Linked Questions

7
votes
3answers
3k views

How to show that $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 5)$ are non-isomorphic? [duplicate]

How to show that $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 5)$ are non-isomorphic as a ring? All I could manage to show is that, for any isomorphism $\phi:\mathbb(\sqrt 2)\to\mathbb(\sqrt 5),$ $\...
4
votes
4answers
7k views

How to show that $\mathbb Q(\sqrt 2)$ is not field isomorphic to $\mathbb Q(\sqrt 3).$ [duplicate]

How to show that $\mathbb Q(\sqrt 2)$ is not field isomorphic to $\mathbb Q(\sqrt 3)?$ My text provides the hint as: Any isomorphism from $\mathbb Q(\sqrt 2)\to\mathbb Q(\sqrt 3)$ is identity when ...
1
vote
2answers
2k views

field isomorphism and isomorphic as a vector space [duplicate]

Possible Duplicate: Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$? I have no idea about field theory much, I am studying it now, I have a doubt as a vector space over $\mathbb{Q}$, $\...
1
vote
2answers
775 views

An element not in a field extension [duplicate]

Possible Duplicate: Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$? Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were ...
5
votes
2answers
411 views

$\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m})$ iff $n=m$ [duplicate]

Page 105 of D. Burton's A First Course in Rings and Ideals reads It is not difficult to show that if (I'll call them $n$ and $m$ instead of $n_1$ and $n_2$) $n$, $m$ are square free integers, then $...
0
votes
1answer
337 views

field isomorphism [duplicate]

Possible Duplicate: Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$? Is it correct that $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are isomorphic as fields?
1
vote
0answers
373 views

Show that Q[x] /(x^2-2) is not isomorphic to Q[x] /(x^2-3) [duplicate]

Show that $$Q[x]/(x^2-2)$$ is not isomorphic to $$Q[x]/(x^2-3)$$ I was practicing some exercises in Hungerford's abstract algebra book, I could make them all but this one. I think I have to verify if ...
1
vote
0answers
113 views

Why is $\mathbb{Q}(\sqrt{2})$ not isomorphic to $\mathbb{Q}(\sqrt{3})$? [duplicate]

The question first asks to show that $\sqrt{3}$ has a root in $\mathbb{Q}(\sqrt{3})=\{a+b\sqrt{3}|a,b \in \mathbb{Q}\}$ but not in $\mathbb{Q}(\sqrt{2})$. This is I solved by assuming, towards a ...
10
votes
4answers
1k views

Prove that the fields $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4 \rangle$ are isomorphic

I have been stuck in this problem for some time now. Prove that $x^2+2$ and $x^2+x+4$ are irreducible over $\mathbb{Z}_{11}$. Also, prove further $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\...
17
votes
1answer
639 views

Examples of non-isomorphic fields with isomorphic group of units and additive group structure

YACP mentions in a comment that: There are examples of non-isomorphic fields $K$ and $L$ with $(K,+)\cong (L,+)$ and $(K^{\times} ,\cdot)\cong (L^{\times},\cdot)$ Can someone provide an example? ...
8
votes
4answers
2k views

Isomorphism of Vector spaces over $\mathbb{Q}$

From this post we see that $\mathbb{R}$ over $\mathbb{Q}$ is infinite dimensional. Similarly $\mathbb{C}$ over $\mathbb{Q}$ is also infinite dimensional, and I rememeber having solved a problem that $\...
2
votes
3answers
457 views

Showing that $\sqrt5$ is not in $\mathbb{Q}(\sqrt7)$

How can I prove that $\sqrt5$ is not in $\mathbb{Q}(\sqrt7)$ ? I can only think of trying to write $\sqrt5 = a+b\sqrt7$ (where $a,b$ are in $\mathbb{Q}$), but I can't think of a good reason that ...
2
votes
4answers
237 views

Are $\mathbb Q\left(\sqrt2\right)$ and $\mathbb Q\left(i\sqrt2\right)$ isomorphic?

I would like to show whether $\mathbb Q\left(\sqrt2\right)$ and $\mathbb Q\left(i\sqrt2\right)$ are isomorphic. I tried noting that $\pm\sqrt2$ and $\pm i\sqrt2$ are the roots of $x^4-4=\left(x^2+2\...
4
votes
2answers
1k views

Show that $\mathbb Q(\sqrt p) \not\simeq\mathbb Q(\sqrt q)$ [duplicate]

I'd like to show that for $p,q$ distinct primes, the extensions $\mathbb Q(\sqrt p),\mathbb Q(\sqrt q)$ are not isomorphic. I don't really have knowledge of the "high-level language" of algebraic ...
0
votes
4answers
307 views

Non-isomorphic quadratic fields

The result I'm trying to prove is: There is a bijection between square-free integers $d\neq 0,1$ and quadratic fields One direction is: If $d$ is a square-free integer $\neq 0,1$, then $\sqrt{d}...

15 30 50 per page