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1/ If $d$ is not a square in $\mathbb{Q}$, show that $\mathbb{Q}[\sqrt{d}]\approxeq\mathbb{Q}[X]/<X^2-d>$ where $(X^2 - d)$ is the principal ideal of $\mathbb{Q}[X]$ generated by $X^2 - d$.

2/ If $d_1, d_2$, and $d_1/d_2$ are not squares in $\mathbb{Q}\backslash\left\{ 0\right\} $, show that $\mathbb{Q}[\sqrt{d_1}]$ and $\mathbb{Q}[\sqrt{d_2}]$ are not isomorphic.

3/ Let $R_1 = \mathbb{Z}_5[X]/<X^2 - 2>$ and $R_2 = \mathbb{Z}_5[X]/(X^2 - 3)$, then the statement $R_1\approxeq R_2$ true or false?.

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Is this homework? (The phrasing makes it sound like it is). What have you tried? –  Tobias Kildetoft Oct 26 '12 at 9:28
    
$\varphi_d:\mathbb{Q}[X]\to\mathbb{Q}[\sqrt{d}]$ give by $\varphi_d(a_0+a_1X+\ldots+a_nX^n)=a_0+a_1\sqrt{d}+\ldots+a_n(\sqrt{d})^n$, then $\varphi_d$ is homomorphism, and $\varphi_d(X^2-d)=0$ then $<X^2-d> \subset Ker \varphi_d$, but why $<X^2-d> = Ker \varphi_d$ –  Muniain Oct 26 '12 at 10:14
    
nobody help me? –  Muniain Oct 26 '12 at 16:30
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1 Answer

Hints:

1) Define

$$\phi:\Bbb Q[x]\to\Bbb Q[\sqrt d]\,\,\,,\,\,\phi(f(x)):=f(\sqrt d)$$

and show the above is a ring homomorphism, find its kernel and use the first isomorphism theorem for rings.

Further hint for this one: You may want to show that for $\,g(x)\in\Bbb [x]\,$ ,

$$g(\sqrt d)=0\Longleftrightarrow (x^2-d)\mid g(x)$$

2) Prove that a ring homomorphism as wanted sends a root of some irreducible polynomial to another root of the same irred. polynomial.

$${}$$

General hint Using the same notation as in the question, $\,\Bbb Q[\sqrt d]=\Bbb Q(\sqrt d)\,$ , i.e.: this ring is actually a field.

$${}$$

3) Prove that if $\,F_p:=\Bbb Z/p\Bbb Z\,=$ the prime with $\,p \,$elements , $\,p\,$ a prime, then

$$f(x)\in\Bbb F_p[x]\;\;\text{is irreducible}\,\,\,\Longleftrightarrow \,\,\text{the ideal}\,\,(f(x))\in\Bbb F_p[x]\,\,\,\text{is maximal}\,$$

$$\Longleftrightarrow \,\Bbb F_p[x]/(f(x))\,\,\,\text{is a field}$$

and show that in this last case, the field has $\,p^{\deg f}\,$ elements . Use then that there exists one single field, up to isomorphism, of any finite order (which, of course, is always a power of a prime)

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I can't imagine what did you say for 2) and 3) –  Muniain Oct 29 '12 at 10:15
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