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Can you give me a way to established a norm on $k[x_1,\ldots,x_2]$ for division with remainder such in $k[x].$

Thanks.

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I'm not sure what you're looking for. Are you looking for an analogue of a degree such that there is a euclidean division with remainder of lesser degree? This doesn't exist. Do you have something else in mind? –  Olivier Bégassat Oct 8 '11 at 4:14
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There isn't such a thing: if there was a division algorithn like the one you want, then all ideals would be principal, but $(x,y)$ is not a principal ideal. The best you can do is read about Groebner bases. For example, Cox et al.'s book "Ideals, varieties, algorithms". –  Mariano Suárez-Alvarez Oct 8 '11 at 4:14
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It exists even less in that case. –  Mariano Suárez-Alvarez Oct 8 '11 at 4:23
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Why do you write $k[x_1,...,x_2]$ instead of $k[x_1,x_2]$? Are you interested in the case of 2 variables or $n$ variables with any $n \geq 2$? If the latter, write $k[x_1,...,x_n]$. But for $n=2$ don't use the dots between $x_1$ and $x_2$. –  KCd Oct 8 '11 at 16:37
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Mariano, would you upgrade your comment to an answer? It is the most satisfying resolution of the search for an analogue of the Euclidean algorithm in several variables. –  KCd Oct 8 '11 at 16:39
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2 Answers 2

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Because $\rm\:k[x_1,\ldots,x_n]\:$ is no longer a PID for $\rm\:n > 1\:,\:$ one requires a generalization of the division algorithm. This is the foundation of various standard basis algorithms, e.g. Gröbner bases, which generalize both the Euclidean algorithm and row reduction. For a very nice introduction see David Bayer's Harvard thesis The Division Algorithm and The Hilbert Scheme, 1982, which was advised by Eisenbud and Mumford. Appended below is an excerpt from the introduction.

Such generalized division algorithms can also be viewed as special cases of critical pair algorithms employed in equational completion "algorithms", e.g. that of Knuth-Bendix. See here for more.

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Realizing that you are constrained by a keyboard without umlauts I took the liberty of adding one to your nice answer :-) –  Jyrki Lahtonen Oct 8 '11 at 20:09
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The more knowledgable commenters apparently won't write an answer, so I will type up a CW summary lest this question end up in the unanswered files.

The answer is 'No'. If a function like a Euclidean norm should exist on an integral domain, then all the ideals of the said ring would have to be principal. But e.g. the ideal $I$ of polynomials $p$ from $k[x,y]$, $k$ any field, such that $p(0,0)=0$ needs two generators, for example $x$ and $y$.

OTOH you can view $k[x_1,x_2]$ as a subring of $k(x_1)[x_2]$. This latter object is the ring of polynomials of $x_2$ with coefficients from the field of rational function $k(x_1)$. Because this is a ring of univariate polynomials with coefficients from a field, it is a Euclidean domain, and hence has a Euclidean norm and a division algorithm. When you apply the division algorithm of this ring to two elements $a,b\neq0$ of the subring $k[x_1,x_2]$ what will often happen is that the division process introduces non-constant polynomials of $k[x_1]$ into the denominators. So the end result (= the $(q,r)$ pair of a quotient and a remainder satisfying the familiar equation $a=qb+r$ with $\deg r < \deg b$) no longer belongs to the subring $k[x_1,x_2]$. Also observe that in the context of the ring $k(x_1)[x_2]$ the degree of a polynomial is determined by looking at powers of $x_2$ only. In this context any power of $x_1$ is just another 'constant coefficient'.

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