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<head><title>bruns -- Returns an ideal generated by three elements whose 2nd syzygy module is isomorphic to a given module</title>
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<div><h1>bruns -- Returns an ideal generated by three elements whose 2nd syzygy module is isomorphic to a given module</h1>
<div class="single"><h2>Synopsis</h2>
<ul><li><div class="list"><dl class="element"><dt class="heading">Usage: </dt><dd class="value"><div><tt>j= bruns M or j= bruns f</tt></div>
</dd></dl>
</div>
</li>
<li><div class="single">Inputs:<ul><li><span><tt>M</tt>, <span>a <a href="../../Macaulay2Doc/html/___Module.html">module</a></span>, a second syzygy (graded) module</span></li>
<li><span><tt>f</tt>, <span>a <a href="../../Macaulay2Doc/html/___Matrix.html">matrix</a></span>, whose cokernel is a second syzygy (graded) module</span></li>
</ul>
</div>
</li>
<li><div class="single">Outputs:<ul><li><span><tt>j</tt>, <span>an <a href="../../Macaulay2Doc/html/___Ideal.html">ideal</a></span>, a homogeneous ideal generated by three elements whose second syzygy module is isomorphic to M, or image f</span></li>
</ul>
</div>
</li>
</ul>
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<div class="single"><h2>Description</h2>
<div><div>This function takes a graded module M over a polynomial ring S that is a second syzygy, and returns a three-generator ideal j whose second syzygy is M, so that the resolution of S/j, from the third step, is isomorphic to the resolution of M. Alternately <tt>bruns</tt> takes a matrix whose cokernel is a second syzygy, and finds a 3-generator ideal whose second syzygy is the image of that matrix.</div>
<table class="examples"><tr><td><pre>i1 : kk=ZZ/32003
o1 = kk
o1 : QuotientRing</pre>
</td></tr>
<tr><td><pre>i2 : S=kk[a..d]
o2 = S
o2 : PolynomialRing</pre>
</td></tr>
<tr><td><pre>i3 : i=ideal(a^2,b^2,c^2, d^2)
2 2 2 2
o3 = ideal (a , b , c , d )
o3 : Ideal of S</pre>
</td></tr>
<tr><td><pre>i4 : betti (F=res i)
0 1 2 3 4
o4 = total: 1 4 6 4 1
0: 1 . . . .
1: . 4 . . .
2: . . 6 . .
3: . . . 4 .
4: . . . . 1
o4 : BettiTally</pre>
</td></tr>
<tr><td><pre>i5 : M = image F.dd_3
o5 = image {4} | c2 d2 0 0 |
{4} | -b2 0 d2 0 |
{4} | a2 0 0 d2 |
{4} | 0 -b2 -c2 0 |
{4} | 0 a2 0 -c2 |
{4} | 0 0 a2 b2 |
6
o5 : S-module, submodule of S</pre>
</td></tr>
<tr><td><pre>i6 : f=F.dd_3
o6 = {4} | c2 d2 0 0 |
{4} | -b2 0 d2 0 |
{4} | a2 0 0 d2 |
{4} | 0 -b2 -c2 0 |
{4} | 0 a2 0 -c2 |
{4} | 0 0 a2 b2 |
6 4
o6 : Matrix S <--- S</pre>
</td></tr>
<tr><td><pre>i7 : j=bruns M;
o7 : Ideal of S</pre>
</td></tr>
<tr><td><pre>i8 : betti res j -- the ideal has 3 generators
0 1 2 3 4
o8 = total: 1 3 5 4 1
0: 1 . . . .
1: . . . . .
2: . . . . .
3: . 3 . . .
4: . . . . .
5: . . . . .
6: . . 5 . .
7: . . . 4 .
8: . . . . 1
o8 : BettiTally</pre>
</td></tr>
</table>
<div>Here is a more complicated example, also involving a complete intersection. You can see that columns three and four in the two Betti diagrams are the same.</div>
<table class="examples"><tr><td><pre>i9 : kk=ZZ/32003
o9 = kk
o9 : QuotientRing</pre>
</td></tr>
<tr><td><pre>i10 : S=kk[a..d]
o10 = S
o10 : PolynomialRing</pre>
</td></tr>
<tr><td><pre>i11 : i=ideal(a^2,b^3,c^4, d^5)
2 3 4 5
o11 = ideal (a , b , c , d )
o11 : Ideal of S</pre>
</td></tr>
<tr><td><pre>i12 : betti (F=res i)
0 1 2 3 4
o12 = total: 1 4 6 4 1
0: 1 . . . .
1: . 1 . . .
2: . 1 . . .
3: . 1 1 . .
4: . 1 1 . .
5: . . 2 . .
6: . . 1 1 .
7: . . 1 1 .
8: . . . 1 .
9: . . . 1 .
10: . . . . 1
o12 : BettiTally</pre>
</td></tr>
<tr><td><pre>i13 : M = image F.dd_3
o13 = image {5} | c4 d5 0 0 |
{6} | -b3 0 d5 0 |
{7} | a2 0 0 d5 |
{7} | 0 -b3 -c4 0 |
{8} | 0 a2 0 -c4 |
{9} | 0 0 a2 b3 |
6
o13 : S-module, submodule of S</pre>
</td></tr>
<tr><td><pre>i14 : f=F.dd_3
o14 = {5} | c4 d5 0 0 |
{6} | -b3 0 d5 0 |
{7} | a2 0 0 d5 |
{7} | 0 -b3 -c4 0 |
{8} | 0 a2 0 -c4 |
{9} | 0 0 a2 b3 |
6 4
o14 : Matrix S <--- S</pre>
</td></tr>
<tr><td><pre>i15 : j1=bruns f;
o15 : Ideal of S</pre>
</td></tr>
<tr><td><pre>i16 : betti res j1
0 1 2 3 4
o16 = total: 1 3 5 4 1
0: 1 . . . .
1: . . . . .
2: . . . . .
3: . . . . .
4: . . . . .
5: . . . . .
6: . . . . .
7: . . . . .
8: . 1 . . .
9: . 2 . . .
10: . . . . .
11: . . . . .
12: . . . . .
13: . . . . .
14: . . . . .
15: . . 1 . .
16: . . 1 . .
17: . . 2 . .
18: . . 1 1 .
19: . . . 1 .
20: . . . 1 .
21: . . . 1 .
22: . . . . 1
o16 : BettiTally</pre>
</td></tr>
<tr><td><pre>i17 : j=bruns M;
o17 : Ideal of S</pre>
</td></tr>
<tr><td><pre>i18 : betti res j
0 1 2 3 4
o18 = total: 1 3 5 4 1
0: 1 . . . .
1: . . . . .
2: . . . . .
3: . . . . .
4: . . . . .
5: . . . . .
6: . . . . .
7: . . . . .
8: . 1 . . .
9: . 2 . . .
10: . . . . .
11: . . . . .
12: . . . . .
13: . . . . .
14: . . . . .
15: . . 1 . .
16: . . 1 . .
17: . . 2 . .
18: . . 1 1 .
19: . . . 1 .
20: . . . 1 .
21: . . . 1 .
22: . . . . 1
o18 : BettiTally</pre>
</td></tr>
</table>
<div>In the next example, we perform the "Brunsification" of a rational curve.</div>
<table class="examples"><tr><td><pre>i19 : kk=ZZ/32003
o19 = kk
o19 : QuotientRing</pre>
</td></tr>
<tr><td><pre>i20 : S=kk[a..e]
o20 = S
o20 : PolynomialRing</pre>
</td></tr>
<tr><td><pre>i21 : i=monomialCurveIdeal(S, {1,3,4,5})
2 2 2 3 2
o21 = ideal (d - c*e, b*d - a*e, c - b*e, b*c - a*d, a*c*d - b e, b - a c)
o21 : Ideal of S</pre>
</td></tr>
<tr><td><pre>i22 : betti (F=res i)
0 1 2 3 4
o22 = total: 1 5 8 5 1
0: 1 . . . .
1: . 4 2 . .
2: . 1 6 5 1
o22 : BettiTally</pre>
</td></tr>
<tr><td><pre>i23 : time j=bruns F.dd_3;
-- used 0.229965 seconds
o23 : Ideal of S</pre>
</td></tr>
<tr><td><pre>i24 : betti res j
0 1 2 3 4
o24 = total: 1 3 6 5 1
0: 1 . . . .
1: . . . . .
2: . . . . .
3: . . . . .
4: . 3 . . .
5: . . . . .
6: . . . . .
7: . . 2 . .
8: . . 4 5 1
o24 : BettiTally</pre>
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<div class="single"><h2>See also</h2>
<ul><li><span><a href="_bruns__Ideal.html" title="Returns an ideal generated by three elements whose 2nd syzygy module agrees with the given ideal">brunsIdeal</a> -- Returns an ideal generated by three elements whose 2nd syzygy module agrees with the given ideal</span></li>
</ul>
</div>
<div class="waystouse"><h2>Ways to use <tt>bruns</tt> :</h2>
<ul><li>bruns(Matrix)</li>
<li>bruns(Module)</li>
</ul>
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