LEC ALG3 3

Recap

Let $I$ be a two sided ideal of $R$.

$$\begin{array}{r}R\mapsto \frac{R}{I}\\ J\mapsto \frac{J+I}{I}.\end{array}$$

Lemma 1(Lemma).

Let $\phi :R\to S$ be a homomorphism and $J\subseteq S$ be an ideal. Then, ${\phi }^{-1}(J)$ is an ideal in $R$.

Lemma 2(Lemma).

Let $R,S$ be rings and $\phi :R\to S$ be a ring homomorphism. Then, $S$ is an $R$ module.

Proof.

$$\begin{array}{rl}& R\times S\to S,\\ & (r,m)\mapsto rm:=\phi (r)m\end{array}$$

Easy to verify that $S$ is an $R$ module.□

Corollary

1. If $I$ is a left ideal in $R$, then $I$ is a left $R$ module.

2. If $I$ is a two sided ideal in $R$, then $R/I$ is an $R$ module.

Proof.

$(1)$: $R\times I\to I$, $(r,m)\mapsto rm:=rm$.

$(2)$: $R\times R/I\to R/I$: $(r,a+I)=r(a+I)\equiv ra+I$. Verify well definedness. Alternatively, use Lemma 2.□

Corollary

Let $\phi :R\to S$ be a homomorphism. Let $I\subseteq R$ and $J\subseteq S$ be ideals such that $\phi (I)\subseteq J$. Then, $\overline{\phi }:R/I\to S/J$ is a homomorphism.

Lemma 3(Lemma).

Let $\phi :R\to S$. Let $I$ be an ideal in $R$. Let $\phi (I)\cdot S$ be the ideal in $S$ generated by $\phi (I)$. From Lemma 1, ${\phi }^{-1}(\phi (I)\cdot S)$ is an ideal in $R$ containing $I$.

If $J$ is an ideal of $S$, $\phi ({\phi }^{-1}(J))\cdot S$ is an ideal in $S$ contained in $J$.

Every ideal $J\subseteq \mathbb{Z}[x]$ contains $n\mathbb{Z}$ for some $n$.