Consider the map det of $ M_n(R)$ into R where det(A) is the determinant of the matrix A for A $\in M_n(R)$. Is det a ring homomorphism? why or why not?
Please, help

Question

Consider the map det of $ M_n(R)$ into R where det(A) is the determinant of the matrix A for A $\in M_n(R)$. Is det a ring homomorphism? why or why not?
Please, help

phi · Answer

ask a mathematician

amistre64 · Answer

id have to start off by asking how you define a ring homomorphism.

anonymous · Answer

probably a linear function that inhabits the axioms of a ring. But first time I have read about this myself.

amistre64 · Answer

im taking abstract algebra this term so im still loose on how to define the terms :)

loser66 · Answer

For rings R and R', a map $\phi$: R$\rightarrow$R' is a homomorphism if the following 2 conditions are satisfied for all a, b $\in$ R
1/ $\phi (a+b) =\phi(a) +\phi(b)$
2/$\phi(ab) = \phi(a)\phi(b)$

loser66 · Answer

oh, I got it, Thanks for reply. Ha!!! quite easy when speaking out the definition.

anonymous · Answer

more about the laws of a determinant then the laws of the homomorphism, but yes it inhabits the structures.

anonymous · Answer

However, wouldn't know if you first have to verify that $\det (M_n(R))$ is a Ring.

loser66 · Answer

no, it's not, because det (a+b) $\neq det(a)+det(b)$

loser66 · Answer

in this case $\phi:R->R' = det :R->R'$

anonymous · Answer

Ah I see, the determinant goes into the expression as a function, I thought they would only consider the $M_n(R)$ Matrices. But surely, including the Determinant it no longer works. You're right.

loser66 · Answer

Thank you