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I prepare my qualifying exams for my Ms.C. and I do a lot of exams, but a few problems in there, I couldn't resolve, I hope can you help me.

1) Prove the following ring isomorphism $\mathbb{C}[x,y] \cong \{ (p(x),q(y)) \in \mathbb{C}[x]\times \mathbb{C}[y] | p(0)=q(0)\}.$ 2) Let $p$ a prime number and $\omega$ a $p$-th root of unit and let $L=\mathbb{Q}(\omega)$. In addition, suppose that $K$ is a subfield of $L$ with $[K:\mathbb{Q}]=\frac{p-1}{2},$ and $K=\mathbb{Q}(\omega+\omega^{-1})$. Then prove that $\sqrt{(-1)^{\frac{p-1}{2}}p} \in L$.

3) Let $F$ subfield of $\mathbb{C}$ that contain a $p$-th unit root, where $p$ is a prime number. Let $K$ a Galois extension of $F$ with $[K:F]=p$. Prove that $K$ is the splitting field of the polynomial $f(x)=x^p-a$ for some $a \in F$.

4) The last one, let $I,J$ ideal of a ring $R$ such that $R=I+J$. Prove that for all pair of elements $a,b \in R$, there exists $x \in R$ such that $x \equiv a \; (mod\; I)$ and $x \equiv a \; (mod \; J)$.

That's all, some this problems i resolve with a very powerful theorems and i think they have a solution very more simple. So, i wait for your answers and good luck with the problems, thank you.

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    @BillCook Mmm... your solution looks good, may be the problem has a type error in the exams. Thank for you solution, i think that you have right.2012-01-04

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In (4) I assume that you really want $x\equiv b\pmod J$. You have $a-b=i+j$ for some $i\in I,j\in J$; just let $x=a-i=b+j$. (This does not require that $R$ be unital.)

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    Nice solution, very elegant. Thank you, that's it what i need.2012-01-04
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(4) is the Chinese remainder theorem, which is proved easily and explicitly: write $1=i+j$ with $i\in I$ and $j\in J$. Then let $x=bi+aj$.