Let $R$ be an integrally closed domain and let $K$ be its fraction field. Let   $L\supseteq K$ be a field. If $\alpha\in L$ is integral over $R$ (i.e.   if it satisfies a monic polynomial in $R[x]$), does its minimal   polynomial over $K$ lie in $R[x]$?   CONTEXT:  I'm trying to prove that the trace $t_{L/K}$ of integral elements lie in $R$ (provided that the extension $L/K$ is finite). I'm trying to use the fact that the trace of $\alpha$ is an integer multiple of certain coefficient of its minimal polynomial, so this trace lies in $R$ if such coefficient does.   Since $\alpha$ satisfies an integral relation $p(\alpha)=0$ over $R$, it's minimal polynomial $q$ over $K$ exists and it divides $p$, does it imply that $q\in R[x]$? if so, then I'd be done.   It's clear that the result is true if $R$ is a UFD. In such a case, it's only a matter of looking at the unique factorization of the polynomial in $R[x]$ and apply Gauss' lemma. However, I don't see a straight proof nor counter example in the general case.
123992
I am having some trouble understanding Proposition 5.15 in Introduction to Commutative Algebra by Atiyah and Macdonald.      Let $A\subset B$ be integral domains, $A$ integrally closed, and $x\in B$ be integral over an ideal $\mathfrak{a}$ of $A$. Then $x$ is algebraic over the field of fractions $K$ of $A$, and if its minimal polynomial over $K$ is $t^n+a_1t^{n-1} + \cdots + a_n$, then the $a_1, a_2, \ldots , a_n$ all lie in the radical of $\mathfrak{a}$.   The proof of the proposition states that the coefficients of the minimal polynomial of $x$ are polynomials in the $x_i$ (the conjugates of $x$). I don't understand this, and am struggling with the rest of the proof as well. Thanks for any help.
75146
If $R$ is a unital integral ring, then its characteristic is either $0$ or prime. If $R$ is a ring without unit, then the char of $R$ is defined to be the smallest positive integer $p$ s.t. $ pa = 0 $ for some nonzero element $a \in R$. I am not sure how to prove that the characteristic of an integral domain without a unit is still either $0$ or a prime $p$. I know that if $p$ is the char of $R$, then $px = 0 $ for all $x \in R$. If we assume $ p \neq 0 $ and $R$ has nonzero char, and $p$ factors into $nm$, then $ (nm) a = 0 $ , which means $ n (ma) = 0 $. Well $ma \neq 0$, because this would contradict the minimality of $p$ on $a$. But I don't know where to go from this point w/o invoking a unit.   Edit: I had left out the assumption that $R$ is assumed to be a integral domain. This has been corrected.
92571
Let $A$ be a ring and let $A[x]$ be the ring of polynomials in an indeterminate $x,$ with coefficients in $A.$ Let $f=a_0 + a_1x+\cdots+a_nx^n \in A[x].$ $f$ is said to be primitive if $(a_0,a_1,\ldots,a_n)=(1).$     Prove that if $f,g\in A[x],$ then $f$ and $g$ are primitive $\Rightarrow$ $fg$ is primitive.   I know Gauss lemma holds true for UFDs, but in this case it isn't required $A$ to be UFD: is it possible to drop the assumption, or is some hypothesis missing here?
35626
If $R$ is an integral domain and a subring $S$ has identity $1_S$, how would you show that $1_S=1_R$ (here $1_R$ is the identity of the ring $R$)? I am unsure about what an integral domain really is and how the subring comes into play here.
112729
Let $R$ is  an integral domain .  Show that If $R[X]$ is Euclidean domain then $R$ is a field .  I'll be waiting for your help. Thank you very much in advance!
42788
How can we show that if $R$ is an infinite commutative ring and $R/I$ is finite for every nonzero $I \unlhd R$, then $R$ is an integral domain?   I tried proceeding by contradiction: assume $a$,$b$ $\in R \backslash \{0\}$ and $ab=0$; then $R/(a)$ and $R/(b)$ must be finite, say $R/(a)=\{k_i + (a) : 1 \leq i \leq m\}$ and $R/(b)=\{l_j + (b) : 1 \leq j \leq n\}$. Does this mean $R$ must be finite? Or what about using the fact that $R/(a,b)$ finite?  Thanks for any help with this!
121367
This is just a textbook problem from Dummit and Foote, but the issue is that our class barely touched on PIDs and the preceding material, so I don't really know or understand much.  Anyway,     Let $R$ be an integral domain.  If $R$ satisfies the following:           i. any two nonzero $a,b \in R$ have a gcd in the form of $d=ra+sb, r,s \in R$          ii. If $a_1, a_2,...$ are nonzero in $R$ such that $a_{i+1}$ divides $a_i$ for every $i$ then $\exists N \in \mathbb{Z}$ such that $a_n$ is an unit times $a_N$  $\forall n \ge N$,          prove that $R$ is a principal ideal domain.   From what I've gathered, the second condition means that $(a_1) \subset (a_2) \subset...$ and $(a_N) = (a_n)$ $\forall n \ge N$.  However, I'm not sure what I should do now. I'm worried that I'm lacking the necessary context here especially since we used Artin pretty much all semester.  In short, my question is: how should I proceed with this proof given what I know?
35956
Let $f,g \in k[t]$, $k$ is a field of characteristic zero, $\deg(f)=n \geq 2$, $\deg(g)=m \geq 2$.     Is it possible to characterize all such $f$ and $g$ for which $k[f,g]$ is integrally closed in its field of fractions $k(f,g)$?   I do not mind to further assume that $k(f,g)=k(t)$ (but I do not want to assume that $k[f,g]=k[t]$, which makes my question trivial).  Thank you very much!  Edit: (i)  is relevant.  (ii) If also $k(f,g)=k(t)$, then being integrally closed is equivalent to $k[f,g]=k[t]$. Indeed: (a) If $k[f,g]=k[t]$ then $k[f,g]$ is integrally closed (since $k[t]$ is). (b) If $k[f,g]$ is integrally closed then, in particular, since $t \in k(t)=k(f,g)$ (= the field of fractions of $k[f,g]$) and obviously $t$ is integral over $k[f,g]$ (for example, $t$ is a root of $f(T)-f \in k[f,g][T]$) hence $t \in k[f,g]$, so $k[f,g]=k[t]$.  For example: $f(t)=t^2+1$, $g(t)=t^3-4t$. We have, $t= \frac{t(t^2-4)}{t^2+1-5}=\frac{g}{f-5}$, so $k(f,g)=k(t)$. By Abhyankar-Moh-Suzuki theorem, $k[f,g] \neq k[t]$, so $k[f,g]$ is not integrally closed. Similarly, for every $f,g$ such that $k(f,g)=k(t)$ and one of the degrees does not divides the other, then $k[f,g]$ is not integrally closed.  (iii) If also $k(f,g)=k(t)$ and $k$ is algebraically closed, then being integrally closed is equivalent to $H(t)=(f(t),g(t))$ is injective and $H'(t)=(f'(t),g'(t)) \neq (0,0)$ for every $t \in k$. Indeed, $k[f,g]=k[t]$ is equivalent to the new conditions (it is important that $k$ is algebraically closed!!).  Edit: What if we replace $k$ by an integral domain $D$ (which is a $k$-algebra)? See also  question.
115861
Let $R=\frac{\mathbb{C}[x,y]}{\langle x^2-y^2-1 \rangle}$. Prove that (1) $R$ is an integral domain. (2) Any nonzero prime ideal of $R$ is maximal.  My idea: (1) The first part is easy. Since $x^2-y^2-1\in \mathbb{C}[y][x]$ is Eisenstein with respect to the prime $y+i$, it is irreducible. Hence the ideal generated by it is a prime ideal. (2) I am struggling with this part. I have one approach (not sure if correct). Approach : Let $I=\langle x^2-y^2-1 \rangle$. Then $R=\mathbb{C}[x,y]/I$.  Any ideal of $R$ is of the form $J/I$ where $J\subseteq \mathbb{C}[x,y]$ is an ideal containing $I$. If we can show that any prime ideal $J$ containing $I$ is maximal, then we are done. How to prove this? Is there any other way to prove the second part? Edit: I am trying to avoid Hilbert Nullstellensatz and the notion of dimension of a ring, as these concepts are not there in the syllabus of the exam from which I found this question.
102588
This is actually not a problem, but it's a statement which is taken for granted and I don't know how to prove it. Hope some one can help me. I really appreciate:          Suppose $A$ is subring of ring $B$, and $A$, $B$ are integral domains. If $B$ is integral over $A$, prove that field of fraction of $B$ is algebraic over field of fraction of $A$. Does the converse remain true?
18876
How can I prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle$ is an integral domain?  Also, I need to prove that its field of fractions is isomorphic to the field of rational functions $\mathbb{Q}(t)$?    (The question is taken from )
116297
I'm working on the following problem...     Suppose that $A$ is a Dedekind domain with fraction field $K$.  $L/K$ is a finite separable extension of $A$ of degree $n$, and $B$ is the integral closure of $A$ in $L$ (so that $B$ is also a Dedekind domain).  Suppose that $\mathfrak{p}$ is a nonzero prime ideal of $A$, and $B\mathfrak{p}=\mathfrak{q}^{e_1}\cdots \mathfrak{q}_r^{e_r}$ is the prime factorization of $B\mathfrak{p}$.  Show that $$e_1f_1+\cdots +e_rf_r=n$$ where $f_i=[B/\mathfrak{q}_i:A/\mathfrak{p}]$.    My first question is, why does integrality over $A$ imply that $B$ is also a Dedekind domain? My second question is, well, how to do the problem.  I need an outline so that I know what I'm looking for. What results will be needed here?
63936
Let $R$ be a Euclidean domain in which the quotient and remainder are always unique. Does it follow that the ring $R$ is either a field or a polynomial ring $F[X]$ for some field $F$?
54881
I am told that $K \leq L$ is a field extension of degree 2 and that $K$ has characteristic not 2. I am then asked to show that $\exists \: \alpha \in L$ such that $L = K(\alpha), \; \alpha^2\in K $  So far I have that:  $\exists \; \alpha_1, \alpha_2 \in L$ such that $L = K( \alpha_1, \alpha_2)$  $\Rightarrow |L:K| = |L:K(\alpha_1)| |K(\alpha):K| $ so $|K(\alpha):K| = 1$ or $2$  $|K(\alpha_1):K| = 2 \Rightarrow L = K(\alpha_1)$ and $deg(f_{\alpha_1}(t)) = 2  $ where $f_{\alpha_1}(t)$ is the minimal polynomial of $\alpha_1$ over $K$   So $f_{\alpha_1}(t) = t^2 + at + b$ for some $a,b \in K$  Also, $|K(\alpha_1):K|=1 \Rightarrow K(\alpha_1) = K$ and $L = K(\alpha_2)$ which means $degf_{\alpha_2}(t) = 2$ where $f_{\alpha_2}(t)$ is the minimal polynomial of $\alpha_2$ over $K$.   So $f_{\alpha_2}(t) = t^2 + at + b$ for some $a,b \in K$  Given that we consider the same polynomial each time, relabel to consider:   $f_{\alpha}(t) = t^2 + at+ b $ for some $a,b \in K$  We have already demonstrated that $L=K(\alpha)$ for some $\alpha \in L$ so now we much show that $\alpha^2 \in K$  $\alpha^2 \in K \Leftrightarrow a = 0_K$  I now have no idea how to show that this coefficient of $f_{\alpha}(t)$ is necessarily $0$.   I know that $f_{\alpha}(t)$ has two roots in $L$ and that because $K$ and thus $L$ has characteristic not 2, they cannot be the same root, but I don't know how to then conclude that they must sum to $0$  Any help would be greatly appreciated.
4074
Let $R$ be a Dedekind domain, $\mathfrak{m}$ be a maximal ideal in $R[x]$ is of the form $\mathfrak{m} = (\mathfrak{p},f(x))$ where $\mathfrak{p}$ is a maximal ideal in $R$, and $f$ is a polynomial in $R[x]$ which is irreducible in $(R/\mathfrak{p})[x]$. Let $g(x) \in \mathfrak{m}$.  I am know that exist a ideal $\mathfrak{a}$ in $R$ such that $(\mathfrak{p},\mathfrak{a}) = 1 $, i.e $\mathfrak{p} + \mathfrak{a} = R$ and $\mathfrak{p}\mathfrak{a} = \mathfrak{p}\cap \mathfrak{a} = (p)$. (chapter 16, corollary 19, page 768, Abstract Algebra, Dummit Foote...)  Easily, since $g(x)\in \mathfrak{m}$, we have $ g(x) = l(x) + f(x).k(x)$ where $l(x) \in \mathfrak{p}[x]$. The questions is: Can we make a "strong" form as to show that there exists an ideal $\mathfrak{a}$ in $R$ such that $g$ can be show as the form $$ag(x) = p h(x) + a f(x) k(x)$$ where $p\in \mathfrak{p}$ such that $(p) = \mathfrak{p} \mathfrak{a}$, $(\mathfrak{p},\mathfrak{a}) = 1$ and $a\in \mathfrak{a} - \mathfrak{a}\mathfrak{p}$.
41252
Let $F$ be a field and $F[x]$ the univariate polynomial ring over $F$. Is it the case that every quotient ring ${F[x]}/{(p(x))}$, where $p(x) \in F[x]$, is a PID (principal ideal domain)?    This is just a minor issue that's been bothering me for the last couple of hours. I'm aware that when the ideal $(p(x))$ is prime (i.e. $p(x)$ irreducible) then ${F[x]}/{(p(x))}$ is a PID.
49955
I'm wondering if anyone has come across the following ring before. Let $R$ be the ring of complex power series $f=\sum_{n \ge 0} a_n t^n$ such that   $a_n \in \mathbb{Q} \: \: \forall \: n$ The function $f: \mathbb{C} \rightarrow \mathbb{C}$ is entire   Any information about it would be welcome but in order to keep the question specific, I'll list a few things I would like to know in particular.   Is $R$ a unique factorisation domain? Are there any elements of $R$ which are known to be prime/irreducible but are not associates of polynomials? If an element $f \in R$ is prime/irreducible, must all its zeroes (in $\mathbb{C}$) be simple? Is every $\alpha \in \mathbb{C}$ a root of some nonzero $f \in R$ ?