Jacobson Radical and Direct Sums - Bland, Proposition 6.1.4 ....

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 6.1 The Jacobson Radical ... ...

I need help with the proof of Proposition 6.1.4 ... Proposition 6.1.4 and its proof read as follows:
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In the above proof from Bland we read:"... ... If $$i_1 \ : \ M_1 \longrightarrow M_1 \oplus M_2$$ and $$i_2 \ : \ M_2 \longrightarrow M_1 \oplus M_2$$ are the canonical injections, then (a) of Exercise 3 shows that

$$i_k \ : \ \text{ Rad}(M_k) \longrightarrow \text{ Rad}(M_1 \oplus M_2) $$... ...

... ... ... "

My question/problem/issue is that I cannot understand the meaning of the statement that (a) of Exercise 3 shows that

$$i_k \ : \ \text{ Rad}(M_k) \longrightarrow \text{ Rad}(M_1 \oplus M_2)$$ ... ...
How do the canonical injections $$i_k$$ apply to $$\text{ Rad}(M_k)$$ and $$\text{ Rad}(M_1 \oplus M_2)$$ ... ...Is it trivially simple in that $$\text{ Rad}(M_k) $$ is a set of elements of $$M_k$$ and so the canonical injection operates as usual? But then... why, exactly, do we need Exercise 3? Can someone explain in simple terms ... ... exactly what is going on ...

Peter
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NOTE ... ... The above post refers to Exercise 3 of Bland, Section 6.1 so I am providing the text of that example ... as follows:
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[Euge]

Hi Peter,

The Exercise was needed in order to establish $\operatorname{Rad}(M_1) \oplus \operatorname{Rad}(M_2) \subset \operatorname{Rad}(M_1\oplus M_2)$. Let $(x,y)$ be an element of $\operatorname{Rad}(M_1) \oplus \operatorname{Rad}(M_2)$. Then $(x,y) = (x,0) + (0,y) = i_1(x) + i_2(y)$. We want to get $i_1(x), i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$, for then the sum $i_1(x) + i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$. By the Exercise, $i_1$ maps $\operatorname{Rad}(M_1)$ into $\operatorname{Rad}(M_1\oplus M_2)$; since $x\in \operatorname{Rad}(M_1)$, then $i_1(x)\in \operatorname{Rad}(M_1\oplus M_2)$. Similarly, $i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$, as desired. As $(x,y)$ was arbitrary, $\operatorname{Rad}(M_1) \oplus \operatorname{Rad}(M_2) \subset \operatorname{Rad}(M_1\oplus M_2)$.

 

[Math Amateur]

Hi Peter,

The Exercise was needed in order to establish $\operatorname{Rad}(M_1) \oplus \operatorname{Rad}(M_2) \subset \operatorname{Rad}(M_1\oplus M_2)$. Let $(x,y)$ be an element of $\operatorname{Rad}(M_1) \oplus \operatorname{Rad}(M_2)$. Then $(x,y) = (x,0) + (0,y) = i_1(x) + i_2(y)$. We want to get $i_1(x), i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$, for then the sum $i_1(x) + i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$. By the Exercise, $i_1$ maps $\operatorname{Rad}(M_1)$ into $\operatorname{Rad}(M_1\oplus M_2)$; since $x\in \operatorname{Rad}(M_1)$, then $i_1(x)\in \operatorname{Rad}(M_1\oplus M_2)$. Similarly, $i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$, as desired. As $(x,y)$ was arbitrary, $\operatorname{Rad}(M_1) \oplus \operatorname{Rad}(M_2) \subset \operatorname{Rad}(M_1\oplus M_2)$.

Thanks Euge ... really appreciate your help ...

just reflecting on the proof and what what you have said ...

Peter

 

[Math Amateur]

Hi Peter,

The Exercise was needed in order to establish $\operatorname{Rad}(M_1) \oplus \operatorname{Rad}(M_2) \subset \operatorname{Rad}(M_1\oplus M_2)$. Let $(x,y)$ be an element of $\operatorname{Rad}(M_1) \oplus \operatorname{Rad}(M_2)$. Then $(x,y) = (x,0) + (0,y) = i_1(x) + i_2(y)$. We want to get $i_1(x), i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$, for then the sum $i_1(x) + i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$. By the Exercise, $i_1$ maps $\operatorname{Rad}(M_1)$ into $\operatorname{Rad}(M_1\oplus M_2)$; since $x\in \operatorname{Rad}(M_1)$, then $i_1(x)\in \operatorname{Rad}(M_1\oplus M_2)$. Similarly, $i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$, as desired. As $(x,y)$ was arbitrary, $\operatorname{Rad}(M_1) \oplus \operatorname{Rad}(M_2) \subset \operatorname{Rad}(M_1\oplus M_2)$.

Hi Euge,

Thanks for the post ... but I need some further help ... sorry if i am not following you exactly ...

You write:

" ... ... By the Exercise, $i_1$ maps $\operatorname{Rad}(M_1)$ into $\operatorname{Rad}(M_1\oplus M_2)$; since $x\in \operatorname{Rad}(M_1)$, then $i_1(x)\in \operatorname{Rad}(M_1\oplus M_2)$. Similarly, $i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$, as desired. ... ... ""I am unsure how exactly Exercise 1 shows that $i_1$ maps $\operatorname{Rad}(M_1)$ into $\operatorname{Rad}(M_1\oplus M_2)$ ... ... what part of Exercise 3 shows this ...??In particular I am unsure regarding exactly why $x\in \operatorname{Rad}(M_1)$ implies that $i_1(x)\in \operatorname{Rad}(M_1\oplus M_2)$ ... ... can you show me why this is true ... ? I mean, obviously if $$x \in M_1$$ then $$i_1(x) \in M_1 \oplus M_2$$ ... ... but why does it also follow for the radicals of $$M_1$$ and $$M_1 \oplus M_2$$ ... ?Hope you can help ...

Peter

 

[Euge]

Hi Euge,

Thanks for the post ... but I need some further help ... sorry if i am not following you exactly ...

You write:

" ... ... By the Exercise, $i_1$ maps $\operatorname{Rad}(M_1)$ into $\operatorname{Rad}(M_1\oplus M_2)$; since $x\in \operatorname{Rad}(M_1)$, then $i_1(x)\in \operatorname{Rad}(M_1\oplus M_2)$. Similarly, $i_2(y)\in \operatorname{Rad}(M_1\oplus M_2)$, as desired. ... ... ""I am unsure how exactly Exercise 1 shows that $i_1$ maps $\operatorname{Rad}(M_1)$ into $\operatorname{Rad}(M_1\oplus M_2)$ ... ... what part of Exercise 3 shows this ...??In particular I am unsure regarding exactly why $x\in \operatorname{Rad}(M_1)$ implies that $i_1(x)\in \operatorname{Rad}(M_1\oplus M_2)$ ... ... can you show me why this is true ... ? I mean, obviously if $$x \in M_1$$ then $$i_1(x) \in M_1 \oplus M_2$$ ... ... but why does it also follow for the radicals of $$M_1$$ and $$M_1 \oplus M_2$$ ... ?Hope you can help ...

Peter

Based on your reply, it would seem that you don't know what the $i_k$. In case you don't, $i_1 : M_1\to M_1 \oplus M_2$ is given by $i_1(x) = (x,0)$, and $i_2 : M_2 \to M_1 \oplus M_2$ is given by $i_2(y) = (0,y)$. The part I used from Exercise 3(a) was just the result in its first sentence.

Take $f = i_k$, $M = M_k$, and $N = M_1 \oplus M_2$ in the first sentence of Exercise 3(a). The result gives $i_k(\operatorname{Rad}(M_k)) \subset \operatorname{Rad}(M_1\oplus M_2)$. In other words, $i_k$ maps $\operatorname{Rad}(M_k)$ into $\operatorname{Rad}(M_1 \oplus M_2)$. So you can see that if $x\in \operatorname{Rad}(M_k)$, then $i_k(x) \in \operatorname{Rad}(M_1\oplus M_2)$, by the very condition that $i_k(\operatorname{Rad}(M_k)) \subset \operatorname{Rad}(M_1\oplus M_2)$.