Abstract Algebra | k[x] is a PID

We prove that every ideal of the polynomial ring k[x], where k is a field, is principal. That is, k[x] is a principal ideal domain.

[adelalomar]

In Case 2: Do you mean r(x) = alpha and not p(x) = alpha.Or I misunderstood something. But Thanks for the proof

[gulnazshalgumbayeva]

Hi I am from Kazakhstan ;) Thanks for your videos, very helpful.

[danielgrunkin]

What about the case when I = (0)? In this case I guess we cannot require that deg(p(x)) >= 1 since that would mean that I /= (0).

[jaewoolee]

Could we have used the theorem for irreducible polynomial for the proof? Since quotient ring of ring over principal ideal generated by prime polynomials is field, principal ideal generated by prime polynomial is also a maximal one. Therefore, no more than one element could be used to generate an ideal thus ring is a PID.

[newwaveinfantry]

6:05 - Why is the ideal generated by alpha equal to k[x] and not just k? Is it because we're working in the context of k[x].

[keepitplainsimple]

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