Polynomials in n indeterminates and UFDs

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SUMMARY

The polynomial ring K[X1, X2, ..., Xn] over a field K is a unique factorization domain (UFD), as established by Ernst Kunz in "Introduction to Plane Algebraic Curves." A proof can be constructed using induction, starting from the case of a single variable polynomial ring K[X]. The induction step confirms that if a ring R is a UFD, then R[X] is also a UFD. For further reading, the article at http://math.harvard.edu/~waffle/ufds2.pdf provides a comprehensive overview of UFDs and examples.

PREREQUISITES
  • Understanding of unique factorization domains (UFDs)
  • Familiarity with polynomial rings, specifically K[X]
  • Knowledge of mathematical induction
  • Basic concepts of irreducible factors in polynomials
NEXT STEPS
  • Study the proof of UFDs in polynomial rings using induction
  • Read the article on UFDs at http://math.harvard.edu/~waffle/ufds2.pdf
  • Explore the graduate algebra book by Auslander and Buchsbaum, particularly Chapter 5
  • Examine the class notes available at http://alpha.math.uga.edu/%7Eroy/844-1.pdf for additional insights
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Mathematicians, algebraists, and students studying unique factorization domains and polynomial rings, particularly those interested in advanced algebra concepts.

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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
 
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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.
 
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fresh_42 said:
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
 
Math Amateur said:
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 ) 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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fresh_42 said:
The induction step is: If a ring ##R## is UFD, so is ##R[X]##.
(See http://math.harvard.edu/~waffle/ufds2.pdf ) 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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