Why is the ring of matrices over a field simple?

Denote by Mn×n(k) the ring of n by n matrices with coefficients in the field k. Then why does this ring not contain any two-sided ideal?

Thanks for any clarification, and this is an exercise from the notes of Commutative Algebra by Pete L Clark, of which I thought as simple but I cannot figure it out now.

Answer

Suppose that you have an ideal I which contains a matrix with a nonzero entry aij. Multiplying by the matrix that has 0‘s everywhere except a 1 in entry (i,i), kill all rows except the ith row; multiplying by a suitable matrices on the right, kill all columns except the jth column; now you have a matrix, necessarily in I, which contains exactly one nonzero entry, namely aij in position (i,j).

Now show that I must contain all matrices in Mn×n(k). This will show that a 2-sided ideal consists either of only the 0 matrix, or must be equal to the entire ring.

Added. Now that you have a matrix that has a single nonzero entry, can you get a matrix that has a single nonzero entry on whatever coordinate you specify, and such that this nonzero entry is whatever element of k you want, by multiplying this matrix (on either left, or right, or both) by suitable elementary matrices? Will they all be in I?

And…

(abcd)=(a000)+

Attribution
Source : Link , Question Author : awllower , Answer Author : Arturo Magidin

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