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Let a magic square (row, columns and diagonals total the same amount) by a 3 by 3 matrix with entries $abc\ def\ ghi$. Let the polynomials inside the ideal all equal zero. Let ideal be generated by the polynomials $$I=\langle a+b+c-d-e-f,a+b+c-g-h-i, d+e+f-g-h-i, a+d+g-b-e-h, a+d+g-c-f-i, a+e+i-c-e-g, b+e+h-c-f-i \rangle \subset \mathbb{Q}[a,\ldots ,i].$$ Show that if $F \in I$, then $F$ is zero on any magic square.

Then show that $$(100a+10b+c)^2+(100d+10e+f)^2+(100g+10h+i)^2-(100c+10b+a)^2-(100f+10e+d)^2-(100i+10h+g)^2 \in I.$$

For the first part I know F would be a linear combination of the generators, but how do I show it zero?

For the second part I need to show it a linear combination of the generators. How would Groebner bases help me here?

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    The question is not correctly stated. For example the polynomial $a+b+c-d-e-f \in I$, and it's not the zero polynomial. I suggest looking up the _actual_ problem statement and then editing your question accordingly. – 2017-01-18
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    There's a small problem: if *any* in $F\in I$ is $0$, then the generators themselves are $0$. – 2017-01-18
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    For the second part, there are various ways. Are you using a software package to compute Groebner bases, or are you doing it by hand? – 2017-01-18
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    Yes I am using a computer software . But how does it help computing the basis ? – 2017-01-18
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    @Bob -- I haven't been following the thread, but looking at it again now, I see you've edited the question (as was necessary for the question to make sense), and I also see an answer as well as a lot of explanatory comments. So what is it that you still don't understand? – 2017-01-19
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    @Bob -- F evaluates to zero on the square since it's a weighted combination of your generators (where the weights are integer polynomials). Since the generators evaluate to zero on the square, so does F. – 2017-01-19
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    @Bob -- I don't know what you're asking. – 2017-01-19
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    For simplicity, I would use an extra variable, $s$, Then the 8 equations are $a+b+c=s$, $d+e+f=s$, ... etc.The ideal $I$ is the ideal generated by the polynomials $a+b+c-s$, $d+e+f-s$, ... etc. – 2017-01-19
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    @Bob --- No, I didn't say the 6 equations are redundant. I said any single one of them can be dropped, leaving the other 5. This is because the sum of all the rows is the same as the sum of all the columns. – 2017-01-19
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    @Bob -- I said you could delete either one row equation or one column equation. I didn't say you could delete either of the two diagonal equations. – 2017-01-19
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    @Bob -- The 3rd equation in your OP is implied by the first two -- can you see why? – 2017-01-19
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    @Bob -- The equations are much more self-evident if you use the extra variable $s$. – 2017-01-19
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    @Bob -- the last equation in your latest comment is redundant. _Please_ write them all down using $s$, as I recommended -- _then_ you can eliminate ones that are implied by the rest. And of course, at that point, if you wish, you can also eliminate the variable $s$. But why not just leave it in? – 2017-01-19
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    @Bob -- Now it's OK. – 2017-01-19
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    @Bob -- Good point. But even if the alphas belong to $Q[a,...,i]$, $F$ would still be a weighted combination of the polynomials $x_1,,,,,x_7$, all of which evaluate to zero on the square. – 2017-01-19
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    @Bob -- you've written so many things, I don't know what answer you're referring to? – 2017-01-19
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    @Bob -- I'll post an answer in a few minutes ... – 2017-01-19
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    @Bob -- Ok, I posted my version of a solution to your problem, using Groebner bases (as the problem specified). – 2017-01-19

2 Answers 2

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Here is an image of a Maple session.

tst.png

I used the symbol $U$ for the ideal generated by the linear polynomials corresponding to the equations based on the specification that your $3{\times}3$ matrix is a magic square. In your OP, you called it $I$, but in Maple, $I$ is a reserved symbol.

I used the symbol $p$ for the polynomial you called $F$ in your OP.

The goal is to show that $p \in U$.

Let $V = (U,p)$ be the ideal generated by $U$ together with $p$.

I computed $U$_bas, the Groebner basis for $U$, and $V$_bas, the Groebner basis for $V$. Since, as it turns out, the resulting Groebner bases are equal, it follows that $V=U$, hence $p \in U$, as was to be shown.

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    Just the choice of variables (I included $s$, you didn't) and the order of the variables. If we use the same variables and the same order, our results would be the same. – 2017-01-19
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    But it doesn't matter -- just let $J = (I,F)$ and see that the basis for $J$ is the same as the basis for $I$. If so, that automatically implies $J=I$, hence $F\in I$. – 2017-01-19
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    @Bob -- yes, it's fine. – 2017-01-19
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    In M2: `R=QQ[a..i,s,MonomialOrder=>Lex]` `I=ideal(a+b+c-s,d+e+f-s,g+h+i-s,a+d+g-s,b+e+h-s,c+f+i-s,a+e+i-s,c+e+g-s)` `((100*a+10*b+c)^2+(100*d+10*e+f)^2+(100*g+10*h+i)^2-(100*c+10*b+a)^2-(100*f+10*e+d)^2-(100*i+10*h+g)^2) % I` `-- 0` – 2017-01-19
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    @Jan-Magnus Økland -- Yes, that's better. No need to compute Groebner bases twice. Just have the Groebner basis package mod out the candidate polynomial by $I$ and see if the result is $0$. – 2017-01-19
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"The equations inside the ideal all equal zero" is confusing. You have a certain ideal $I$ of polynomials in indeterminates $a,b,\ldots,i$ (which for convenience I'll write as $X_1, \ldots, X_9$). There are no "equations inside the ideal", just polynomials, and only one of them is $0$. If this is over a field $k$ (perhaps $\mathbb C$ in your case), then the polynomials can be mapped to functions on $k^9$, and there is the affine variety $V(I)$ which is the set of $9$-tuples $(x_1, \ldots, x_9)$ such that $F(x_1,\ldots,x_9) = 0$ for all $F \in I$. I think the first question is to show that $(x_1, \ldots, x_9) \in V(I)$ if $F(x_1,\ldots,x_9) = 0$ for each of the generators of $I$.

The second question is wrong, unless you're working over a field of characteristic $2$, $3$ or $11$. For example, $$[a=-1, b=2, c=0,d=2,e=-1,f=0,g=0,h=0,i=1]$$ makes all your generators $0$, but $$\left( 100\,a+10\,b+c \right) ^{2}+ \left( 100\,d+10\,e+f \right) ^{2 }+ \left( 100\,g+10\,h+i \right) ^{2}- \left( 100\,c+10\,b+a \right) ^ {2}- \left( 100\,f+10\,e+d \right) ^{2}- \left( 100\,i+10\,h+g \right) ^{2} = 32076 $$

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    The second part was to show: if we read the rows as numbers forwards or backwards, we get the same sum of squares e.g. the 3 by 3 matrix $816 357 492 : 816^2+357^2+492^2=618^2+753^2+294^2$ – 2017-01-18
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    The counterexample still stands. Or if you want to make the entries positive, consider the "magic square" $$ \pmatrix{1 & 4 & 2\cr 4 & 1 & 2\cr 2 & 2 & 3\cr}$$. – 2017-01-19
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    I think your "magic square" needs the sums of rows, columns **and diagonals** equal. – 2017-01-19
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    Then it will work. Note that your generators are all linear equations, so you can solve them using linear algebra (with, in this case, $3$ arbitrary variables); substitute the result into the new polynomial and simplify. – 2017-01-19
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    No need for Groebner here. Just linear algebra. – 2017-01-19
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    You **can** do it with Groebner, but for linear polynomials that's using a sledgehammer to crack a nut. – 2017-01-19
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    The generators are the polynomials you wrote down in $I = \langle \ldots \rangle$. The new polynomial is the one you want to show is in $I$. – 2017-01-19
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    You left out polynomials telling you about diagonal sums. If $B$ is your Groebner basis and $Q$ your polynomial, in Maple I would use simplify(Q,B); – 2017-01-19