We remarked here that the composition of derivations of a ring need not be a derivation. In this post, we prove this simple yet interesting result that if 
Example 1. Consider the ring 
Now let’s show that the converse is true if
Example 2. Let 
Proof. We have 
Let’s now get to the subject of this post.
Lemma 1. Let
for all
Proof. Since
On the other hand, since
So we get from 
Replacing 
Corollary. Let 
Proof. Suppose that 
Lemma 2. Let 
Proof. Since 
So 
Remark. Lemma 2 remains true if we replace the condition 
Theorem (Edward C. Posner, 1957). Let 
Proof. First note that, by Example 2, the condition 
But by the identity 
Thus, by Lemma 2, either 
Adding the above identity to the identity 
Example 3. The condition ![R:=\mathbb{Z}_2[x],](https://s0.wp.com/latex.php?latex=R%3A%3D%5Cmathbb%7BZ%7D_2%5Bx%5D%2C&bg=f9f9f9&fg=333333&s=0&c=20201002)
Note. Examples and the Corollary in this post are mine. I have also slightly simplified Posner’s proof of the Theorem.
is its center. We denote by 
the set of idempotents of 
such that 
Also, ![R[x]](https://s0.wp.com/latex.php?latex=R%5Bx%5D&bg=f9f9f9&fg=333333&s=0&c=20201002)
is the ring of polynomials in 
with coefficients in 
the left-multiplication map 
is defined by 
for all 
are clearly right 
for all 
But for which 
i.e. every idempotent of 
of 
is multiplicatively closed,
the set 
is a subring of 
if and only if 
![\mathcal{L}(R) =R \cap \mathcal{L}(R[x])](https://s0.wp.com/latex.php?latex=%5Cmathcal%7BL%7D%28R%29+%3DR+%5Ccap+%5Cmathcal%7BL%7D%28R%5Bx%5D%29&bg=f9f9f9&fg=333333&s=0&c=20201002)
and if ![\mathcal{L}(R[x])=\mathcal{L}(R),](https://s0.wp.com/latex.php?latex=%5Cmathcal%7BL%7D%28R%5Bx%5D%29%3D%5Cmathcal%7BL%7D%28R%29%2C&bg=f9f9f9&fg=333333&s=0&c=20201002)
for any family of rings 
i.e. 
for all 
gives 
Conversely, if 
then 
if and only if 
for all 
i.e 
and 
and so 
Thus 
then 
and so 
Then choosing 
in i) gives 
and so 
and 
and so 
by v). Suppose now that 
and so, by i), 
for all 
and so 
So let 
we are done because that means 
and so, since 
we get that
holds for any ring 
Then, by i), 
and so 
which gives 
or 
![r \in \mathcal{L}(R[x])](https://s0.wp.com/latex.php?latex=r+%5Cin+%5Cmathcal%7BL%7D%28R%5Bx%5D%29&bg=f9f9f9&fg=333333&s=0&c=20201002)
if and only if 
for all ![f(x) \in R[x]](https://s0.wp.com/latex.php?latex=f%28x%29+%5Cin+R%5Bx%5D&bg=f9f9f9&fg=333333&s=0&c=20201002)
if and only if 
for all ![R[x],](https://s0.wp.com/latex.php?latex=R%5Bx%5D%2C&bg=f9f9f9&fg=333333&s=0&c=20201002)
if and only if 
by i).
and so, since the polynomial ring ![\mathcal{L}(R[x])=\mathcal{I}(R[x])=\mathcal{I}(R)=\mathcal{L}(R).](https://s0.wp.com/latex.php?latex=%5Cmathcal%7BL%7D%28R%5Bx%5D%29%3D%5Cmathcal%7BI%7D%28R%5Bx%5D%29%3D%5Cmathcal%7BI%7D%28R%29%3D%5Cmathcal%7BL%7D%28R%29.&bg=f9f9f9&fg=333333&s=0&c=20201002)
![\mathcal{I}(R[x])=\mathcal{I}(R)](https://s0.wp.com/latex.php?latex=%5Cmathcal%7BI%7D%28R%5Bx%5D%29%3D%5Cmathcal%7BI%7D%28R%29&bg=f9f9f9&fg=333333&s=0&c=20201002)
is true by ![f(x)=\sum_{i=0}^nr_ix^i \in \mathcal{L}(R[x]), \ n \ge 1.](https://s0.wp.com/latex.php?latex=f%28x%29%3D%5Csum_%7Bi%3D0%7D%5Enr_ix%5Ei+%5Cin+%5Cmathcal%7BL%7D%28R%5Bx%5D%29%2C+%5C+n+%5Cge+1.&bg=f9f9f9&fg=333333&s=0&c=20201002)
So 
for all 
on both sides of the equality gives 
Thus 
and so, since 
contradiction. This proves that 
and hence 
giving ![\mathcal{L}(R[x]) = \mathcal{L}(R).](https://s0.wp.com/latex.php?latex=%5Cmathcal%7BL%7D%28R%5Bx%5D%29+%3D+%5Cmathcal%7BL%7D%28R%29.&bg=f9f9f9&fg=333333&s=0&c=20201002)
be a commutative domain, and let 
upper triangular matrices with entries in 
Then, using the first part of the Problem, it is easy to see that
![\mathcal{L}(R[x]) \neq \mathcal{L}(R)](https://s0.wp.com/latex.php?latex=%5Cmathcal%7BL%7D%28R%5Bx%5D%29+%5Cneq+%5Cmathcal%7BL%7D%28R%29&bg=f9f9f9&fg=333333&s=0&c=20201002)
because ![f(x)=r+sx \in \mathcal{L}(R[x]),](https://s0.wp.com/latex.php?latex=f%28x%29%3Dr%2Bsx+%5Cin+%5Cmathcal%7BL%7D%28R%5Bx%5D%29%2C&bg=f9f9f9&fg=333333&s=0&c=20201002)
where
for some positive integer 
is the ring of 
matrices with entries from 
which satisfies this condition: if 
for some ideals 
of 
or 
By weakening this condition, we get a set of ideals of 
we call these ideals semiprime. 
of a ring 
of 
implies that 
for all ideals 
is equivalent to the condition 
for all 
for all 
the radical of 
If 
or, equivalently, 
then we say that 
implies that 
for some 
Then 
Thus 
which gives 
for some ideal 
Then 
which proves that 
is a semiprime ideal of a commutative ring 
i.e. if 
for some 
then 
(Example 1). However the converse is not true (Example 2). If 
then 
implies that 
for some 
and so 
because 
over a field 
is a semiprime ring which is not reduced. 
for all 
is invertible. Thus 
Then 
is semiprime if and only if 
is semiprime.
for some ideal 
then 
where 
implies that 
implies that 
or 
In general ring theory, we have the following similar result.
implies that 
for some 
Then 
Thus either 
which gives 
or 
which gives 
We need to show that 
Then 
and so 
because 
Hence 
which proves that 
implies that 
or 
It is true that if 
then 
or 
implies that 
for some 
and so either 
because 
So 
where 
for some ideals 
of 
We have 
for some ideals 
and hence 
because 
or 
and so 
is prime.
and 
then 
To see this, suppose that 
is nilpotent and 
If 
giving 
which contradicts minimality of 
is odd, then 
gives 
and so 
which again contradicts minimality of 
and hence 
i.e. 
with the standard basis 
is never reduced because, for example, 
for some 
then 
As a result, the left and the right annihilators of an element of 
we have 
and thus 
Suppose that 
then 
and thus, since 
or 
So 
Then, by Remark 1, 
and so, since 
or 
To prove that reduced rings are semiprime, suppose that 
Then 
and so 
and let 
See that
in that intersection. Since 
for all 
Let
and so 
Thus, by Zorn’s lemma, 
has a maximal element, say 
To get a contradiction, we only need to show that 
Since 
we get 
Thus 
for some 
But then 
which is absurd. 
the set of prime ideals of 
. Clearly 
because 
Let 
be a totally ordered subset of 
Since 
Zorn’s lemma gives us a prime ideal 
which is maximal in 
i.e. 
then 
Then from 
we get 
and thus 
Hence for every 
we’ll have 
which gives us 
Also, since 
which means 
for some 
Thus
Therefore 
which proves 
Conversely, let 
and 
. Since 
we have 
and thus 
where 
is direct iff 
is direct.
be a nilpotent ideal of 
is an essential left ideal of 
of 
of 
of 
So we only need to prove the first part of the claim. Let 
for some left ideal 
of 
is essential in 
i.e. 
Thus 
for some 
So 
and 
Thus 
We have 
because 
and 
for some 
and 
We can write 
for some 
Then 
So 
is a right ideal of 
Similarly 
and hence 
As a result, 
is a regular submonoid which is left Ore. So 
exists and 
, then 
and if 
then 
Also 
Then 
and so 
Hence 
Similarly 
For the last part, if 
then 
and so 
then 
and 
It is also true that 
where 
and 
and 
such that 
and thus 
where 
For the last part, suppose that 
where 
by the second part. So we have 
and thus 
we have 
where 
We have 
and thus 
for some 
Then, since 
we get 
which gives us 
is a unit in 
and 
and 
Then 
For the second part, let 
Then 
and thus 
So, since 
such that 
and thus 
Therefore 
for all 
Hence 
and so 
)
for some 
Then 
for all 
which gives us 
i.e. 
Therefore either 
or 
which means either 
or 
This also proves that if 
be a faithful simple left 
is a division ring and 
in the usual way. Let 
and define 
by 
for all 
Then 
is a well-defined ring homomorphism. Also 
Now 
is a submodule of 
or 
If 
then we get 
which is nonsense. Finally, if 
then we will have 
Thus 
and so 
a contradiction! 
be a composition series of 
Therefore 
is a simple left 
We also let 
Then each 
is an ideal of 
Thus 
because 
for some 
because 
is a faithful simple left 
, as usual, be the 
. So suppose that 
and choose 
Then 
because 
Also either 
, or 
then 
for some 
Thus 
which gives us the contradiction 
because 
is invertible in 
Abstract Algebra 