properties of the Jacobson radical

Theorem:
Let $R,T$ be rings and $\phi :R\to T$ be a surjective homomorphism. Then $\phi (J(R))\subseteq J(T)$.

Proof:
We shall use the characterization of the Jacobson radical as the set of all $a\in R$ such that for all $r\in R$, $1-ra$ is left invertible.

Let $a\in J(R),t\in T$. We claim that $1-t\phi (a)$ is left invertible:

Since $\phi$ is surjective, $t=\phi (r)$ for some $r\in R$. Since $a\in J(R)$, we know $1-ra$ is left invertible, so there exists $u\in R$ such that$u(1-ra)=1$. Then we have

$$\phi (u)\left(\phi (1)-\phi (r)\phi (a)\right)=\phi (u)\phi (1-ra)=\phi (1)=1$$

So $\phi (a)\in J(T)$ as required.

Theorem:
Let $R,T$ be rings. Then $J(R\times T)\subseteq J(R)\times J(T)$.

Proof:
Let ${\pi }_1:R\times T\to R$ be a (surjective) projection.By the previous theorem, ${\pi }_1(J(R\times T))\subseteq J(R)$.

Similarly let ${\pi }_2:R\times T\to T$ be a (surjective) projection. We see that ${\pi }_2(J(R\times T))\subseteq J(T)$.

Now take $(a,b)\in J(R\times T)$. Note that $a={\pi }_1(a,b)\in J(R)$ and $b={\pi }_2(a,b)\in J(T)$. Hence $(a,b)\in J(R)\times J(T)$ as required.