Specifically what does stuff like $\mathbb{Q}[x] / \langle x^{2} - 1\rangle$ generally means? I thought it meant the set of polynomials with rational coefficients divided by $x^{2} - 1$, but when I see stuff like $\mathbb{Z}[x]/\langle x^{2} + 1, 5\rangle$ I have no idea what it means. Also, how do you generally find what $R/I$ is isomorphic to?
120588
In general, I struggle working with quotients of polynomial rings, so I was hoping someone might be able to help with the following exercise:  Show that $\mathbb{Q}[x]/(x^2-2x)$ is isomorphic to $\mathbb{Q}[x]/(x^2-1)$, but the latter is not isomorphic to $\mathbb{Q}[x]/(x^2)$.  Mostly I want to understand what these quotients all &quot;look like.&quot; We did few examples in class and few homework problems with such quotients, and the ones we do work with are much simpler (typically involving $\mathbb{Z}_2$ or $\mathbb{Z}_3$ or $\mathbb{R}$ instead of $\mathbb{Q}$). Any help is appreciated, thank you!
80032
Why must $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ -- the set of all polynomials in $\sqrt{2}$ and $\sqrt{3}$ with rational coefficients  -- contain multiplicative inverses?   I have gathered that every element of $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ takes form $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{2}\sqrt{3}$ for some $a,b,c,d \in \mathbb{Q}$, and I have shown that $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are both fields, but it's not clear to me why this allows for general multiplicative inverses to exist in $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.
55249
I've seen similar questions asked here, but I've not been able to find a comprehensive answer.  I know that for a ring $R$, $R[X]$ denotes the ring of polynomials over $R$ and $R(X)$ denotes the field of fractions of $R[X]$.  But if $\alpha \in S$, where $S \supseteq R$ are rings, what is the distinction between $R[\alpha]$ and $R(\alpha)$?  It is my understanding that if $R$ is a field then $R[\alpha] \cong R(\alpha)$, but that this is not generally true for any ring.  Is this correct?  Adding to my confusion is the fact that $\mathbb Z[i]$ and $\mathbb Z(i)$ are used interchangeably, despite $\mathbb Z$ not being a field.  However, it's clear to me that they are equivalent ( i.e. $\mathbb Z[i] = \mathbb Z(i) = \{ a + bi | a,b \in \mathbb Z \}$); is this because $i$ is algebraic over $\mathbb Z$?  Or for some other reason?  Thanks for your help!
86430
Problem: Consider the subfield $$E = \mathbb{Q} \big(\left\{\sqrt[n]{2} \mid n \in \mathbb{N} \setminus \left\{0\right\} \right\} \big)$$ of $\mathbb{R}$. Prove that $[E: \mathbb{Q}]$ is not finite.  Attempt: I proved that $E$ is algebraic over $\mathbb{Q}$. Because let $u \in E$. Then there is a $n \in \mathbb{N} \setminus \left\{0\right\}$ such that $u = \sqrt[n]{2}$. Then $f(x) = x^n - 2 \in \mathbb{Q}[x]$ is the minimal polynomial of $u$ over $\mathbb{Q}$ such that $f(u) = 0$. Hence $E$ is algebraic over $\mathbb{Q}$.  I now want to prove the degree of extension is not finite. I assume it is finite. Suppose then $[E: \mathbb{Q}] = k \in \mathbb{N}$. I don't know how to derive a contradiction from this.   I know that if $u$ is algebraic over $\mathbb{Q}$ and $f$ is the minimal polynomial of $u$ over $\mathbb{Q}$, then $[\mathbb{Q}(u) : \mathbb{Q}] = \deg(f)$.   Help/suggestions are appreciated.
18653
I'm trying to show that $\langle1-i\rangle$ is a maximal ideal of $\mathbb{Z}[i]$.  I started by assuming there is some ideal $A$ that properly contains $\langle1-i\rangle$, and then I want to show that $1 \in A$, and then I will have that $A= \mathbb{Z}[i] $, so $\langle1-i\rangle$ is maximal.  Is this the right approach?  Thanks  Second Attempt: Because $\forall a+bi \in \mathbb{Z}[i]$, we have $a+bi = (a+b) -b(1-i)$, then we have $a+bi - (a+b) = -b(1-i) \in \langle1-i\rangle$, which implies $a+bi +\langle1-i\rangle = a+b +\langle1-i\rangle$.  So every coset is of the form $c +\langle1-i\rangle$, where $c\in \mathbb{R}$.  Furthermore, $-i(1-i)^2=2 \in \langle1-i\rangle$. Write $c = 2q + r$, where $q,r\in \mathbb{R}$ and $r=0,1$.  Then because $2 \in \langle1-i\rangle$, we have $2q\in \langle1-i\rangle$ and therefore $2q= 2q+r-r = c-r\in \langle1-i\rangle$, so $c+\langle1-i\rangle = r+\langle1-i\rangle$, and every coset is of the form $0 + \langle1-i\rangle$ or $1 +\langle1-i\rangle$. Thus, $1 +\langle1-i\rangle$ is the only nonzero element, and is invertible with itself as its inverse.  Thus, $\mathbb{Z}[i]/\langle1-i\rangle$ is a field with two elements, and by a theorem, $\langle1-i\rangle$ must be maximal.  Is this correct?
75449
Just would like to see if I'm right.  Proof: Let $G$ be generated by $x$ (Let $G=\langle x \rangle$ for some $x$ in $G$.) We wish to show that there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$. First, we will show that for all $x^n$ in $G$, $f(x^n)=f^n(x)$. We proceed by induction.   Base case: $n=1$. $f(x*e_G)=f(x)f(e_G)=f(x)$. Now, assume that $f(x^k)=f^k(x)$ for some arbitrary positive integer k. We show that $f(x^{k+1})=f^{k+1}(x)$. Observe that $f(x^{k+1})=f(x^kx)=f(x^k)f(x)=f^k(x)f(x)=f^{k+1}(x).$   Next we will show that for all $x^{-n}$ in $G$, $f(x^{-n})=f^{-n}(x)$. We proceed by induction. Base case: $n=-1$. $f(x^{-1}*e_G)=f(x^{-1})f(e_G)=f(x)$. Now, assume that $f(x^{-k})=f^{-k}(x)$ for some arbitrary positive integer k. We show that $f(x^{-k-1})=f^{-k-1}(x)$. Observe that $f(x^{-k-1})=f(x^{-k}x)=f(x^{-k})f(x)=f^{-k}(x)f(x)=f^{-k-1}(x).$ Because $f$ is onto, $Im(f)=G'$ and therefore $Im(f)=(...,f^{-2}(x),f^{-1}(x),e_G,f(x),f^2(x),...)=G'$.  Thus, there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$ and the proof is complete by the principle of mathematical induction. Q.E.D.  I have a feeling the proof shouldn't be this long. Any suggestions for a shorter, cleaner proof?
99271
Let $d\in \Bbb{Z}$ be a square free in, I need to find an isomorphism between $\Bbb{Z}[\sqrt{d}]$ and a quotient of $\Bbb{Z}[X].$ I Would guess it's $$\Bbb{Z}[\sqrt{d}]\simeq\Bbb{Z}[X]/\langle X^2-d\rangle.$$ But I need to prove it. So, as $X^2+1$ is monic, I can perform the euclidean algorithm: all polynomial $\Bbb{Z}[X]$  can be written $$P=(X^2-d)Q+(aX+b)$$ for $Q\in\Bbb{Z}[X].$ We have $P(\sqrt{d})=0$ if and only if $a+\sqrt{d}b=0.$  If $d&lt;0$ then $\sqrt{d}=i\sqrt{-d}$ so $a=b=0.$      If $d&gt;0$ I am not sure how can I prove that $a=b=0$, I can do it for some special ("easy") case.    More than a solution, I need to understand what's going on to get the idea of the proof. I can conclude, but
124076
Let $V$ be the vector space over $\mathbb{R}$ consisting of all polynomials of degree $\leqslant n$. If $f,g\in V$, let $$\langle f,g \rangle = \int_\limits{0}^{1}f(t)g(t)dt$$ Find the matrix of this scalar product with respect to the basis $\{1,t,\dots,t^n\}$.  What does it mean to find a matrix of the scalar product? Thanks in advance!
18159
In this  we saw isomorphism of vector spaces over $\mathbb{Q}$. Just came across this question:   Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?   I know these as $\mathbb{Q}$-Vector spaces, are isomorphic from the  linked post. But as fields are they isomorphic? I neither know how to prove it nor how to disprove it.
109765
For some of you, this question is going to seem extremely basic.  I think I understand what an ideal is. Such as $\langle 2, 1 + \sqrt{-5} \rangle$, it consists of all numbers in this ring of the form $2a + (1 + \sqrt{-5})b$.  But then what is $\langle 2, 1 + \sqrt{-5} \rangle^2$? My first thought was $\langle 4, -4 + 2 \sqrt{-5} \rangle$, but that seems wrong somehow. I also had something in the back of my mind saying $\langle 2 \rangle$, but I'm not sure about that one either.  Then I thought about trying to figure out $\langle 2, 1 + \sqrt{-5} \rangle \langle 2, 1 + \sqrt{-5} \rangle$ when I realized I don't actually understand how to multiply ideals to begin with.  I'm only using $\langle 2, 1 + \sqrt{-5} \rangle$ as an example (though that does draw in one question identified as similar that looks much more relevant than all the questions identified as "Questions that may already have your answer"). In a principal ideal domain, would it be correct to think that $\langle a \rangle \langle b \rangle = \langle ab \rangle$?  Any help would be much appreciated.
81649
I found an argument online that the two fields are not isomorphic but I can't make sense of the argument. You find it .  These are the things I'm confused about when it comes to the argument.     It says that the field $\mathbb Q[x]/(x^2+1)$ contains a root of the polynomial $x^2+1$. I suppose this is the congruence class $[x]$? But this is not the same as the zero element of the field $\mathbb Q[x]/(x^2+1)$, right?   It says that every isomorfism takes $0$ to $0$. I interpret this as the zero-element in $\mathbb Q[x]/(x^2+1)$ (which I interpret to be $[0]$) is taken to the zero-element in $\mathbb Q[x]/(x^2+2)$ denoted $[0]_2$. But in the argument it says that this means that if $\mathbb Q[x]/(x^2+1)$ has a root of the polynomial, then also $\mathbb Q[x]/(x^2+2)$ must have a root of this same polynomial. Why is that?     Greatful for any help!
15497
This problem is from a practice exam I was working on.     What is the cardinality of the quotient $\mathbb{Z}[x]/(x^2-3,2x+4)$ ?   Thoughts. If I find a ring that is easier to handle then this then I can go from there. So I think this is isomorphic to $\mathbb{Z}_2[x]/(x^2-3,x+2)$. And then I am stuck.  Am I correct so far? What should I be looking to do from here?  Thanks for your time and your answers.
77054
I am trying to show the isomorphism  $$\mathbb Z^n/\langle (a,...,a) \rangle \cong \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle.$$  I've tried to define $\psi:\mathbb Z^n \to \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$ an epimorphism with $\ker(\psi)=\langle (a,...,a) \rangle$ so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.
101011
Let $(\mathbb{Z}[x],+)$ be the additive group of polynomials with integer coefficients and $ (\mathbb{Q}_{&gt;0},\cdot)$ the multiplicative group of positive rationals. Show these groups are isomorphic.   Thanks
69869
What is the general form of a rational function which has absolute value $1$ on the circle $|z|=1$? In particular, how are the zeros and poles related to each other?   So, write $R(z)=\dfrac{P(z)}{Q(z)}$, where $P,Q$ are polynomials in $z$. The condition specifies that $|R(z)|=1$ for all $z$ such that $|z|=1$. In other words, $|P(z)|=|Q(z)|$ for all $z$ such that $|z|=1$. What can we say about $P$ and $Q$?
86616
I am trying to prove that $\mathbb Z[i]/ \langle 1+2i \rangle$ is isomorphic to  $\mathbb  Z_5$.   The only thing that came to my mind was trying to apply the first isomorphism theorem using an appropiate function. If I consider the euclidean function $N: Z[i] \setminus \{0 \}  \to  \mathbb N$ defined as $N(a+bi)=a^2+b^2$, then I can express any element $z$ in $\mathbb Z[i]$ as $z=(1+2i)q+r$ with $N(r)=1,2,3,4$ or $r=0$.   If I define $f:\mathbb Z[i] \to \mathbb  Z_5$ as $f(x=(1+2i)q_x+r_x)=\overline{r_x}$, then it is clear that $f(x)=0$ if and only if $x \in  \langle 1+2i \rangle$. The problem with this function is that it doesn't satisfy $f(x+y)=f(x)+f(y)$ and it is not surjective for if $r_x=f(x)=3 (5)$, then if $r_x=a+bi$, we have $a^2+b^2=3$, which is absurd.  I don't know what else to do, any suggestions would be appreciated.
73025
So I want to find the isomorphism $\phi$ that takes $F = \mathbb{Z}_3/\langle x^3 - x - 1\rangle$ to $E = \mathbb{Z}_3/\langle x^3 - x + 1\rangle$.  I understand that these are both finite fields of size 3^3 since $x^3 - x - 1$ and $x^3 - x + 1$ are both irreducible polynomials of degree 3.  I feel like a good start would be to take the subfield $\mathbb{Z_3}$ in $F$ to the subfield $\mathbb{Z_3}$ in $E$.  Or maybe find the right generator of each multiplicative group of size $3^3-1$ of each field and send those to each other.