).$$ So $\theta$ induce an isomorphism $\theta':\frak{a}/\frak{a}^2\longrightarrow$$k.$   Now let $\frak b$ be the ideal of $Y$ in $A,$ and let $f_1,\dots,f_n,$ be a set of generators of $\frak b.$ Then the rank of the Jacobian matrix $\left(\frac{\partial f_i}{\partial x_j}\right)_{i,j}$ is just the dimension of $\theta(\frak b)$ as a subspace of $k^n.$ Using the isomorphism $\theta'$ this is the same as the  dimension of the subspace $\frak a^2+b/a^2$ of $\frak a/a^2,$ if $\frak m$ is the maximal ideal of $\mathcal{O}_Р,$ we have $$\frak m/m^2\cong a/(b+a^2).$$   I don't understand how I can use $\theta'$ to get $\dim_k \theta(\frak b)$$=\dim_k(\frak a^2+b/a^2)$ and why  $$\frak m/m^2\cong a/(b+a^2).$$   Thank you for any help.
101812
Question. Let $k$ be an algebraically closed field, an let $\mathbb{P}^n$ be projective $n$-space over $k$. Why is it true that every regular map $\mathbb{P}^n \to \mathbb{P}^m$ is constant, when $n &gt; m$?  I can't see any obvious obstructions: there are certainly homomorphisms of function fields (giving rise to the dominant rational maps), and we're not demanding the map be injective or anything. While it is clear that $(F_0 : \cdots : F_m)$ cannot define a regular map on its own unless $F_0, \ldots, F_m$ are all constants, I don't see why it should be impossible to extend $(F_0 : \cdots : F_m)$ by choosing some other $(G_0 : \cdots  G_m)$ which agrees with $(F_0 : \cdots : F_m)$ on the intersection of their domains. Is there something conceptual I'm missing?
32922
Let $X$ be topological space and $Y\subset X$. The goal is to show that $\dim Y\leq \dim X$.   Here we use this definition of dimension.  Let $Y_0\subsetneq Y_1\subsetneq \cdots \subsetneq Y_{n}$ be a chain of strictly increasing closed irreducible subsets of $X$ then $\dim X$ is the supremum over all such chains.   This is what I've got so far. I'm just stuck on one last bit. A hint would be nice.   Let $Y_0\subsetneq Y_1\subsetneq \cdots \subsetneq Y_n$ be a chain of irreducible closed subsets of $Y$. Since each $Y_i$ is irreducible, we just need to close them in $X$. There exists $F_i$ closed in $X$ such that $Y_i=F_i\cap Y$. Then if we closed each set in $X$, $\overline{F_0\cap Y}\subset\overline{F_1\cap Y}\subset \cdots \subset \overline{F_n\cap Y}$ is a chain of irreducible closed subsets of $X$. I just need to show that they are distinct. I really don't know where to proceed. A hint would be appreciated. Thanks.
2836
I have started working through the textbook Ideals, Varieties, and Algorithms by Cox, Little, and O'Shea and I am stuck on one part of an introductory question.  The question begins by getting one to show   (1) If $f(t)$ and $g(t)$ are polynomials of degree $\le n$ in $t$, then for $m$ large enough, the monomials  $$[f(t)]^e[g(t)]^f$$ with $e+f\le m$ form a linearly dependent set in $k[t]$.  The next part of the question asks one to deduce      (2) If $C\,:\,x=f(t),\,y=g(t)$ is any polynomial parametic curve in $k^2$, then $C$ is contained in $\textbf{V}(F)$ for some $F\in k[x,y]$.   I have demonstrated (1), but I am at a loss as to how to use that fact to prove (2). It doesn't seem like it is supposed to be a hard problem, but never the less I am having trouble wrapping my head around it. Additionally I assume that for (2) we are not interested in the trivial solution $F=0$.  [Added for completeness]  To do part (1), we recognize that if $\deg(f),\deg(g)\le n$, then $$\deg\left([f(t)]^e[g(t)]^f\right)\le ne+nf\le nm$$ Hence if we have more than $nm+1$ polynomials of the form $[f(t)]^e[g(t)]^f$ then we have a dependent set (since $d+1$ is the dimension of the vector space of polynomials of degree less than or equal to $d$). Now we simply count the number of polynomials in this form (this was an exercise before the problem) - it is $\binom{m+2}{2}=\frac{1}{2}(m+1)(m+2)$ from a standard application of . Ensuring $m$ large enough (specifically $m&gt;2n-3$) gives us the conclusion.
107787
Suppose $X$ is a scheme of finite type over $\mathbb Z$. I want to prove that:  (1) The residue fields of closed points of $X$ are finite;  (2) For a given $q=p^n$ with $p$ prime, there is only a finite number of closed points of $X$ whose residue field is $\mathbb F_q$.  For (1), I see that one has to use some form of Nullstellensatz. First, we suppose $X=\operatorname{Spec}(A)$ affine, and $A=\mathbb{Z}[X_1,...,X_n]/I$. We need to show that if $m$ a maximal ideal of $A$, then $A/m$ is finite. Note $p=\mathbb Z\cap m$. If $p$ is maximal, we get that $A/m$ is of finite type over $\mathbb F_p$ thus a finite extension by Nullstellensatz and we are done. But $p$ could be the $(0)$ ideal in which case I neither conclude nor get a contradiction.  For (2), we can again argue locally, so we need to show that $A$ contains only a finite number of maximal ideals $m$ with $A/m$ of fixed cardinality $q=p^n$. I get an impression that to count these ideals its the same as to count the number of automorphisms of $\mathbb{F}_{q}$ which fix $\mathbb F_p$, but the corrspondence is not bijective, so we only get an upper bound which is enough. Is this idea correct?
20951
I'm having trouble with the exercise in the title, even with part (a), which asks to prove that if $X$ is a closed subset of $\mathbb{P}^n_k$ of dimension at least 1 and $H$ is a non-empty hypersurface, then $X$ and $H$ meet.   Following the hint, I've done the following: By assumption, $X$ is given by $V(I)$ for some homogenous prime ideal $I\subset k[x_0,\dots x_n]=:A$ and $H$ by $V(f)$ for some homogenous polynomial $f\in A$. We wish to prove that there is a homogenous prime ideal $\mathfrak{p}$ (that's not the irrelevant one) containing both $I$ and $f$.  Consider the affine cone $\mathrm{Spec}\ A/I$ of $X$ in $\mathrm{Spec}\ A$. Being homogenous, both $I$ and $f$ are contained in the ideal $(x_0,\dots x_n)$. By Krull's principal ideal theorem, every minimal prime ideal containing $f\mod I$ has codimension one. $X$ having dimension at least one implies that there are homogenous prime prime ideals $\mathfrak{q}\supsetneq \mathfrak{r}\supset I$. This implies that $(x_o,\dots x_n)$ has codimension at least two. Thus there are prime ideals containing $f\mod I$ that are contained in $(x_o,\dots x_n)$. However, I'd need to show that there are homogenous such prime ideals, right? I'd rather get hints than full solutions.
40688
Is there a (proper) subfield $K$ of $\mathbb{R}$ such that $\mathbb{R}$ is an algebraic extension of $K$?  From this question, , it is clear that for such an $K$, $[\mathbb{R}:K]=\infty$.  This question seems to be non-trivial, and I suspect any existence result will be non-constructive. Any references will be appreciated.
35633
I've got to show that any irreducible conic in $\mathbb{P^2(\mathbb{K})}$ where $\mathbb{K}$ is algebraically closed is isomorphic as a quasiprojective variety to the projective line. I wrote the conic as equivalent to the zero locus of $ x^2 +y^2+z^2$ in $\mathbb{P^2(\mathbb{K})}$. I tried to use the standard affine charts to cover this simpler variety and parametrize it but i didn't manage to do that. What am i missing?
44262
My question is about  about solving Hartshorne's Exercise I-1.9 by using exercise I-1.8:  Hartshorne's Chapter 1, exercise 1.9: Irreducible components of $Z(\mathfrak a)$ have dimension $\geq n-r$ if $\mathfrak a$ is an ideal generated by $r$ elements. Hartshorne's Chapter 1, exercise 1.8: Let $Y$ be an affine variety of dimension $r$ in $\mathbf A^n$. Let $H$ be a hypersurface in $\mathbf A^n$, and assume that $Y \nsubseteq H$. Then every irreducible component of $Y \cap H$ has dimension $r-1$.  Suppose $\mathfrak{a}$ can be generated by $f_1, \dots, f_r$ and no one is in the ideal generated by the rest of them (note they are not necessarily irreducible. But it seems to me that I don't need to assue $H$ in Ex. 1.8 is the zero set of an irreducible polynomial. Let me know if I am wrong). We'll induct on $r$. I can prove that any irreducible component $Y$ of $Z(\mathfrak{a})$ is contained in an irreducible component $Y'$ of $Z(f_1, \dots, f_{r-1})$. But, to use 1.8, I have to show that $Y'$ is not contained in $Z(f_r)$ and that $Y$ is an irreducible component of $Y' \cap Z(f_r)$ Question 1: does the assumption that $f_r\notin (f_1,\dots, f_{r-1})$ imply that $Y'$ is not contained in $Z(f_r)$? If not, how do we remedy this? Question 2: how to prove that $Y$ is an irreducible component of $Y' \cap Z(f_r)$ (something might be missing here and I can't prove it at least)? It is irreducible but why it appears in the irreducible decomposition of $Y' \cap Z(f_r)$?
98625
Let $K$ be an algebraic extension field (not necessarily finite) of $\mathbb{Q}$. Let $\mathscr{O}_K$ be the integral closure of $\mathbb{Z}$ in $K$.  Then, is $\mathscr{O}_K$ Noetherian?  If $K$ is a finite extension of $\mathbb{Q}$, then $\mathscr{O}_K$ can be shown to be Noetherian. However, if we do not assume that finiteness condition, is it still true?  (It can be shown that $\mathscr{O}_K$ is an integrally closed domain of Krull-dimension $1$. Hence, if it is Noetherian, it is a Dedekind domain.)
36949
Consider square matrices over a field $K$. I don't think additional assumptions about $K$ like algebraically closed or characteristic $0$ are pertinent, but feel free to make them for comfort. For any such matrix $A$, the set $K[A]$ of polynomials in $A$ is a commutative subalgebra of $M_n(K)$; the question is whether for any pair of commuting matrices $X,Y$ at least one such commutative subalgebra can be found that contains both $X$ and $Y$.  I was asking myself this in connection with frequently recurring requests to completely characterise commuting pairs of matrices, like . While providing a useful characterisation seems impossible, a positive anwer to the current question would at least provide some answer.  Note that in many rather likely situations one can in fact take $A$ to be one of the matrices $X,Y$, for instance when one of the matrices , or more generally if its  (so coincides with the characteristic polynomial). However this is not always possible, as can be easily seen for instance for diagonal matrices $X=\operatorname{diag}(0,0,1)$ and $Y=\operatorname{diag}(0,1,1)$. However in that case both will be polynomials in $A=\operatorname{diag}(x,y,z)$ for any distinct values $x,y,z$ (then $K[A]$ consists of all diagonal matrices); although in the example in  the matrices are not both diagonalisable, an appropriate $A$ can be found there as well.   I thought for some time that any maximal commutative subalgebra of $M_n(K)$ was of the form $K[A]$ (which would imply a positive answer) for some $A$ with minimal polynomial of degree$~n$, and that a positive answer to my question was in fact instrumental in proving this. However I was wrong on both counts: there exist (for $n\geq 4$) commutative subalgebras of dimension${}&gt;n$ (whereas $\dim_KK[A]\leq n$ for all $A\in M_n(K)$) as shown in , and I was forced to correct  in the light of this; however it seems (at least in the cases I looked at) that many (all?) pairs of matrices $X,Y$ in such a subalgebra still admit a matrix $A$ (which in general is not in the subalgebra) such that $X,Y\in K[A]$. This indicates that a positive answer to my question would not contradict the existence of such large commutative subalgebras: it would just mean that to obtain a maximal dimensional subalgebra containing $X,Y$ one should in general avoid throwing in an $A$ with $X,Y\in K[A]$. I do think these large subalgebras easily show that my question but for three commuting matrices has a negative answer.  Finally I note that  to the cited MO question mentions a result by Gerstenhaber that the dimension of the subalgebra generated two commuting matrices in $M_n(K)$ cannot exceed$~n$. This unfortunately does not settle my question (if $X,Y$ would generate a subalgebra of dimension${}&gt;n$, it would have proved a negative answer); it just might be that the mentioned result is true because of the existence of $A$ (I don't have access to a proof right now, but given the formulation it seems unlikely that it was done this way).  OK, I've tried to build up the suspense. Honesty demands that I say that I do know the answer to my question, since a colleague of mine provided a convincing one. I will however not give this answer right away, but post it once there has been some time for gathering answers here; who knows somebody will prove a different answer than the one I have (heaven forbid), or at least give the same answer with a different justification.
92660
Let $K$ be a field and $f(x)\in K[X]$ be a polynomial of degree $n$. And let $F$ be its splitting field. Show that $[F:K]$ divides $n!$.   Here $[F:K]$ denotes the dimension of $F$ over $K$ as a vector space.  I already know it is true for $K=F$, so the problem here is when $f$ is irreducible in $K[X]$.
114160
I'm trying to do the following exercise from the book An Invitation to Algebraic Geometry:     Show that if $X \to Y$ is a surjective morphism of affine algebraic varieties, then the dimension of $X$ is at least as large as the dimension of $Y$.   Could anyone give me a hint? I'm looking for a proof that avoids commutative algebra since it is not mentionned in the book at this point.