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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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Are you sure about (1)? Consider the map $\varphi(f(x,y))=(f(x,0),f(0,y))$. This is obviously a ring homomorphism (preserves +,$\times$, sends 1 to 1). Onto isn't hard: $(g(x),h(y))$ such that $g(0)=h(0)$ then $g(x)=g'(x)x+c$ and $h(y)=h'(y)y+c$ so $f(x,y)=g'(x)x+h'(y)y+c$ maps to $(g(x),h(y)$. Finally $f(x,y)$ maps to $(0,0)$ iff $f(x,0)=0$ and $f(0,y)=0$. Thus $f(x,y)=xyf'(x,y)$. So $\mathbb{C}[x,y]/(xy) \cong \{(p(x),q(y))\in \mathbb{C}[x] \times \mathbb{C}[y] \;|\; p(0)=q(0)\}$. –  Bill Cook Jan 4 '12 at 1:21
@BastianGalasso-Diaz Two people have already told you how to do number (4). In fact using those methods mentioned prove that $I,J$ are ideals of a ring $R$ such that $I + J = R$, then $IJ = I \cap J$ (I assume you know what the ideal $IJ$ looks like). –  user38268 Jan 4 '12 at 4:23
For (3), some relevant phrases are "Kummer theory", "cyclic extension", and "Hilbert Theorem 90". More later if no one takes this up during my walking of the dog. –  Dylan Moreland Jan 4 '12 at 5:02
@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. –  Bastian Galasso-Diaz Jan 4 '12 at 15:01

2 Answers 2

up vote 4 down vote accepted

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. –  Bastian Galasso-Diaz Jan 4 '12 at 14:53

(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 $x=bi+aj$.

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