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I am currently studying for my Algebra comprehensive exam which will be given at the end of August. As part of my materials, I have several old exams, minus the answers. I would like to post my solutions here to see if they are accurate.
I am posting here rather than in the homework section, because this is not homework, and I don't want to take attention away from those people asking real homework questions. That being said, if you do see a mistake in my work, please do not give me the solution. All constructive comments are welcome. Thanks!
1. Let f:R \rightarrow S be a ring homomorphism, let I be an ideal in R, and J be an ideal in S.
a. Show that the inverse image f^{-1}(J) is an ideal of R that contains the kernal of f.
b. Prove that if f is surjective, then f(I) is an ideal in S. Give an example showing that if f is not surjective then f(I) need not be an ideal in S.
SOLUTION
a. I will show that f^{-1}(J) is nonempty, closed under subtraction and multiplication (making it a subring of R), and absorbs products from R.
Since 0 \in J and f(0) = 0, (since f is a ring homomorphism), f^{-1}(J) is nonempty.
Suppose x,y \in f^{-1}(J). Then f(x) \in J and f(y) \in J. Then, f(x - y) = f(x) - f(y) (since f is a ring homomorphism). And, f(x) - f(y) \in J (since J is an ideal, and therefore closed under subtraction).
Likewise, f(xy) = f(x)f(y) \in J.
Finally, for any r \in R, f(rx) = f(r)f(x) \in J (since J is an ideal, f(r) \in S, f(x) \in J).
Thus, we have established that f^{-1}(J) is an ideal of R. To see that it contains the kernal of f, let k \in ker(f). Then f(k) = 0 \in J. So, ker(f) \subseteq f^{-1}(J).
b. Similarly to part a, I will show that f(I) is nonempty, closed under subtraction and multiplication, and absorbs products from S.
Since I is an ideal, 0 \in I. Since f is a ring homomorphism, f(0) = 0 \in S. So, f(I) is nonempty.
Suppose x,y \in f(I). Then there exist a,b \in I such that f(a) = x and f(b) = y. Then, x - y = f(a) - f(b) = f(a - b) (since f is a ring homomorphism). And, f(a - b) \in f(I) (since I is an ideal, and therefore closed under subtraction).
Likewise, xy = f(ab) = f(a)f(b) \in f(I).
Finally, suppose s \in S. Since f is surjective, there exists r \in R such that f(r) = s. Then, sx = f(r)f(a) = f(ra) \in f(I) (since I is an ideal, r \in R, a \in I).
Thus, we have established that f(I) is an ideal of S. To show that the surjectivity of f is required, consider the identity map f: \mathbb{Z} \rightarrow \mathbb{Q}, which is clearly not surjective. n \mathbb{Z} is an ideal of \mathbb{Z} for any n \in \mathbb{Z}, but f(n \mathbb{Z}) = n \mathbb{Z} is not an ideal of \mathbb{Q} (being a field, \mathbb{Q} has no properly contained ideals).
I am posting here rather than in the homework section, because this is not homework, and I don't want to take attention away from those people asking real homework questions. That being said, if you do see a mistake in my work, please do not give me the solution. All constructive comments are welcome. Thanks!
1. Let f:R \rightarrow S be a ring homomorphism, let I be an ideal in R, and J be an ideal in S.
a. Show that the inverse image f^{-1}(J) is an ideal of R that contains the kernal of f.
b. Prove that if f is surjective, then f(I) is an ideal in S. Give an example showing that if f is not surjective then f(I) need not be an ideal in S.
SOLUTION
a. I will show that f^{-1}(J) is nonempty, closed under subtraction and multiplication (making it a subring of R), and absorbs products from R.
Since 0 \in J and f(0) = 0, (since f is a ring homomorphism), f^{-1}(J) is nonempty.
Suppose x,y \in f^{-1}(J). Then f(x) \in J and f(y) \in J. Then, f(x - y) = f(x) - f(y) (since f is a ring homomorphism). And, f(x) - f(y) \in J (since J is an ideal, and therefore closed under subtraction).
Likewise, f(xy) = f(x)f(y) \in J.
Finally, for any r \in R, f(rx) = f(r)f(x) \in J (since J is an ideal, f(r) \in S, f(x) \in J).
Thus, we have established that f^{-1}(J) is an ideal of R. To see that it contains the kernal of f, let k \in ker(f). Then f(k) = 0 \in J. So, ker(f) \subseteq f^{-1}(J).
b. Similarly to part a, I will show that f(I) is nonempty, closed under subtraction and multiplication, and absorbs products from S.
Since I is an ideal, 0 \in I. Since f is a ring homomorphism, f(0) = 0 \in S. So, f(I) is nonempty.
Suppose x,y \in f(I). Then there exist a,b \in I such that f(a) = x and f(b) = y. Then, x - y = f(a) - f(b) = f(a - b) (since f is a ring homomorphism). And, f(a - b) \in f(I) (since I is an ideal, and therefore closed under subtraction).
Likewise, xy = f(ab) = f(a)f(b) \in f(I).
Finally, suppose s \in S. Since f is surjective, there exists r \in R such that f(r) = s. Then, sx = f(r)f(a) = f(ra) \in f(I) (since I is an ideal, r \in R, a \in I).
Thus, we have established that f(I) is an ideal of S. To show that the surjectivity of f is required, consider the identity map f: \mathbb{Z} \rightarrow \mathbb{Q}, which is clearly not surjective. n \mathbb{Z} is an ideal of \mathbb{Z} for any n \in \mathbb{Z}, but f(n \mathbb{Z}) = n \mathbb{Z} is not an ideal of \mathbb{Q} (being a field, \mathbb{Q} has no properly contained ideals).
