4

In my text, there was only given proofs for commutative ring with unity, then I found that the same arguments work for just commutative ring by some tricks.

Here are what I have proven:

Let $R$ be a commutative ring (Not necessarily with unity)

Let $A,B\in M_n(R)$.

Then, $\det(AB)=\det(A)\det(B)$ and $\det(A^t)=\det(A)$

Is this true? (This is a yes or no question.. To make it sure)

(It's too tedious and long to post the actual proof for this kind of problems.. So please don't blame me for not posting a proof..)

Community
  • 1
Rubertos
  • 12,491
  • 1
    The definition of determinant in terms of signed products of entries still makes sense, and the effects of adding a multiple of one row to another, exchanging two rows, and scaling a row by a multiple also work out at before. However lacking a multiplicative unit in the ring prevents us from speaking of matrix inverses, and it complicates the definition of "row echelon forms". – hardmath Sep 28 '14 at 02:23
  • Of course you need $n$ to be positive now... Also, the notion of row echelon forms is mostly not-very-useful for rings with unity, too. They exist and are unique over fields; maybe the uniqueness generalizes to some weaker assumptions but the existence definitely does not. – darij grinberg Sep 28 '14 at 02:47
  • Actually, the uniqueness of the row-reduced echelon form of a matrix (in the sense that two row-reduced echelon matrices $A$ and $B$ satisfying $A = RB$ for some invertible matrix $R$ must be identical) holds over any nontrivial commutative ring. But these matrices exist rarely. – darij grinberg Sep 28 '14 at 03:05

1 Answers1

5

If $R$ is a ring without unit, you can consider the ring $S=\mathbb Z\oplus R$ with multiplication defined so that $(a,r)\cdot(b,s)=(ab,as+rb+rs)$, which has a unit. The inclusion $i:R\to S$ is an injective map of (non-unital) rings.

If $A$ and $B$ are matrices with entries in $R$, then $i(A)$ and $i(B)$ (apply $i$ to each entry)) are matrices with entries in $S$, and you know that $\det(i(A)i(B))=\det i(A)\cdot\det i(B)$. You can easily show that $\det(i(A)i(B))=i(\det(AB))$ and that $\det i(A)\cdot\det i(B)=i(\det A\cdot\det B)$. Since $i$ is injective, what you want follows at once from this.

Mariano Suárez-Álvarez
  • 135,076