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The question is:
Let f : R ! S be a homomorphism. Let
T = { f(r) : r 2 R}.
T is called the image of f. Show that T is a subring of S.
Hint: In this problem, you can use the following facts:
f(0R) = 0S,f(−x) = −f(x) for x 2 R, and f(x − y) = f(x) − f(y) for x, y 2 R. These facts were stated in class for isomorphisms only but the same properties (with the same proofs) hold
for homomorphisms.
Thanks very much. :confused:
You forgot to mention it, but R and S must be rings.Quote:
Originally Posted by suedenation
T is, of course, a subset of S.
By definition of the homomorphism, f(a+b)=f(a)+f(b) and f(a*b)=f(a)*f(b).
If a set T is a ring, it must have the following properties:
1. a+b=b+a
2. (a+b)+c=a+(b+c)
3. An identity (0) for addition exists.
4. An additive inverse exists for all elements of T.
5. (a*b)*c=a*(b*c)
6. An identity (1) for multiplication exists.
7. * distributes over +.
You can use the homomorphism to assist you. For example, the first property:
f(a+b)=f(a)+f(b)
f(b+a)=f(b)+f(a)
Since f(a+b)=f(b+a) we have that f(a)+f(b)=f(b)+f(a). So addition in T is commutative.
The second property follows the same pattern.
The third property: f(a+0)=f(a)+f(0)=f(a), so f(0)=0T exists.
The fourth property: f(a+-a)=f(a)+f(-a)=f(0)=0T, so f(-a)=-f(a) exists.
The fifth property:
f((a*b)*c)=(f(a)*f(b))*f(c)
f(a*(b*c))=f(a)*(f(b)*f(c))
and the two of these are equal.
The sixth property f(a*1)=f(a)*f(1)=f(a) so f(1)=1T exists.
The seventh property: f(a*(b+c))=f(a)*(f(b)+f(c)) etc.
And, of course, you should put in a statement that 0T and 1T correspond to 0S and 1S respectively, which is easy to show.
-Dan