Ideals of a ring of matrixes

[Alex224]

Hi.
I have this ring of matrixes: R = { \[ \begin{pmatrix} a & 0\\ b & c\\ \end{pmatrix} \]}while a,b,c is from some field F.

now, I need to find all the ideals of this ring. I found five ideals. here there are:i1 = { \[ \begin{pmatrix} 0 & 0\\ b & 0\\ \end{pmatrix} \]}i2 = { \[ \begin{pmatrix} a & 0\\ b & 0\\ \end{pmatrix} \]}i3 = { \[ \begin{pmatrix} 0 & 0\\ b & c\\ \end{pmatrix} \]}i4 = R

i5 = {0}

now, I am kind of stuck to explain why there cannot be a six's ideal. I know intuitively why there cannot be another ideal but its like I can't figure out how formally explain it. I feel like I am dancing around the answer for hours but can't make it right on the spot.

any help?
thank you!

 

[jvicens]

Thank you for sharing your findings about the ideals of the ring R with us. It seems like you have correctly identified five of the six possible ideals in this ring. The reason why there cannot be a sixth ideal can be explained using the definition of an ideal.

An ideal of a ring is a subset that satisfies two properties: closure under addition and closure under multiplication by elements of the ring. In other words, if we take any two elements from the ideal and add them, the result must also be in the ideal. Similarly, if we take any element from the ideal and multiply it by any element from the ring, the result must also be in the ideal.

Now, let's consider the ring R and its five ideals that you have identified. Each of these ideals satisfies the two properties mentioned above. For example, in i1, if we take any two elements and add them, we get a result that is also in i1. Similarly, if we take any element from i1 and multiply it by any element from R, the result is still in i1.

However, if we try to construct a sixth ideal, it will either violate the closure under addition property or the closure under multiplication property. Let's say we try to add two elements from i1 and i3, the result will not be in either of these ideals as it will have a non-zero entry in the top left corner. Similarly, if we try to multiply an element from i1 with an element from R, the result will not be in i1 as it will have a non-zero entry in the top left corner.

Hence, we can conclude that there cannot be a sixth ideal in the ring R. I hope this helps to clarify your understanding. If you have any further questions, please feel free to ask. Keep up the good work in exploring the properties of this ring!