Part of proving R[x] is not a Euclidean domain given R is not a field

Question

Given: Let $R$ be an integral domain. Assume that $R$ is not a ﬁeld.

Previous part of the question asked me to prove that $(a)$ is not equal to $R$, which I have already done.

Let $I = (a, X)$ be the ideal of $R[X]$ generated by $a$ and $X$. Then $I =\{P\cdot a+Q\cdot X: P,Q\in R[X]\}$. Assume that $I$ is principal, so there is some $p\in R[X]$ such that $I = (p)$. Using the fact that $a\in (p)$, show that $p$ has degree $0$. (So $p$ is an element of R.)

So far, I have: since $a\in (p)$, $a$ is also an element of $I$. Therefore for some $P,Q ∈ R[X]$, $a=P\cdot a+Q\cdot X$.

From here, how do I show that $a$ has degree $0$, hence, $p$ has degree $0$. Am I on the right track?

Oh, I see, $(a)$ is the ideal generated by $(a)$, and you've proved that for some $a$, $(a)\neq R$? — Thomas Andrews, Nov 12 '16 at 02:55
$a\in I$ is obvious: $a=1\cdot a + 0\cdot X$. Also, $a$ has degree $0$ by definition - how is degree defined in $R[X]$? — Thomas Andrews, Nov 12 '16 at 02:57
@ThomasAndrews deg(pq)=deg(p)+deg(q). So, since deg(a)=0, and p divides a, deg(p)=0? — Monica, Nov 12 '16 at 03:02
from here, how do I show that p is a unit using the fact that X is an element of (p)? — Monica, Nov 12 '16 at 03:03
This may help you: http://math.stackexchange.com/questions/940443/r-is-a-commutative-integral-ring-rx-is-a-principal-ideal-domain-imply-r — Xam, Nov 12 '16 at 03:04
No, deg(pq) is not always the sum of the degrees. That's only true if the ring is an integral domain. — Thomas Andrews, Nov 12 '16 at 03:05
@ThomasAndrews yes and we're working in an integral domain as stated above. — Monica, Nov 12 '16 at 03:06

score 1 · Answer 1 · answered Nov 12 '16 at 03:09

To show: $R$ is not a field $\implies R[x]$ not a Euclidean Domain

is equivalent to show that $R[x]$ is a Euclidean Domain $\implies R$ is a field.

Proof:

$R[x]$ is a Euclidean Domain $\implies R[x]$ is a Principal Ideal Domain $\implies R $ is a field

Part of proving R[x] is not a Euclidean domain given R is not a field

1 Answers1