Hw12 Problem 3

Problem: (+) Consider the map $\det$ of $M_n(\mathbb{R})$ onto $\mathbb{R}$ where det($A$) is the determinant of the matrix $A$. Is $\det$ a ring homomorphism? Why or why not?

Solution: This is not a ring homomorphism. From linear algebra, we know that $\det(AB) = \det(A)\det(B)$, satisfying condition 2 to be a ring homomorphism. However, condition 1 is not upheld. Let $A = \begin{vmatrix}1&2\\3&1\end{vmatrix}$ and let $B = \begin{vmatrix}2&1\\1&4\end{vmatrix}$. Also, $A + B = \begin{vmatrix}3&3\\ 4&5\end{vmatrix}$. Now,

(1)
\begin{align} \det(A) = (1*1) - (2*3) = -5 \end{align}
(2)
\begin{align} \det(B) = (2*1) - (1*4) = -2 \end{align}
(3)
\begin{align} \det(A+B) = (3*5) - (3*4) = 3 \end{align}

We can see that $\det(A + B) = 3 \not= -7 = \det(A) + \det(B)$, hence condition 1 is not satisfied. Therefore $\det$ is not a ring homomorphism.