Understanding Determinants of Matrices over Commutative Rings

[Bleys]

There is a small outline in a book about finding the determinant of a matrix over an arbitrary commutative ring. There are a few things I don't understand; here it is:

'Let R be a commutative ring with a subring K which is a field. We consider the matrix X = [itex](x_{ij})[/itex] whose entries are independent indeterminates over K (that is, elements of the ring S = K[x_{11),...,x_{nn}]). Now S is an integral domain so we can compute the determinant of X in Quot(S). Now we obtain the determinant of an arbitrary matrix over R by sustituting elements of R for the indeterminantes (this substitution is a ring homomorphism from S to R)'

The part that confuses me is how are you able to use the substitution homomorphism for R? Isn't K a subring of R (possibly proper), and S is over K not R? I might be rusty on my field theory... Why is the substitution homomorphism even from S to R and not from Quot(S) to R?

Any help is appreciated

 

[Hurkyl]

This is a... weird way to go about things. Maybe it would make more sense in context.


But anyways, the main point is that the determinant is a polynomial expression in the entries of the matrix (with integer coefficients!), so it makes sense for any ring at all.
(Even for rings that don't contain fields, like the integers)

 

[lavinia]

There is a small outline in a book about finding the determinant of a matrix over an arbitrary commutative ring. There are a few things I don't understand; here it is:

'Let R be a commutative ring with a subring K which is a field. We consider the matrix X = [itex](x_{ij})[/itex] whose entries are independent indeterminates over K (that is, elements of the ring S = K[x_{11),...,x_{nn}]). Now S is an integral domain so we can compute the determinant of X in Quot(S). Now we obtain the determinant of an arbitrary matrix over R by sustituting elements of R for the indeterminantes (this substitution is a ring homomorphism from S to R)'

The part that confuses me is how are you able to use the substitution homomorphism for R? Isn't K a subring of R (possibly proper), and S is over K not R? I might be rusty on my field theory... Why is the substitution homomorphism even from S to R and not from Quot(S) to R?

Any help is appreciated

I think that they just mean that if you assign a specific ring element to each indeterminate you get a homomorphism from the polynomial ring into R.

 

[Bleys]

This is a... weird way to go about things. Maybe it would make more sense in context.

Well the chapter is on vector spaces, so all matrices worked so far have been over fields. The determinant function has been defined (by giving it some properties) and showed to exist (explicit) and be unique, using some row operations facts. Then there is this really small section without much detail about determinants over commutative rings. It outlines a couple of methods to this:
1) If R is an integral domain it goes to Quot(R)
2) If R is arbitrary it says you can define the determinant formula just as before but properties of it have to be reproved
3) if R contains a field, which is the case I'm asking about

Are there other methods you had in mind, Hurkyl; maybe those would help me understand better.

But anyways, the main point is that the determinant is a polynomial expression in the entries of the matrix (with integer coefficients!), so it makes sense for any ring at all.
(Even for rings that don't contain fields, like the integers)

I'm sorry, I don't understand. Why are there integer coefficients? Aren't the coeffiecient coming from K?

if you assign a specific ring element to each indeterminate you get a homomorphism from the polynomial ring into R.

If you have a polynomials over some field, say K[x], and R is a ring containing K, then the substitution homomorphism still works? I think that makes sense. I've never really thought about that, since I've only seen it as f : K[x] -> K (EDIT: ok, I was confused because I kept thinking R was some random commutative ring possibly not related to K in any way, so i was wondering about the compatibility with the coefficients; but K<R so... nevermind about that last part :) )

 

[trambolin]

I still think about this. It appears also in behavioral control theory. Suppose you have something like this
[tex]
R(x) = \begin{pmatrix}x^3 &-1-x &4\\-x^2+x &0 &x^3\end{pmatrix}
[/tex]
where x is the indeterminate over the R[x] for representing the differential operator d/dt i.e.
[tex]
R(\frac{d}{dt})w = 0
[/tex]
Now if you check for the solutions of this ODE triplet, you can talk about the rank of the polynomial matrix etc. Hence there are some examples which I think are related to "Einsetzunghomöomorphismus" or evaluation homomorphism. I have asked a version of what you asked in this thread

https://www.physicsforums.com/showthread.php?t=224920

 

[Office_Shredder]

I'm sorry, I don't understand. Why are there integer coefficients? Aren't the coeffiecient coming from K?

If you expand the determinant formula by minors, eventually you have a polynomial in the entries of the polynomial. For example, if yo8u have a matrix
[tex]A=\left( \begin{array}{cc}
a & b \\
c & d \end{array} \right)[/tex]

then det(A)=ad-bc. If a, b, c and d are unknowns this is a polynomial with four indeterminants, and the coefficients of this polynomial are integers (in this case there's a 1 and a -1 next to ad and bc respectively).

This isn't a 2 dimensional phenomenon, you can expand any nxn determinant to get a polynomial in the entries of the determinant. In all cases, the coefficients of the polynomial are integers, which means that the polynomial makes sense over any ring (with the integers just being defined as repeated addition of the 1 element)