While exploring concepts related to field extensions, I came across the following statement:

"Let $K$ be an extension field of $F$ and $u\in K$ an algebraic element over $F$. Consider the homomorphism $F[x]\to K$ defined by evaluation of a polynomial at $u$. Since the image is a subring of a field, the kernel is a prime ideal in the PID $F[x]$"

How does one prove the final sentence?


Hint: Every subring of a field is a domain.The image is isomorphic to $F[x]/kernel$.

  • $\begingroup$ How does that imply that the kernel is a prime ideal? What am I missing? $\endgroup$ – Romain S Apr 7 '17 at 0:41
  • $\begingroup$ If $\;R\;$ is a unitary commutative ring, then $\;R/I\;$ is an integral domain iff $\;I\le R\;$is a prime ideal. $\endgroup$ – DonAntonio Apr 7 '17 at 0:52

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