Does "K" a Field?

Question

Does the group $$ K =\begin{matrix}\pmatrix{ a & 0 \\ 0 & 0\\ } \end{matrix} $$

is a Field relatively to additive and multiply methods about Matrices?

I tried to show that it's NOT by showing this: $$ \begin{matrix}\pmatrix{ a & 0 \\ 0 & 0\\ } \end{matrix} * \begin{matrix}\pmatrix{ a^-1 & 0 \\ 0 & 0\\ }\end{matrix} \ne \begin{matrix}\pmatrix{ 1 & 0 \\ 0 & 1\\ }\end{matrix} $$ But I don't know if "1" means it's the Identity matrix$( a * a^-1 = 1)$. Any suggestions how can I prove K is a field?

Answer 1

If by $K$ you mean the set of all matrices $\left[ \begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right]$ where $a$ is in some field, say $F$, then you can do the following:

Clearly $\left[ \begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right]+\left[ \begin{matrix} b & 0 \\ 0 & 0 \end{matrix}\right]=\left[ \begin{matrix} a+b & 0 \\ 0 & 0 \end{matrix}\right]$ and $\left[ \begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right] \left[ \begin{matrix} b & 0 \\0 & 0 \end{matrix}\right]=\left[ \begin{matrix} ab & 0 \\ 0 & 0 \end{matrix}\right]$, so we can define a ring homomorphism $f:K\rightarrow F$ by sending $\left[ \begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right]$ to $a$. The above computations show that this is a ring homomorphism, and this is obviously not the zero homomorphism, so that it must injective. It's also rather clear that it is surjective, so we have an isomorphism, and thus $K$ is a field.

Direct Proof that $K$ is a field: Denote $\left[ \begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right]$ by $[a]$.

• We showed above that $K$ is closed under addition and multiplication, and more specifically that $[a]+[b]=[a+b]$ and $[a][b]=[ab]$.
• $[0]+[a]=[a]+[0]=[a+0]=[a]$, so $[0]$ is the additive identity.
• $[1][a]=[1a]=[a]=[a1]=[a][1]$, so $[1]$ is the multiplicative identity.
• Because $F$ is a field, we see that $$[a]+\left( [b]+[c]\right)=[a]+[b+c]=[a+(b+c)]=[(a+b)+c]=[a+b]+[c]=\left( [a]+[b]\right)+[c]$$ and $$[a]+[b]=[a+b]=[b+a]=[b]+[a]$$ so we have associativity and commutativity of the addition, respectively.
• Again, because $F$ is a field, we see that $$[a]\left( [b][c]\right)=[a][bc]=[a(bc)]=[(ab)c]=[ab][c]=\left( [a][b]\right)[c]$$ and $$[a][b]=[ab]=[ba]=[b][a]$$ so we have associativity and commutativity of the multiplication, respectively.
• $[a]\neq [0]$ if and only if $a\neq 0$, so we set $[a]^{-1}=[a^{-1}]$ (which we can do because $F$ is a field and $a\neq 0$), and clearly $[a][a]^{-1}=[1]$, as $[a][a]^{-1}=[a][a^{-1}]=[aa^{-1}]=[1]$.
• Finally, because $F$ is a field, $$[a]\left( [b]+[c]\right)=[a][b+c]=[a(b+c)]=[ab+ac]=[ab]+[ac]=[a][b]+[a][c]$$ so we have distributivity.

Answer 2

Since $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ is not even an element of $K$, it cannot possibly be the "$1$" of your perhaps-a-field.

However, there might be another matrix that can satisfy that role? Remember that $B\in K$ is a multiplicative identity in $K$ if $AB=B=BA$ for all $A\in K$ -- what it does to things outside $K$ is irrelevant.