I want to prove that $\bar{K}[V]/M_p \simeq \bar{K}$ where $K$ is a field, $\bar{K}$ is its algebraic closure and $$\bar{K}[V]=\bar{K}[x_1,...,x_n]/I_V,$$ where $I_V$ is the ideal attached to a variety, so it is prime, and $$M_p=\{f \in \bar{K}[V] : f(p)=0\}.$$  The thing is that I think that there is something wrong with this, because Can one mod out more than once?.  My intuition tells me that in $\bar{K}[V]/M_p$ we are killing all the polynomials that vanish in the variety so I was trying to use this to map a polynomial to its coefficients in $\bar{K}$ and doing this the other way around to get a bijection, but I think there has to be a better way since I am not quite sure that this will work.  Can someone help me with this issue?  Thanks in advance.
120875
I want to prove that $\bar{K}[V]/M_p \simeq \bar{K}$ where $K$ is a field, $\bar{K}$ is its algebraic closure and   $$\bar{K}[V]=\bar{K}[x_1,...,x_n]/I_V,$$   where $I_V$ is the ideal attached to a variety, so it is prime.    Now, on the other hand:  $$M_p=\{f \in \bar{K}[V] : f(p)=0\}.$$  So we have to find a bijective function from this two fields (may be there are better ways to do this, but I am not quite sure if in field theory we can find that machinery), but my intuition says that in $\bar{K}[V]/M_p$ we don't have any vanishing polynomial at $V$ so I can't figure out how to find this bijection.  Can someone help me with this problem?  Thanks a lot in advance.
3524
This is a follow up question to my previous question  Let $k$ be a field and $V \subseteq \Bbb{A}^n$ and $W \subseteq \Bbb{A}^m$ be algebraic sets. Then it should be true that $I(V \times W ) = I(V) + I(W)$ where by $I(V)$ here we mean the extension of $I(V)k[x_1,\ldots,x_{m+n}]$. Now I believe I have proven this (see the proof at the bottom of my question) but when I look at Martin's answer , it is instead claimed that we  actually have  $$I(V \times W )= \sqrt{I(V) + I(W)},$$  and for $I(V) + I(W)$ to be a radical ideal we need $k$ to be algebraically closed.           My question is: What's going on here? I believe my claim is true even without the assumption that $k$ is algebraically closed.           Here's a proof user Sanchez told me of, which I have simplified:       First it is clear that we always have $I(V) + I(W) \subseteq I(V \times W)$. For the reverse inclusion consider a polynomial $f \in I(V \times W)$. Then we can always write $$f = \sum_{i=1}^n f_ig_i$$ where $f_i \in k[x_1,\ldots,x_n]$ and $g_i \in k[x_{n+1},\ldots,x_{m+n}]$. Now take any $b' \in W$. If for all $i$ we have $g_i(b') = 0$ then since $b'$ is arbitrary, $g_i \in I(W)$ for all $i$. Then $f \in I(W) \subseteq I(V ) + I(w)$ and we are done. Otherwise suppose there exists $b \in W$  and $i$ such that $g_i(b) \neq 0$. Then wlog we may suppose that  $g_1(b) \neq 0$.   Next, $\sum f_ig_i(b) = 0 $ on all of $V$ by assumption of $f \in I(V \times W)$. So $\sum f_i g_i(b) = p$ for some $p \in I(V)$. Now write $$f_1 = \frac{ p - g_2(b) f_2 + \ldots g_n(b)f_n}{g_1(b)}.$$  Substituting this for $f_1$ in $\sum f_ig_i$, followed by taking things mod $I(V)$ we get an  expression with only $n-1$ terms $\mod{I(V)}$. Continuing this process we will finally get an expression with $0$ terms $\mod{I(V)}$ so that $f \in I(V)$. This shows $I(V \times W) \subseteq I(V) + I(W)$ which completes the proof.
110004
Theorem in question is:     Let $\mathfrak o$ be a complete local ring with maximal ideal $\mathfrak m$. Let $$ f(X) = \sum_{i = 0} a_iX^i $$be a power series in $\mathfrak o [[X]]$ such that not all $a_i$ lie in $\mathfrak m$.   Let $a_0, \ldots, a_{n-1} \in \mathfrak m$ and $a_n \in \mathfrak o^\times$ is a unit.   Given $g \in \mathfrak o [[X]]$ we can solve the equation   $$g = qf + r$$   uniquely with $q \in \mathfrak o [[X]], r \in \mathfrak o [X], \deg r &lt; n$.   For those who have access to book, this is chapter IV (Polynomials), subchapter 9 (Power series). For those without a book, I'll transcribe Lang's proof:  Let $\alpha$ and $\tau$ be projections on the beginning and tail end of the series, given by: $$ \alpha : \sum b_i X^i \to b_0 + b_1 X + \ldots b_{n-1}X^{n-1},$$ $$ \tau : \sum b_i X^i \to b_n + b_{n+1} X + b_{n+2}X^2 + \ldots .$$  Notice:   $\tau(X^n h) = h$ for all $h \in \mathfrak o[[X]]$. $h$ is a polynomial of $&lt; n$ if and only if $\tau(h) = 0$.   The existence of $q, r$ is equivalent with a condition that there exists $q$ such that $$ \tau(g) = \tau(qf).$$  We can write $f = \alpha(f) + X^n \tau(f)$, hence our problem is  equivalent with solving $$ \tau(g) = \tau(qf) = \tau(q\alpha(f)) + \tau(q X^n \tau(f))  = \tau(q \alpha(f)) + q\tau(f),$$ by using (1) in the last equality.  Notice that $\tau(f)$ is invertible. Writing $Z = q\tau(f)$, the equation above is equivalent with:  $$ \tau(g) = \tau \left(Z \frac{\alpha(f)}{\tau(f)}\right) + Z = \left(I + \tau \circ \frac{\alpha(f)}{\tau(f)}\right)Z$$  Here comes my issue with this proof.  Note that $\alpha(f) \in \mathfrak m[[X]]$, so that image of $\tau \circ \frac{\alpha(f)}{\tau(f)}$ is in $\mathfrak m[[X]]$.  We can therefore invert to find $Z$, namely $$ Z = \left(I + \tau \circ \frac{\alpha(f)}{\tau(f)}\right)^{-1}\tau(g).$$    It is not at all clear to me why this remark about image of $\tau \circ \frac{\alpha(f)}{\tau(f)}$ implies invertibility of the function $\bigl(I + \tau \circ \frac{\alpha(f)}{\tau(f)}\bigr)$. Could we write the inverse in a more concrete way?
102881
Does there exist a polynomial $p(x) \in \mathbb C[x]$ such that $p(x) \notin \mathbb R[x]$ and $p(x)p(-x)=p(x^2)$ ?  I have noticed that if $a_n$ is the leading coefficient of $p(x)$ then $a_n=(-1)^n $, so the leading coefficient cannot be imaginary; also $p(0)^2=p(0)$ so the constant term also is real. Please help. Thanks in advance.
120249
I need help proving this theorem:     Given the field extension: $\mathbf{K} \subseteq \mathbf{L}$, for $\alpha \in \mathbf{L}$ and $g(x) \in \mathbf{K}[x]$, $\alpha$'s minimal polynomial over $K$,   and $f(x) \in \mathbf{L}[x]$, $\alpha$'s minimal polynomial over $L$,   then the degree of $g$ is bigger than the degree of $f$ and $f(x)$ divides $g(x)$.
6963
Ex. $2.3.3$ in Algebraic Number Theory by Neukirch is the following:     Let $k$ be a field and $K = k(t)$ the function field in one variable. Show that the valuations $v_{\mathfrak p} $ associated to the prime ideals $\mathfrak p = (p(t))$ of $k[t]$, together with the degree valuation $v_\infty$, are the only valuations of $K$, up to equivalence. What are the residue class fields?   By valuation, I mean the exponential valuation. For instance the 2-valuation on $\Bbb Z$ is $$v(2^sn) = s , (2,n) = 1$$  Take $k = \Bbb Q$ and a non archimedean valuation $v$ on it and extend it to $k[x]$ by $$f(t) = a_0+a_1t+ \dots +a_nt^n \in K, v(f) = \min\{v(a_0),\dots, v(a_n)\}$$ and to $k(x)$ by $$v(f/g) = v(f)-v(g).$$ This is proven to be a valuation earlier on in the chapter. Which of the valuations mentioned in the problem is this equivalent to?
56500
Motivation for the question: in Neukirch - Algebraic Number Theory (Chapter II - Section 6 - Henselian Fields) the author discusses the context of a nonarchimedean valued field $K$ and its completion $\widehat{K}$, and considers the separable closure $K_v$ of $K$ in $\widehat{K}$, so that $K \subseteq K_v \subseteq \widehat{K}$.  Then he writes the following:     When $K_v$ is algebraically closed in $\widehat K$ - hence in particular $\text{char}(K) = 0$ - this is immediately obvious [...]   (emphasis mine).  My question is about the implicit claim that if $K_v$ is algebraically closed,  then $\text{char}(K) = 0$.  Now we know that $K_v$ is algebraically closed if and only if $K$ is perfect (since every $K_v$ will be the algebraic closure of $K$ and so every algebraic extension of $K$ is separable).  So we're asking whether it is true that if $K$ is perfect, then it must have characteristic $0$.  I know that all fields of characteristic $0$ are perfect. I also know that we have the example of finite fields which are also perfect fields but have prime characteristic. This does not contradict the claim above because the finite fields have only the trivial valuation.  So is there something about having a nonarchimedean valuation on a perfect field that forces its characteristic to be zero?
82956
I'm trying to work out the following past paper question and I've got stuck.  $R$ is an integral domain and $S = R[t]$, the polynomial ring in one variable over $R$. We have that $Q$ is a prime ideal strictly contained in another prime ideal $P$ with  $$ P \cap R = Q \cap R  = P_0. $$ We also have $Y = R\setminus P_0$.  I have shown that $P_0S$ is a prime ideal, and that after localisation we have $$ \frac{SY^{-1}}{P_0SY^{-1}} \simeq F[t]    $$ where  $$ F = \frac{RY^{-1}}{P_0Y^{-1}}. $$  This is a field since $P_0 = R\setminus Y$, so $P_0Y^{-1}$ is a maximal ideal of $RY^{-1}$.   What I now have to show is that $P_0S  =  Q$, and the question says that i should do this using the above isomorphism of polynomial rings, but I can't figure it out.  A hint or solution would be great, thanks!
111102
Let $$\phi:\mathbb Z[x]\longrightarrow \mathbb R$$, where $$\phi(f(x))=f(\sqrt 2)$$  find $$\operatorname{Ker}(\phi)=\{f(x)\in \mathbb Z[x]\mid f(\sqrt2)=0\}$$  1st) I wanted to use isomorphism theorem since we have $\mathbb R$-field so $Z[x]/\ker\phi\simeq \mathbb R$ so we find that $\ker\phi$ maximal ideal so it must be irreducible minimal polynomial which has root $\sqrt2$ that is $x^2-2$ but $\phi$ is not surjective.  2nd) Let $f=f_0+f_1x+f_2x^2+f_3x^3+f_4x^4+..$ since $\\f(\sqrt2)=0$ we have $$\\f_0+2f_2+4f_4+8f_6+...=0\\ f_1+2f_3+4f_5+...=0$$but how to show $x^2-2\mid f(x)$
88414
$F$ is an extension field of a field $K$. Let $F$  be an algebraic closure of $\mathbb{Z}_p $ ($p$ prime). Show that $(i)$ $F$ is algebraic Galois over $\mathbb{Z}_p$ $(ii)$  The map $\alpha:F\rightarrow F$ given by $u\mapsto u^p$ is a nonidentiy $\mathbb{Z}_p$-automorphism of $F$. $(iii)$  The subgroup $H=\langle \alpha \rangle$ is a proper subgroup of Aut$(F/\mathbb{Z}_p)$ where the fixed field is $\mathbb{Z}_p$, which is also the fixed field of Aut$(F/\mathbb{Z}_p)$ by $(i).$    So, here is my attempt for (i). Let  $S \subset \mathbb{Z}_p[x])$ of monic polynomials of the form $x^{p^n}-x$.  Then for all $f\in S$, gcd$(f,f^{\prime})=1$(i.e, the polynomials are separble) and $F=\mathbb{Z}_p(a\in F:f(a)=0)$. So $(F/\mathbb{Z}_p)$ is Galois. I'd be glad if I could get assistance for (ii) and (iii) as well.  Thanks.  ADDED:  Attempt at (iii).   but I know that since $F/\mathbb{Z}_p$ is a finite Galois extension, its fixed field is $\mathbb{Z}_p$ hints...  The field $\mathbb{Z}_p$ must be contained in $F$. For $a\in \mathbb{Z}_p$, $\alpha(a)=a^p=a$. Thus the polynomial $x^p-x$ has $p$ zeros in $F$, namely, the elements of $\mathbb{Z}_p$. But the elements fixed under $\alpha$ are precisely the zeros in $F$ of $x^p-x$. Hence the fixed field of $\alpha$ is $\mathbb{Z}_p$ which is also the fixed field of Aut$(F/\mathbb{Z}_p)$. Left to show that $H$ is a proper subgroup... If it helps I know that the order of $\langle \alpha \rangle $ is $n$...
26230
Suppose that $k$ is an algebraically closed field. Then what do the prime ideals in the polynomial ring $k[x,y]$ look like?   As far as I know, the maximal ideals of $k[x,y]$ are of the form $(x-a,y-b)$ where $a,b\in k$. What can we say about the prime ideals? Are there similar results? And what about $k[x,y,z], k[x,y,z,w]$ and so on. Would someone be kind enough to give me some hints or referrence on this topic? Thank you very much!
43255
I have to prove the following question from "A course in complex analysis and riemann surfaces":     Let $p(z)=\sum_{i=0}^na_iz^u$ be a polynomial with all roots inside the (open )unit disk. Denote by $\bar{p}(z)$ the polynomial $\sum_{i=0}^n\bar{a_i}z^i$. Prove that the roots of $p(z)+z^n\bar{p}(\frac{1}{z})$ lie on the unit circle.   Now it's not hard to see that if $r$ is a root of $p(z)$, then $\frac{1}{\bar{r}}$ is a root of $z^n\bar{p}(\frac{1}{z})$. But that's about as far as I got. I would like a hint on how to prove this.
47674
I'm stuck on a question in an exercise on p-adic numbers. Actually, let $p$ a prime number, $F/\mathbb{Q}_p$ a finite extension of fields, $\mathcal{O}$ the integral closure of $\mathbb{Z}_p$ on $F$, $\mathfrak{p}$ the maximal ideal of $\mathcal{O}$.   We have already shown that : $\forall l \in \mathbb{N}_{\geq 2}$, $\gcd(l,p)=1 \;\forall x \in \mathfrak{p} \; \exists y \in \mathcal{O} \quad y^l=1+x$, using Hensel's lemma.    Now, the next question is :      Deduce from this that for any non trivial valuation $w$ on $F$, if we are   noting $\mathcal{O}_w$ its valuation ring, we have  : $\mathcal{O} \subset O_w$.   (We know as well that if we are noting : $v(x) = v_p(N_{F/\mathbb{Q}_p}(x))$, then $\mathcal{O} = \{ x \in F \; | \; v(x) \geq 0\}$)  To do this, I thought to show that $\mathbb{Z}_p \subset O_w$, as $O_w$ is integrally closed on $F$, we would have $\mathcal{O} \subset O_w$.  Now, if $x \in \mathbb{Z}_p$, $x = \frac{a}{b}$, where $\gcd(p,b)=1, \gcd(a,b)=1$. We also know that the restriction to $\mathbb{Q}$ of $|.|_w$ (which is the n.a absolute value corresponding to the valuation) is either trivial or $|.|_q$ for a prime $q$, by Ostrowski's theorem. So : $w(x) = w(\frac{a}{b}) = w(a)-w(b)$. If $w$ is trivial over $\mathbb{Q}$, $w(x) = 0$ and we have the result. Otherwise, $w(x) = v_q(a) - v_q(b)$. But then I'm stuck, cause actually the result I have to prove indicates that $q=p$, but there is no way (I found) from here to conclude this (and it's actually quite normal, cause this is what implies the next question).  So, anyone could help me, please ?  Thank you !
5528
This result appears to be ubiquitous as an algebra exercise. How do you prove this result?     Let $A$ be an integral domain with field of fractions $K$, and let $A_{\mathfrak{m}}$ denote the localisation of $A$ at a maximal ideal $\mathfrak{m}$ considered as a subring of $K$. Prove that $$A = \bigcap_{\mathfrak{m}} A_{\mathfrak{m}}\,,$$ where the intersection is taken over all maximal ideals $\mathfrak{m}$ of $A$.
19236
Show that the polynomial $X^5 + X^3 + \bar{1}$ in $(\mathbb{Z}/2\mathbb{Z})[X]$ is irreducible. (Hint: if it were reducible, it would either have a root or be of the form $g(X) \cdot h(x)$, where deg$g(X) = 2$ and deg$h(X) = 3$) Recall that $\bar{a}$ is shorthand for the coset $a + 2\mathbb{Z} \space $in$ \space (\mathbb{Z}/2\mathbb{Z})[X]$ (for any $a \in \mathbb{Z}$)  This question is worth $7$ marks out of a possible $75$, this is my solution..  $f(X) = X^5 + X^3 + \bar{1}$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$ iff it has no roots in $\mathbb{Z}/2\mathbb{Z}$. Since $f(\bar{0}) = \bar{1}$ and $f(\bar{1}) = \bar{1}$, it follows that $f(x)$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$.  Would this be a sufficient answer? am I missing something?
12952
Given the ring $k[x,y,z]$, where $k$ is a field, and an ideal $I=(xy,x-yz)$, find a primary decomposition of $I$.    I tried to draw the graph of the variety of $I$ and get a decomposition of $(x,y)\cap(x,z)$ (the two prime ideals corresponds to the irreducible components of the variety), but apparently these two ideals are not the primary decomposition of $I$, and I don't know how to fix this.  I would really appreciate it if someone can help me. thx
22880
Let $\mathbb{Z}[X]$ be the ring of polynomials in one variable over $\Bbb Z$.  My question: Is every prime ideal of $\mathbb{Z}[X]$ one of following types? If yes, how would you prove this?  $(0)$.  $(f(X))$, where $f(X)$ is an irreducible polynomial.  $(p)$, where $p$ is a prime number.  $(p, f(X))$, where $p$ is a prime number and $f(X)$ is an irreducible polynomial modulo $p$.