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:

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 $y \in I_2 \ \ ( y = 0 + y$ 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

 

[andrewkirk]

An arbitrary element of $(I_1+I_2)/I_2$ is $(a+b)+I_2$, where $a\in I_1,b\in I_2$.
But $(a+b)+I_2$ is equal to $a+I_2$ because
$$((a+b)+I_2) - (a+I_2) \triangleq ((a+b)-a)+I_2 = b+I_2=I_2=0_{(I_1+I_2)/I_2}$$
where the second last equality follows from the fact that $b\in I_2$.
Therefore
$$(a+b)+I_2=a+I_2 = f(a)\in f(I_1)$$
so $f$ is epimorphic.

 

[Math Amateur]

Thanks Andrew ...

Appreciate the help ...

Peter