MHB Understanding One-Sided Ideals in Mathematics

[cbarker1]

Dear Everyone,

I am reading the Abstract Algebra Book by Dummit and Foote. I am confusing with this example for one-side ideals. So here is the example:
Let $R$ be a commutative ring with $1 \ne 0$ and let \( n\in \mathbb{Z} \) with $n\ge 2$. For each $j\in \{1,2,\dots, n\}$, let $L_j$ be the set of all $n \times n$ matrices in $M_n(R)$ with arbitrary entries in the jth column and zeroes in all other columns. It is clear that $L_j$ is closed under subtraction. It follows directly from the definition of matrix multiplication that any matrix $T \in M_n(R)$ and $A \in L_j$, the product $TA$ has zero entries in the ith column for all $i\ne j$. This shows $L_j$ is a left ideal of $M_n(R)$. Moreover, $L_j$ is not a right ideal.

What does this example look like with math symbols?

Thanks,
Cbarker1

 

[GJA]

Hi Cbarker1,

It's not entirely clear to me what your question is exactly. If you mean you'd like to see verification of the left ideal/non-right ideal claim, that would look something like this.

Proof $L_{j}$ is a Left Ideal
Let $A\in L_{j},$ $T\in M_{n}(R),$ and let $a_{j}$ be the $j$th column of $A$. Then $$TA = T\left[\begin{array}{c|c|c|c|c} 0 & \ldots & a_{j} & \ldots & 0 \end{array}\right] = \left[\begin{array}{c|c|c|c|c} 0 & \ldots & Ta_{j} & \ldots & 0 \end{array}\right],$$ which shows that $TA\in L_{j}$. Hence, $L_{j}$ is a left ideal over $M_{n}(R).$

Proof $L_{1}$ is not a Right Ideal
Let $a_{1} = \begin{bmatrix}1\\ 0\\ \vdots\\ 0 \end{bmatrix},$ $A = \left[\begin{array}{c|c|c|c} a_{1} & 0 &\ldots & 0 \end{array} \right],$ and $T = \left[\begin{array}{c|c|c|c|c} a_{1} & a_{1} & 0 &\ldots & 0 \end{array} \right].$ Then $AT = T\notin L_{1}.$ Hence, $L_{1}$ is not a right ideal over $M_{n}(R).$ This example can be generalized to any $j$, if desired.

 

[cbarker1]

I am trying to see the symbols. Like this: \[ \begin{pmatrix} 1 & 2 & 3\\ a & b & c \end{pmatrix} \].

 

[cbarker1]

I am trying to see the symbols. Something Like this: \[ \begin{pmatrix} 1 & 2 & 3\\ a & b & c \end{pmatrix} \].

 

[GJA]

OK, let's see if this helps.

A matrix in $L_{j}$ would look be written as $$A = \begin{bmatrix} 0 & \ldots & a_{1j} & \ldots & 0\\ 0 & \ldots & a_{2j} & \ldots & 0\\ \vdots & \ddots & \vdots & \ddots & \vdots\\ 0 & \ldots & a_{nj} & \ldots & 0\end{bmatrix},$$ and a matrix $T\in M_{n}(R)$ would have the form $$T = \begin{bmatrix} t_{11} & \ldots & t_{1j} & \ldots & t_{1n}\\ t_{21} & \ldots & t_{2j} & \ldots & t_{2n}\\ \vdots & \ddots & \vdots & \ddots & \vdots\\ t_{n1} & \ldots & t_{nj} & \ldots & t_{nn}\\\end{bmatrix},$$ where all the $a$'s and $t$'s are elements of $R$. Using these forms for $A$ and $T$, the product $TA$ would be $$TA = \begin{bmatrix}0 & \ldots & t_{11}a_{1j} + t_{12}a_{2j}+ \ldots + t_{1n}a_{nj} & \ldots & 0\\ 0 & \ldots & \vdots & \ldots & 0\\ \vdots & \ddots & \vdots & \ddots & \vdots\\ 0 & \ldots & \ldots & \ldots & 0 \end{bmatrix},$$ where I have left rows $2$ through $n$ of the $j$th column of $TA$ for you to try to fill in for yourself.

Does this answer your question?

 

[cbarker1]

yes.