- Hilbert's basis theorem
In

mathematics ,**Hilbert's basis theorem**states that every ideal in the ring of multivariate polynomials over a field is finitely generated. This can be translated intoalgebraic geometry as follows: everyalgebraic set over a field can be described as the set of common roots of finitely many polynomial equations. The theorem is named for the Germanmathematician David Hilbert who first proved it in1888 .Hilbert produced an innovative proof by contradiction using

mathematical induction ; his method does not give analgorithm to produce the finitely many basis polynomials for a given ideal: it only shows that they must exist. One can determine basis polynomials using the method of Gröbner bases.**Proof**A slightly more general statement of Hilbert's basis theorem is: if "R" is a left (respectively right)

Noetherian ring , then thepolynomial ring "R" ["X"] is also left (respectively right) Noetherian.Proof: For $f\; in\; R\; [x]$, if $f=sum\_\{k=0\}^na\_kx^k$ with $a\_n$ not equal to "0", then $deg\; f\; :=\; n$ and $a\_n$ is the leading coefficient of "f". Let "I" be an ideal in "R [x] " and assume "I" is not finitely generated. Then inductively construct a sequence$f\_1,f\_2,...$ of elements of "I" such that $f\_\{i+1\}$ has minimal degree among elements of $Isetminus\; J\_i$, where $J\_i$ is the ideal generated by $f\_1,...,f\_i$. Let $a\_i$ be the leading coeffecient of $f\_i$ and let "J" be the ideal of "R" generated by the sequence $a\_1,a\_2,...$. Since "R" is Noetherian there exists "N" such that "J" is generated by $a\_1,...,a\_N$. Therefore $a\_\{N+1\}\; =\; sum\_\{i=1\}^Nu\_ia\_i$ for some $u\_1,...,u\_N\; in\; R$. We obtain a contradiction by considering $g\; =\; sum\_\{i=1\}^Nu\_if\_ix^\{n\_i\}$ where $n\_i\; =\; deg\; f\_\{N+1\}\; -\; deg\; f\_i$, because $deg\; g\; =\; deg\; f\_\{N+1\}$ and their leading coefficients agree, so that $f\_\{N+1\}\; -\; g$ has degree strictly less than $deg\; f\_\{N+1\}$, contradicting the choice of $f\_\{N+1\}$. Thus "I" is finitely generated. Since "I" was an arbitrary ideal in "R [x] ", every ideal in "R [x] " is finitely generated and "R [x] " is therefore Noetherian.

or a constructive proof:

Given an ideal J of R [X] let L be the set of leading coefficients of the elements of J. Then L is clearly an ideal in R so is finitely generated by a(1),...,a(n) in L, and there are f(1),...,f(n) in J with a(i) being the leading coefficient of f(i). Let d(i) be the degree of f(i) and let N be the maximum of the d(i)'s. Now for each k=0,...,N-1 let L(k) be the set of leading coeficients of elements of J with degree atmost k. Then again, L(k) is clearly an ideal in R so is finitely generated by a(k,1),...,a(k,m(k)) say. As before, let f(k,i) in J have leading coefficient a(k,i). Let H be the ideal in R [X] generated by the f(i)'s and f(k,i)'s. Then surely H is contained in J and assume there is an element f in J not belonging to H, of least degree d, and leading coefficient a. If d is larger or equal to N then a is in L so, a=r(1)a(1)+...+r(1)a(n) and g= $r(1)X^\{d-d(1)\}f(1)+...+r(n)X^\{d-d(n)\}f(n)$ is of the same degree as f and has the same leading coefficient. Since g is in H, f-g is not,which contradicts the minimality of f. If on the other hand d is strictly smaller than N, then a is in L(d), so a=r(1)a(d,1)+...+r(m(d))a(d,m(d)). A similar construction as above again gives the same contradiction. Thus, J=H, which is finitely generated. QED.

**Other**The Mizar project has completely formalized and automatically checked a proof of Hilbert's basis theorem in the [

*http://www.mizar.org/JFM/Vol12/hilbasis.html HILBASIS file*] .**References*** Cox, Little, and O'Shea, "Ideals, Varieties, and Algorithms", Springer-Verlag, 1997.

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