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.(adsbygoogle = window.adsbygoogle || []).push({});

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 [tex]f:R \rightarrow S[/tex] be a ring homomorphism, let [tex]I[/tex] be an ideal in [tex]R[/tex], and [tex]J[/tex] be an ideal in [tex]S[/tex].

a. Show that the inverse image [tex]f^{-1}(J)[/tex] is an ideal of [tex]R[/tex] that contains the kernal of [tex]f[/tex].

b. Prove that if [tex]f[/tex] is surjective, then [tex]f(I)[/tex] is an ideal in S. Give an example showing that if [tex]f[/tex] is not surjective then [tex]f(I)[/tex] need not be an ideal in [tex]S[/tex].

SOLUTION

a. I will show that [tex]f^{-1}(J)[/tex] is nonempty, closed under subtraction and multiplication (making it a subring of [tex]R[/tex]), and absorbs products from [tex]R[/tex].

Since [tex]0 \in J[/tex] and [tex]f(0) = 0[/tex], (since [tex]f[/tex] is a ring homomorphism), [tex]f^{-1}(J)[/tex] is nonempty.

Suppose [tex]x,y \in f^{-1}(J)[/tex]. Then [tex]f(x) \in J[/tex] and [tex]f(y) \in J[/tex]. Then, [tex]f(x - y) = f(x) - f(y)[/tex] (since [tex]f[/tex] is a ring homomorphism). And, [tex]f(x) - f(y) \in J[/tex] (since [tex]J[/tex] is an ideal, and therefore closed under subtraction).

Likewise, [tex]f(xy) = f(x)f(y) \in J[/tex].

Finally, for any [tex]r \in R[/tex], [tex]f(rx) = f(r)f(x) \in J[/tex] (since [tex]J[/tex] is an ideal, [tex]f(r) \in S[/tex], [tex]f(x) \in J[/tex]).

Thus, we have established that [tex]f^{-1}(J)[/tex] is an ideal of [tex]R[/tex]. To see that it contains the kernal of [tex]f[/tex], let [tex]k \in ker(f)[/tex]. Then [tex]f(k) = 0 \in J[/tex]. So, [tex]ker(f) \subseteq f^{-1}(J)[/tex].

b. Similarly to part a, I will show that [tex]f(I)[/tex] is nonempty, closed under subtraction and multiplication, and absorbs products from [tex]S[/tex].

Since [tex]I[/tex] is an ideal, [tex]0 \in I[/tex]. Since [tex]f[/tex] is a ring homomorphism, [tex]f(0) = 0 \in S[/tex]. So, [tex]f(I)[/tex] is nonempty.

Suppose [tex]x,y \in f(I)[/tex]. Then there exist [tex]a,b \in I[/tex] such that [tex]f(a) = x[/tex] and [tex]f(b) = y[/tex]. Then, [tex]x - y = f(a) - f(b) = f(a - b)[/tex] (since [tex]f[/tex] is a ring homomorphism). And, [tex]f(a - b) \in f(I)[/tex] (since [tex]I[/tex] is an ideal, and therefore closed under subtraction).

Likewise, [tex]xy = f(ab) = f(a)f(b) \in f(I)[/tex].

Finally, suppose [tex]s \in S[/tex]. Since [tex]f[/tex] is surjective, there exists [tex]r \in R[/tex] such that [tex]f(r) = s[/tex]. Then, [tex]sx = f(r)f(a) = f(ra) \in f(I)[/tex] (since [tex]I[/tex] is an ideal, [tex]r \in R[/tex], [tex]a \in I[/tex]).

Thus, we have established that [tex]f(I)[/tex] is an ideal of [tex]S[/tex]. To show that the surjectivity of [tex]f[/tex] is required, consider the identity map [tex]f: \mathbb{Z} \rightarrow \mathbb{Q}[/tex], which is clearly not surjective. [tex]n \mathbb{Z}[/tex] is an ideal of [tex]\mathbb{Z}[/tex] for any [tex]n \in \mathbb{Z}[/tex], but [tex]f(n \mathbb{Z}) = n \mathbb{Z}[/tex] is not an ideal of [tex]\mathbb{Q}[/tex] (being a field, [tex]\mathbb{Q}[/tex] has no properly contained ideals).

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# Prep for Algebra Comprehensive Exam #1

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