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I Basic Question about a Ring Homomorphisms

  1. Oct 11, 2016 #1
    I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

    I need help to clarify a remark of B&K regarding ring homomorphisms from the zero or trivial ring ...

    The relevant text from B&K reads as follows:


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    ?temp_hash=93b8052059b87992d1f7fd12348dde3e.png



    In the above text from B&K's book we read ...

    "... ... This follows from the observation that the obvious map from the zero ring to ##R## is not a ring homomorphism (unless ##R## itself happens to be ##0##). ... ... "



    I do not understand the above statement that the obvious map from the zero ring to ##R## is not a ring homomorphism (unless ##R## itself happens to be ##0##) ... ...


    What, indeed do B&K mean by the obvious map from the zero ring to ##R## ... ... ?


    It seems to me that the obvious map is a homomorphism ... ..


    Consider the rings ##T, R## where ##T## is the zero ring and ##R## is any arbitrary ring ... so ##T = \{ 0 \}## where ##0 = 1## ...

    Then to me it seems that the "obvious" map is ##f( 0_T) = 0_R## ... ... which seems to me to be a ring homomorphism ...

    ... BUT ... this must be wrong ... but why ... ?

    Can someone please clarify the above for me ...

    Some help will be very much appreciated ...

    Peter
     
  2. jcsd
  3. Oct 11, 2016 #2

    fresh_42

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    If you require (RH2) to hold, then you need a ##1_0## which ##\{0\}## doesn't have.
    You cannot say ##0=1##. There is no ##1## in ##T=\{0\}##. But your definition of ##f : T \rightarrow R## is correct.
    The only wrong here is ##1 \notin \{0\}## and thus (RH2) is violated. (In strict logic, the definition doesn't apply to ##\{0\}## at all, because it is not a ring with unity.)

    What can be done with the usage of (RH1) is the following:
    ##f(r') = f(1_R \cdot r') = f(1_R) \cdot f(r')## so ##f(1_R) = 1_{f(R)}##.

    Now this is the crucial point. We cannot conclude ##1_{f(R)} = 1_S## as we do in fields and groups.

    E.g. in ##\mathbb{Z_9}## the ring of remainders of division by nine, there is ##1\,\cdot\,3 = 3 = 4 \,\cdot\, 3## but ##1 \neq 4##.
    Therefore we need the additional property (RH2) to guarantee ##f(1) = 1##.
     
  4. Oct 12, 2016 #3

    micromass

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    It does have one. The zero ring ##\{0\}## has ##0## as ##1_0##, hence ##1=0##. In the definition of a ring with unit, there is usually never the requirement that ##1\neq 0##. That requirement usually is present in fields though.
     
  5. Oct 12, 2016 #4

    fresh_42

    Staff: Mentor

    Thanks. I'm still thinking of an example of a mapping ##f : R \rightarrow S## with ##f(r + s) =f(r)+f(s)## and ##f(rs)=f(r)f(s)## but ##f(1) \neq 1## which is not as simple as ##R=\{0\}##.
     
  6. Oct 12, 2016 #5

    micromass

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    ##f:\mathbb{R}\rightarrow \mathbb{R}[X]/(X^2-X):\alpha\rightarrow \alpha X##
     
  7. Oct 12, 2016 #6

    fresh_42

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    Yes. ##r \rightarrow \alpha r## came to my mind, but to get rid of the square this way didn't come to my mind. Silly me, now I feel riled.
     
  8. Oct 13, 2016 #7
    Interesting ring R[X]/(X2 - X). Let's see — if there is a term [c Xn] (c ∈ R, n ≥ 2) in R[X]/(X2 - X) then we can add

    [0] ≡ [-c Xn + c Xn-1] ≡ [-c (Xn - Xn-1)] ≡ [-cXn-2(X2 - X)] ≡ [-cXn-2]⋅[X2 - X]​

    without changing it, to get (the equivalence class of) a monomial of lower degree. Since any element of R[X] contains only finitely many powers of X in the first place, we can reduce any term to a linear polynomial aX + b with a finite number of such operations.

    So the elements of R[X]/(X2 - X) may be thought of by those coefficients (a, b), i.e., as elements of the plane R2. It looks provable that any element of R[X]/(X2 - X) has a unique representation in this form. Then addition is just what you might expect, i.e.,

    (a, b) + (c, d) = (a+c, b+d).​

    But what is the multiplication rule? I get that (using this representation):

    (a, b) ⋅ (c, d) = (ad + bc + ac, bd),
    which may be thought of as a map

    mult: R2R2.​

    Cool!

    Question: Is it just a coincidence that (in this representation)

    the sum of (the coefficients of the product) = (ad + bc + ac) + bd​

    equals

    the product of (the sums of coefficients) = (a + b) ⋅ (c + d) ?​
     
    Last edited: Oct 13, 2016
  9. Oct 13, 2016 #8
    Thanks to fresh_42, micromass and zinq for the guidance and help ...

    BUT ... I am still puzzling over my original question ...

    Can you help with the original issue ...

    Thanks again for your posts so far ...

    Peter
     
  10. Oct 13, 2016 #9
    -----
    In the above text from B&K's book we read ...

    "... ... This follows from the observation that the obvious map from the zero ring to R is not a ring homomorphism (unless R itself happens to be 0). ... ... "

    I do not understand the above statement that the obvious map from the zero ring to R is not a ring homomorphism (unless R itself happens to be 0)
    -----

    Hint: There are two parts to the book's definition of a ring homomorphism. What happens when you try to prove each one of them to a map from the zero ring to an arbitrary ring? (I did not know where this went wrong until I tried it.) Remember that in the zero ring 0 = 1.
     
  11. Oct 13, 2016 #10

    fresh_42

    Staff: Mentor

    In short: You cannot conclude ##f(1_R) = 1_S## by only the condition (RH1) given.

    Homomorphisms of any kind means basically: preserving the structure. Here we have rings.
    Thus ##f(r_1+r_2) = f(r_1)+f(r_2)## and ##f(r_1\,\cdot\, r_2) = f(r_1)\,\cdot\,f(r_2)## has to hold.

    But what to do with the two neutral elements?
    The ##0_R## is no problem, since ##f(0_R)=0_S## already follows from the above.
    However, this can't be said about ##1_R##.
    But to preserve / respect the whole structure from one ring ##R## to the other ring ##S##, one would expect to have ##f(1_R) = 1_S## as well. As it cannot be proven by (RH1), it is required for ##f## being a ring homomorphism: (RH2).

    The examples above (I think all, but I didn't check) map ##1_R \longmapsto 0_S##.
    This is something we would not like to be called a ring homomorphism as soon as ##S## has a ##1_S##, too.

    But remember post #3. Therefore ##T= \{0\}## counts as an example, where ##1_T = 0_T##. In this case, ##f(1_R)=1_T=0_T## is allowed. There is a unit in both rings and (RH1) and (RH2) hold, even if it happens that in one ring the neutral elements coincide.

    Edit: Homomorphism = homo (equal) morph (structure)
    That's why, e.g. linear maps - ##L(\alpha x + \beta y) = \alpha L(x) + \beta L(y)## - are also called homomorphisms, or monomorphisms (injective), or epimorphisms (surjective), or isomorphisms (bijective), or automorphisms (bijective within the same underlying set). The same is true for groups, fields, algebras and rings and modules, although they all carry a different structure and therefore a different definition. But the structure itself is preserved by morphisms (name for it if categories are involved).
     
    Last edited: Oct 13, 2016
  12. Oct 14, 2016 #11

    mathwonk

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    one suggestion i have is to just forget about the zero ring as it essentially never arises in practice. At least I have spent my whole 40 year career without encountering it significantly. i.e. in my experience, when it arises it is uninteresting and can be ignored. actual math is usually not about such trivial things. at most you might be troubled now and then to check that your ring you are dealing with is not the zero ring, e.g. that the ideal you are dealing with in a quotient is not the whole ring. i.e. there is a reason it is called the trivial ring.

    But this somewhat frivolous comment is just a late night throwaway. If you are enjoying solving the logical puzzle here, go ahead, it may be fun, and even enlightening. But don't lose a lot of sleep over it. It would be more useful to learn the Hilbert nullstellensatz, i.e. a criterion for when an ideal is the unit ideal, (so that if so, then the quotient is the zero ring).
     
    Last edited: Oct 14, 2016
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