Calculate the ring of integers of quadratic number field $\mathbb{Q}(\sqrt{d})$   Solution:  Let $F$ be an algebraic number field. Then an element $b\in F$ is integral iff it monic irreducible polynomial has integer coefficients.  For example, $\sqrt{d}$ for integer $d$ is integral.  If $d\equiv 1 \mod 4$, then the monic irreducible polynomial of $(1+\sqrt{d})/2$ over $\mathbb{Q}$ is $x^2 -x + (1-d)/4 \in \mathbb{Z}[x]$, so $(1+\sqrt{d})/2$ is integral.  Thus the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt{d})$ contains the subring $\mathbb{Z}[\sqrt{d}]$, and the subring $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1 \mod 4$. We show that there are no other integral elements.  An element $a+b\sqrt{d}$ with rational $a$ and $b\neq0$ is integral iff its monic irreducible polynomial $x^2 -2ax +(a^2 -db^2)$ belongs to $\mathbb{Z}[x]$. Therefore, $2a$,$2b$ are integers. If $a=(2k+1)/2$, for $k\in\mathbb{Z}$, then it is easy to see that $a^2 - db^2 \in \mathbb{Z}$ iff $b=(2l+1)/2$ for some $l\in\mathbb{Z}$, and $(2k+1)^2 - d(2l+1)^2$ is divisible by 4. The latter implies that d is quadratic residue modulo 4, i.e. $d\equiv 1 \mod 4$. In turn, if $d\equiv 1 \mod 4$ then every element $(2k+1)/2 +(2l+1)\sqrt{d}/2$ is integral.  Thus, integral elements of $\mathbb{Q}(\sqrt{d})$ are equal to $\mathbb{Z}[\sqrt{d}]$ if $d\not \equiv 1 \mod 4$, and $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1 \mod 4$.    Can someone explain why $\sqrt{d}$ for integer $d$ is integral, and why the monic irreducible polynomial of an integral element $a+b\sqrt{d}$ is of the form $x^2 -2ax +(a^2 -db^2)$?
41135
Quadratic integer ring $\mathcal{O}$ is defined by \begin{equation}    \mathcal{O}=\begin{cases}       \mathbb{Z}[\sqrt{D}] &amp; \text{if}\ D=2,3\ \pmod 4\\       \mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right]\  &amp; \text{if}\ D=1\pmod 4    \end{cases} \end{equation} where $D$ is square-free.   I understand $\mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right]$ is not closed under multiplication if $D=2,3\pmod 4$. But still, isn't it more natural to define $\mathcal{O}=\mathbb{Z}[\sqrt{D}]$ for all square-free $D$? (in that case, it really seems like 'quadratic integer') I wonder what is the motivation of this definition.
70708
I would like to verify the validity of the following line of thought:  Let $K \subset E$ be a field extension. Let $K[x]$ be the polynomial ring over $x$ and denote $K(x)$ its field of quotients (the rational functions over $K$).  Now assume $f(x)=g(x)h(x)$ with $f(x),g(x) \in K[x]$ and $h(x) \in E[x]$ and all $f$, $g$, $h$ are different than zero. Then $\frac{f(x)}{g(x)} = h(x)$ in $E(x)$. So $h(x) \in K(x)$. But $h(x)$ is a polynomial, so $h(x) \in K[x]$.  Is there any other simpler argument than that?  Added:  Let me be more specific. Let $x^p - a \in K[x]$, with $p$ being prime and $char(K)=p$. Let $b$ be the unique root of $x^p - a$ in its splitting field $K(b)$. Then $x^p - a=(x-b)^p \in K(b)[x].$ Let $m_b(x) \in K[x]$ be the minimal polynomial of $b$ over $K$. Then $m_b(x)$ divides $(x-b)^p$ and so $m_b(x)=(x-b)^l$ and $l$ is the smallest positive integer for which $(x-b)^l \in K[x]$. Let $p = lq+r$ where $0 \ge &lt;l$. Then $x^p-a=(m_b(x))^q (x-b)^r$. Can i conclude that $(x-b)^r \in K[x]$ and so $r=0$ by the minimality of $l$?  Thanks.
15237
I've just learned what the integral closure is.     I would like to find what is the intergral closure of $\mathbb{Z}$ in $\mathbb{Q}[i]$.   Let $\mathcal{R}$ the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[i]$. To determine $\mathcal{R}$ I started to note that $$\mathbb{Z}[i]\subset \mathcal{R}.$$ Indeed if $\alpha = x+iy\in \mathbb{Z}[i]$, $\alpha$ is a root of the monic polynomial $$f(X)=X^2-2xX+x^2+y^2\in\mathbb{Z}[X].$$  And we know that since $\mathbb{Z}$ and $\mathbb{Q}[i]$ are rings so is $\mathcal{R}$.  From there I'm not sure, is it true that there exist no ring $A$ such that $$\mathbb{Z}[i] \subsetneq A \subsetneq \mathbb{Q}[i] \text{ ?}$$ I've assumed that and show that $1/2\notin \mathcal{R}$ for example, otherwise it exists a monic polynomial $f\in \mathbb{Z}[X]$ such that  $$f\left(\frac{1}{2}\right) =\frac{1}{2^n}+a_{n-1}\frac{1}{2^{n-1}}+\cdots+a_1\frac{1}{2}+a_0 =0$$  So  $$2^n f\left(\frac{1}{2}\right) =0 \iff -1=2\underbrace{(a_{n-1}+\cdots +2^{n-2}a_1+2^{n-1}a_0)}_{\in \mathbb{Z}}$$  which is impossible.  In conclusion $\mathbb{Q}[i]\not\subset\mathcal{R}$ so $\mathcal{R}=\mathbb{Z}[i]$.     Is my proof correct ? Is there an other (easier) way to prove that ? And the most important, is there exist a general method to find the integral closure please ?   Thank you for your help.
66930
Let $\mathbb M$ be one of the following rings: $\mathbb{R}, \mathbb{Q},  \mathbb{F_9}, \mathbb{C} $. Let $I$ be the ideal generated by $x^4+2x-2$. Is the ring $\mathbb M[x]/I$ a field for some of the above choices for $R$?   I have been trying to work my way around some problems involving ideals, but I get really confused when they get mixed with rings and fields. I'm stuck and don't know how to proceed.
42042
Let $D$ be an integer such that it is not a perfect square in   $\mathbb{Z}$.       If $D \equiv 1$ mod $4$, the set $$T:= \left\{\left( \begin{matrix} a &amp; b  \\ \frac{(D-1)b}{4} &amp; a+b \\ \end{matrix} \right) : a, b \in  \mathbb{Z} \right\}$$ is a subring of $M_2(\mathbb{Z})$ which is isomorphic   to the ring $$O := \left\{a+b\frac{1+\sqrt{D}}{2}: a, b \in \mathbb{Z} \right\}.$$   My problem is to find an isomorphism. However, I find this relation : $$\det \left( \begin{matrix} a &amp; b  \\ \frac{(D-1)b}{4} &amp; a+b \\ \end{matrix} \right)$$ $$= a(a+b)-\frac{b^2(D-1)}{4}$$ $$= \left(a+b\frac{1+\sqrt{D}}{2}\right)\left(a+b\frac{1-\sqrt{D}}{2}\right)$$  and $D \equiv 1$ mod $4$ $\iff$ $\left(\frac{1+\sqrt{D}}{2}\right)\left(\frac{1-\sqrt{D}}{2}\right) = -k$ for some $k \in \mathbb{Z}$  Does someone could give me a little hint to help me complete this problem?
27251
Let $D$ be a squarefree integer. I am trying to prove that $\mathbb Q[\sqrt D]$ is contained in a Galois extension of $\mathbb Q$ with Galois group $\mathbb Z/4$ if and only if $D$ is the sum of two squares, $D = a^2 + b^2$ with $a,b\in \mathbb Q$.   The hint in the exercise in Dummit and Foote suggests considering the extension $\mathbb Q[\sqrt{s + s\sqrt D}]$ for the forward direction. However, I am unsure how to show that this extension is Galois, much less that its Galois group is cyclic of order 4.   For the reverse direction, I expressed the given Galois extension as $\mathbb Q[\sqrt{a+b\sqrt D}]$ for some $a,b\in \mathbb Q$. I showed that this extension has $\mathbb Q[\sqrt{c^2 - d^2 D}]$ as a subfield for some $c,d \in \mathbb Q$. From this, I deduced that the $\sqrt D$ and $\sqrt{c^2 - d^2 D}$ generate the same extension, so $\sqrt D = x \sqrt{c^2 - d^2 D}$ for some $x$, and hence that $D = \frac{x^2 c^2}{1 + x^2 d^2} $. It seems that there should be a way to decompose this as a sum of two squares, but I did not see how.   I did not tag this as homework as I am studying it independently.
29348
I don't see how to solve the following problem:     Let $R$ be a commutative and unitary ring. If there exists a monic polynomial $f(x) \in R[x]$ so that $R[x]/(f(x))$ is a field, show that $R$ is a field. What about if $f(x)$ isn't monic?   I don't have a clue about how proving it. I see that if $R$ is a field and $f(x)$ a monic and irreducible polynomial in $R[X]$ then $\frac{R[x]}{(f(x))}$ is a field, but that is not the question!  Thank you in advance for your help
54698
What are the conditions for integers $D_1$ and $D_2$ so that $\mathbb{Q}[\sqrt {D_1}] \simeq \mathbb{Q}[\sqrt {D_2}]$ as fields.  Here $\mathbb{Q}[\sqrt {D}] := \{a + b \sqrt D \mid a,b \in \mathbb{Q} \}$  Really not sure where to begin with this sort of problem. I was thinking that I should split into cases where the integer is a square or not.
52324
I've tried to construct examples of rings that match all except one of the properties in the definition of a Dedekind domain. (This is an old number theory qual question from Berkeleys MGSA website).  The only starting point that I can think of would be  $R:=K[X_1,X_2,\ldots]$  for $K$ a field. This is clearly integrally closed as any polynomial relation is contained in $K(X_1,\ldots,X_n)$ for some large enough $n$ and $K[X_1,\ldots,X_n]$ is integrally closed being a UFD.  The problem is then to get rid of enough ideals from this ring $R$, so that its dimension would be 1, but I can't seem to figure out what to do.  Anyone have any other ideas for rings that could work as a starting point? The polynomial ring with an infinite number of variables is pretty much the only example that I know of for how to construct a non-noetherian ring that's still a domain.
54804
I want to prove that for any set of distinct integers $a_1,\ldots,a_n$, the polynomial $$h = (x-a_1)\cdots(x-a_n) + 1$$ is irreducible over the field $\mathbb{Q}$, except for the following special cases which are reducible:  $$\left.\begin{cases} a_1 = a\\ a_2 = a+2 \end{cases}\right\} \implies h = (x-a-1)^2$$  and $$\left.\begin{cases} a_1 = a\\ a_2 = a+1\\ a_3 = a+2\\ a_4 = a+3 \end{cases}\right\} \implies h = ((x-a-1)(x-a-2)-1)^2$$
38516
Let $A\subset B$ be an integral extension. If $F$ and $E$ are the fields of fractions of $A$ and $B$, respectively, I want to show that $E$ is an algebraic extension of $F$.  I know that since $A \subset E$, it follows that $F \subset E$. Now I consider an arbitrary element $b_1/b_2\in E$. I would like to show that $b_1/b_2$ is algebraic over $F$. Since $B$ is algebraic over $A$, there exist monic polynomial $f_1(x), f_2(x)\in A[x]$ such that $f_1(b_1) = f_2(b_2) = 0$. But how do I obtain a polynomial $f(x) \in F[x]$ such that $f(b_1/b_2) = 0$?
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?
66378
Suppose $p$ and $q$ primes and $p$ is odd. Then, are there nice and clever ways to factorize polynomials of the form $$f(x)=1+x+\cdots +x^{p-1}$$ in the ring $\mathbb{F}_q[x]$ ? What about the case when $q=2$ ?   I know that there are factorization algorithms but they are too general. I want to know if there are clever ways to do this for these special type of polynomials. Like for example, in $\mathbb{F}_2$ one might add terms $x^r+x^r$ and rearrange them so that $f(x)$ gets factorized.  In case there are no good methods to factorize, are there nice ways to check whether $f(x)$ is irreducible in $\mathbb{F}_q[x]$ ?
24054
Trying to solve a problem, I got stuck on the following question.     If $P\in\mathbb R[X]$ is an irreducible polynomial and $\operatorname {deg} P=2$, then is it true that $$\mathbb{R}[X]/(P)\cong\mathbb C?$$   I believe it must be true from the bits I've been hearing throughout my studies, but of which I haven't seen proofs. I've been unable to find anything on the internet other than . It tells me that there is something called Weierstrass' theorem which should be relevant, but can't find even the precise statement of the theorem, let alone the proof.  Here's what I have tried.   First of all, I can assume that $P$ is monic, because for any polynomial of degree $2$ there's a monic polynomial generating the same ideal in $\mathbb R[X].$ So I have  $$P(x)=x^2+bx+c$$  for some real numbers $b,c$ such that $b^2&lt;4c$.   I want to assume there is an isomorphism $\phi:\mathbb R[X]/(P)\to\mathbb C$ and work out what it should look like. I will denote by $[S]$ the coset of $S\in \mathbb R[X]$ in $\mathbb R[X]/(P).$ I'd like to find $S$ such that $\phi([S])=i.$  I have   $$ [x^2+bx+c-1]\mapsto -1, $$  So the coset that is mapped to $-1$ has to be a square root of $[x^2+bx+c-1]$ in $\mathbb R[X]/(P).$ Such a coset must be represented by a linear polynomial $kx+l$. Thus $$(kx+l)^2-(x^2+bx+c-1)\in (P),$$  that is $$ (kx+l)^2-(x^2+bx+c-1)\equiv a(x^2+bx+c) $$  for some real number $a$, and further  $$ (k^2-1-a)x^2+(2kl-b-ab)x+l^2-c-ac+1\equiv 0. $$  This gives   $$ \begin{cases} k^2=1+a\\ l^2=(1+a)c+1\\ 2kl=(1+a)b \end{cases} $$  The problem is that solving this gives formulas that are very difficult to handle. And I need to handle them if I want to prove that what I'm constructing is indeed an isomorphism. Is there a better way of doing it?
99074
Let $\mathbb{F}_p$ be the finite field with $p$ elements, where $p$ is a prime number. Let $x$ and $y$ be transcendental and algebraically independent over $\mathbb{F}_p$. The extension $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is finite of degree $p^2$. To see this consider the polynomials $T^p-x^p$  and $T^p-y^p\in\mathbb{F}_p(x^p,y^p)[T]$.  So, if $K$ is any (proper) intermediate field of this extension, we must have $[K:\mathbb{F}_p(x^p,y^p)]=p$. For each prime $q\neq p$, consider the polynomial $T^p-x^{qp}\in\mathbb{F}_p(x^p,y^p)[T]$. It is an irreducible polynomial of degree $p$ with $x^q$ as root. So the extension $\mathbb{F}_p(x^p,y^p)(x^q)/\mathbb{F}_p(x^p,y^p)$ has degree $p$ and hence, $\mathbb{F}_p(x^p,y^p)(x^q)$ is a proper intermediate field of the original extension. Note that we can do the same with $y$ instead of $x$. Since there are infinite primes $\neq p$, there will be infinite (proper) intermediate fields of the original extension. These fields are distinct, because they are obtained by adjoining a root of an irreducible polynomial, and their irreducible polynomials are distinct.  This shows that $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is finite but not simple, hence, it is not separable.      My question is: are there other intermediate fields other than that pointed out previously?   I first tried to show that $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is not simple by supposing the existence of a primitive element then trying to get a contradiction, specifically, that $x$ and $y$ are algebraically dependent. The problem is, they are algebraically dependent over $\mathbb{F}_p(x^p,y^p)$, because they vanish this (non-zero) polynomial $T_1^p-x^p+T_2^p-y^p\in\mathbb{F}_p(x^p,y^p)[T_1,T_2]$. If you know someway of proving that $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is not simple by contradiction, please answer it too.
57975
For which integer $d$ is the ring $\mathbb{Z}[\sqrt{d}]$ norm-Euclidean?   Here I'm referring to $\mathbb{Z}[\sqrt{d}] = \{a + b\sqrt{d} : a,b \in \mathbb{Z}\}$, not the ring of integers of $\mathbb{Q}[\sqrt{d}]$.  For $d &lt; 0$, it is easy to show that only $d = -1, -2$ suffice; but what about $d&gt;0$?  Thanks.
67273
I had to calculate the number of elements of this quotient ring:   $$R = \mathbb{Z}[X]/(X^2-3, 2X+4).$$   This is what I've got by myself and by using an internet source:  Writing the ring $R = \mathbb{Z}[X]/(X^2−3, 2X+4)$ as $\mathbb{Z}[X]/I$, we note that $I $ contains $2(X^2−3)−(X−2)(2X+4) = 2$. We then note that the generator $2X + 4$ is actually superfluous since $2X + 4 = 2(X + 2)$.   Now we can write $R = \mathbb{Z}[X]/(X^2 − 3, 2) = \mathbb{Z}[x]/((X+1)^2, 2)$, because $(X+1)^2=X^2+2X+1=X^2-3+2X+4$. The internet source states now the following:  $$ R \cong (\mathbb{Z}/2\mathbb{Z})[\alpha] \quad \text{with} \quad \alpha = X+1$$ I guess that $\mathbb{Z}/2\mathbb{Z}[\alpha]$ represents the set of dual numbers of the field $\mathbb{Z}/2\mathbb{Z}$. I see that $\alpha^2=0$, but what exactly implies the isomorphism? And does this mean that the quotient ring $R$ contains four elements?  If necessary, you can take at the site I used. (It's about page 5, exercise 4.3a.)    I thank you in advance for your answers.