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

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.

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.
• so what you mean is that $A[t]$ is eucledean iff $A[t]$ is principal ? Because one direction is obvious (the direct one) and the other is implied by what you said. – Marine Galantin Jun 9 at 10:37
• So the end of your answer implies the three statement are equivalent right? – Marine Galantin Jun 9 at 11:04
• Yes, the three are equivalent. – user10354138 Jun 9 at 11:08