$A$ is a field iff $A[t]$ is euclidean.

Question

I'm almost sure the question has already been asked but i don't know the english terminologies...

I have in my lecture that :

$A$ a ring.

$A$ is a field iff $A[t]$ is principal.

I'm anoyed because I think we can do better. It seems that $ \mathbb R [t] $ is euclidean. So, shoudln't be that theorem stated like that :

$A$ is a field iff $A[t]$ is euclidean.

what do you think ?

I think that in my lectures, we are dealing with rings with a unity, commutative and integral.

Answer 1

Assuming $t$ is transcendence over $A$.

Stated as

$A$ is a field iff $A[t]$ is Euclidean.

is weaker than

$A$ is a field iff $A[t]$ is PID.

since we lost the ability to go $A[t]$ PID $\implies A$ a field.

I think the solution you are looking at is to have three equivalent statements

• $A$ is a field.
• $A[t]$ is Euclidean.
• $A[t]$ is principal.