# Math Help - Algebra, Problems For Fun (40)

1. ## Algebra, Problems For Fun (40)

This is the last problem of this series of problems that I've been posting for a while now:

Let $F$ be a field and $R=\left \{\begin{pmatrix}x & 0 & y \\ 0 & x & z \\ 0 & 0 & x \end{pmatrix}: \ \ x,y,z \in F \right \}.$ It's easy to see that $R$ is a commutative ring with identity element. Find all maximal ideals of $R.$

Hint:
Spoiler:
$R$ has only one maximal ideal.

2. Clearly $S=\left\{\begin{pmatrix}x & 0 & 0\\
0 & x & 0\\
0 & 0 & x
\end{pmatrix}:x\in F\right\}$
is an ideal of $R$ and the quotient ring $R/S$ is a commutative ring with identity $I_3+S.$

Let’s let $A(x,y,z)$ denote the element $\begin{pmatrix}x & 0 & y\\
0 & x & z\\
0 & 0 & x
\end{pmatrix}+S$
in $R/S.$

Then if $x\ne0,$ $A(x,y,z)\cdot A(x^{-1},-yx^{-2},-zx^{-2})=I_3+S.$ Hence $R/S$ is a field and so $S$ is a maximal ideal.

3. Originally Posted by TheAbstractionist
Clearly $S=\left\{\begin{pmatrix}x & 0 & 0\\
0 & x & 0\\
0 & 0 & x
\end{pmatrix}:x\in F\right\}$
is an ideal of $R$ and the quotient ring $R/S$ is a commutative ring with identity $I_3+S.$

Let’s let $A(x,y,z)$ denote the element $\begin{pmatrix}x & 0 & y\\
0 & x & z\\
0 & 0 & x
\end{pmatrix}+S$
in $R/S.$

Then if $x\ne0,$ $A(x,y,z)\cdot A(x^{-1},-yx^{-2},-zx^{-2})=I_3+S.$ Hence $R/S$ is a field and so $S$ is a maximal ideal.
$S$ contains the identity element of $R.$ so if it was an ideal of $R,$ it would be equal to $R.$ but it's not! so $S$ is not an ideal of $R.$

you can also see that by checking that $S$ is not closed under multiplication by elements of $R.$

4. Originally Posted by NonCommAlg
This is the last problem of this series of problems that I've been posting for a while now:

Let $F$ be a field and $R=\left \{\begin{pmatrix}x & 0 & y \\ 0 & x & z \\ 0 & 0 & x \end{pmatrix}: \ \ x,y,z \in F \right \}.$ It's easy to see that $R$ is a commutative ring with identity element. Find all maximal ideals of $R.$

Hint:
Spoiler:
$R$ has only one maximal ideal.
(I've used $(x,y,z) = \begin{pmatrix}x & 0 & y \\ 0 & x & z \\ 0 & 0 & x \end{pmatrix}$, with the equivalent multiplication, because it's quicker to type...)

The subring with $x=0$ is clearly an ideal: $(x,y,z).(a,b,c) = (xa, xb+ya, xc+za)$ $\Rightarrow (0,y,z).(a,b,c) = (0, ya, za)=(a,b,c).(0,y,z)$.

However, this ideal is also maximal. This is because if $I$ is an ideal containing an element $(x,y,z)$, $x \neq 0$, then $[(x,y,z)(x^{-1},0,0)](1, -yx^{-1}, -zx^{-1}) = id$ and so $I=R$.

Subsequently this ideal contains all other proper ideals, as if it didn't then there would be an ideal with $x \neq 0$, a contradiction. Thus it is the only maximal ideal.