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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.

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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.
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  • $\begingroup$ 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. $\endgroup$ – Marine Galantin Jun 9 at 10:37
  • $\begingroup$ So the end of your answer implies the three statement are equivalent right? $\endgroup$ – Marine Galantin Jun 9 at 11:04
  • $\begingroup$ Yes, the three are equivalent. $\endgroup$ – user10354138 Jun 9 at 11:08

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