Third Isomorphism Theorem for Rings .... Bland Theorem 3.3.16 .... ....

[Math Amateur]

I am reading "The Basics of Abstract Algebra" by Paul E. Bland ... ...

I am currently focused on Chapter 3: Sets with Two Binary Operations: Rings ... ...

I need help with Bland's proof of the Third Isomorphism Theorem for rings ...

Bland's Third Isomorphism Theorem for rings and its proof read as follows:https://www.physicsforums.com/attachments/7973
In the above proof by Bland we read the following:

" ... ... The mapping \(\displaystyle f \ : \ I_1 \rightarrow ( I_1 + I_2 ) / I_2\) given by \(\displaystyle f(x) = x + I_2\) is a well-defined ring epimorphism with kernel \(\displaystyle I_1 \cap I_2\). ... ... "I cannot see how \(\displaystyle f\) can be an epimorphism as it does not seem to be onto \(\displaystyle ( I_1 + I_2 ) / I_2\) ... ...

My reasoning (which I strongly suspect is faulty) is as follows:... ... The domain of \(\displaystyle f\) is \(\displaystyle I_1\), so \(\displaystyle x \in I_1\) ...

Now there exists elements \(\displaystyle y \in I_1 + I_2\) such that \(\displaystyle y \in I_2 \ \ ( y = 0 + y \text{ where } 0 \in I_1, y \in I_2) \)

For such \(\displaystyle y\) there is a coset in \(\displaystyle ( I_1 + I_2 ) / I_2\) of the form \(\displaystyle y + I_2\) that is not in the range of \(\displaystyle f\) ...

... so \(\displaystyle f\) is not an epimorphism ...
It seems certain to me that my reasoning is wrong somewhere ... can someone please point out the error(s) in my analysis above ...

Peter

 

[GJA]

Hi, Peter.

I cannot see how \(\displaystyle f\) can be an epimorphism as it does not seem to be onto \(\displaystyle ( I_1 + I_2 ) / I_2\) ... ...

My reasoning (which I strongly suspect is faulty) is as follows:

... ... The domain of \(\displaystyle f\) is \(\displaystyle I_1\), so \(\displaystyle x \in I_1\) ...

Now there exists elements \(\displaystyle y \in I_1 + I_2\) such that \(\displaystyle y \in I_2 \ \ ( y = 0 + y \text{ where } 0 \in I_1, y \in I_2) \)

For such \(\displaystyle y\) there is a coset in \(\displaystyle ( I_1 + I_2 ) / I_2\) of the form \(\displaystyle y + I_2\) that is not in the range of \(\displaystyle f\) ...

... so \(\displaystyle f\) is not an epimorphism ...

Recall that ideals are (sub)rings and thus have an additive group structure. In particular, $y+I_{2}=I_{2}$ for all $y\in I_{2}.$ Hence, $f$ is onto because for any $y\in I_{2}$, $x+y+I_{2}=x+I_{2}$.

 

[Math Amateur]

Hi, Peter.
Recall that ideals are (sub)rings and thus have an additive group structure. In particular, $y+I_{2}=I_{2}$ for all $y\in I_{2}.$ Hence, $f$ is onto because for any $y\in I_{2}$, $x+y+I_{2}=x+I_{2}$.

Thanks GJA ...

Appreciate your help ...

Peter

 

[castor28]

Hi Peter,

There is a drawing I like very much and which helps remembering the statement of the theorem (it may also help understanding the proof):

\begin{tikzpicture}
\draw (0,0) rectangle (6,6);
\draw (2,0) -- (2,6);
\draw (0,2) -- (6,2);
\draw (1,1) node {$I_1\cap I_2$};
\draw (1,5) node {$I_1$};
\draw (5,1) node {$I_2$};
\draw (5,5) node {$I_1+I_2$};
\end{tikzpicture}

Basically, the theorem states that what the drawing suggests is actually true;).

You may enjoy making a similar drawing for the second isomorphism theorem. (By the way, many books interchange the numbers of these two theorems).