# Polynomials in n indeterminates and UFDs

Gold Member
MHB
In the introduction to Chapter 1 of his book "Introduction to Plane Algebraic Curves", Ernst Kunz states that the polynomial ring ##K[ X_1, X_2, \ ... \ ... \ , X_n]## over a field ##K## is a unique factorization domain ... ... but he does not prove this fact ...

Can someone demonstrate a proof of this proposition ... or point me to a text or online notes that contain a proof ...

Help will be appreciated ... ...

Peter

Mentor
2021 Award
You can show it for one variable and proceed by induction, because ##K[X_1,X_2, \dots , X_n] = K[X_1,X_2, \dots , X_{n-1}][X_n].##
For ##K[X]## you know that you can apply the Euclidean algorithm (division) to find all irreducible factors of a polynomial.

Math Amateur
Gold Member
MHB
You can show it for one variable and proceed by induction, because ##K[X_1,X_2, \dots , X_n] = K[X_1,X_2, \dots , X_{n-1}][X_n].##
For ##K[X]## you know that you can apply the Euclidean algorithm (division) to find all irreducible factors of a polynomial.

Hmm ... yes, get the general idea ... but not quite sure how the induction is set up and how exactly it proceeds ... Thinking ...

Thanks for the help ...

Peter

Mentor
2021 Award
Hmm ... yes, get the general idea ... but not quite sure how the induction is set up and how exactly it proceeds ... Thinking ...

Thanks for the help ...

Peter
The induction step is: If a ring ##R## is UFD, so is ##R[X]##.
(See http://math.harvard.edu/~waffle/ufds2.pdf [Broken]) This brief article also contains a pretty good overview on some frequent classes of rings. I think you should read it to gain a feeling for the concepts and a pool of examples (14 pages).

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Gold Member
MHB
The induction step is: If a ring ##R## is UFD, so is ##R[X]##.
(See http://math.harvard.edu/~waffle/ufds2.pdf [Broken]) This brief article also contains a pretty good overview on some frequent classes of rings. I think you should read it to gain a feeling for the concepts and a pool of examples (14 pages).

Thanks fresh_42 ... ... most helpful ... appreciate the help ...

Peter

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