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Prep for Algebra Comprehensive Exam #1

  1. Jul 20, 2007 #1
    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 [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].


    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).
  2. jcsd
  3. Jul 21, 2007 #2
    I think you should use this definition of ideal: http://en.wikipedia.org/wiki/Ideal_(ring_theory)

    since it's more common. This also means you don't have to show closure under multiplication. Everything else looks fine, I didn't notice anything off. Goodluck.
    Last edited: Jul 21, 2007
  4. Jul 21, 2007 #3


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    If it absorbs products from the ring, then it's automatically closed under multiplication, so even using the OP's definition, proving explicitly that the sets are closed under multiplication is redundant.

    Also, the word is kernel. :wink:
  5. Jul 23, 2007 #4
    Thanks you both for your responses. I agree that I was redundant, and I will look for this in later proofs. As for the misspelling of kernel - well, I have no excuse for that!

    I will post new questions/solutions under different headings (#2, #3, etc.)
  6. Jul 23, 2007 #5
    Here is a major hint on (b).

    Theorem: Given [tex]f:R\to S[/tex] be a ring homomorphism. And [tex]I\triangleleft R[/tex] then [tex]f\triangleleft f[/tex]. Now if [tex]f[/tex] is surjective then [tex]f=S[/tex]. Complete this proof.
  7. Jul 23, 2007 #6


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    In your class, rings are not required to have a unit?
  8. Jul 24, 2007 #7
    No they aren't. But I've seen books with both defenitions. We're using Dummit & Foote (3rd Edition).
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