1. ## Simple Abstract Algebra

Looking for some help/explanations for these problems.

1. R is a UFD => LCM's exist.

2. R is Euclidean. If a, b are associates, then phi(a)=phi(b).

3. R is a PID and S is an integral domain
phi: R--->S is onto. show either phi is an isomorphism or S is a field.

4. R is a commutative ring with a 1. Show R[x] is PID iff R is a field.

5. Prove that if R is a UFD, then R[x_1,x_2,....,X_k] (polynomials in k indeterminates) is a UFD. Exhibit a nonprincipal ideal.

2. Originally Posted by mathisthebestpuzzle
1. R is a UFD => LCM's exist.
Let $S$ be a set of representatives for each irreducible element.
For $a\in R$ be non-zero and $\pi \in S$, define $d_{\pi}(a)=n$ so that $\pi^n |a$ but $\pi^{n+1} \not | a$.
Then $a = \alpha \prod_j \pi_j ^{d_{\pi_j}(a)}$ where $\alpha$ is unit.
Let $b$ be non-zero so $b= \beta \prod_j \pi_j^{d_{\pi_j}(b)}$.
Where the product is over $\pi_j \in S$.

Notice that $\prod_j \pi_j^{m_j}$ where $m_j = \max\{ d_{\pi_j}(a), d_{\pi_j}(b)\}$ is a LCM.

2. R is Euclidean. If a, b are associates, then phi(a)=phi(b)

3. R is a PID and S is an integral domain
phi: R--->S is onto. show either phi is an isomorphism or S is a field.
What is $\phi$?

4. R is a commutative ring with a 1. Show R[x] is PID iff R is a field.
It should be a known result that $R$ being a field implies that $R[x]$ is a PID.
Now if $R[x]$ is a PID, let $a\not = 0$ be in $R$.
For the other direction see this.

5. Prove that if R is a UFD, then R[x_1,x_2,....,X_k] (polynomials in k indeterminates) is a UFD.
If $R$ is an UFD then $R[x]$ is an UFD[/tex].
Once you know this you can induct.

Exhibit a nonprincipal ideal
Consider $\mathbb{Z}[x,y]$. And consider the set of all polynomials that have constant term equal to zero. This set is an ideal but it is a non-pricipal ideal.