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

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

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

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

... so $$f$$ 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 $$f$$ can be an epimorphism as it does not seem to be onto $$( I_1 + I_2 ) / I_2$$ ... ...

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

... ... The domain of $$f$$ is $$I_1$$, so $$x \in I_1$$ ...

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

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

... so $$f$$ 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).