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Every field contains a copy of Z_p

  1. Feb 26, 2015 #1
    So, every field contains a 'copy' of Z_p. I'm a bit confused by this.
    F is a field so it has characteristic p, and therefore contains a copy of the field Z_p.

    Here are some of my thoughts, are they correct? Do you have anything to add? (I'm trying to see a bigger picture here!):
    -The character of Z_p refers to the amount of times you need to add the multiplicative identity to itself to get to the additive identity, zero.
    -If |F| = p^n, then F* is a group with p^n - 1 elements. Is |Z_p| = p, and is Z_p* a group with p-1 elements? The * means we are taking all non-units, so in this case the only thing we are taking out is the zero.

    Ah so I had a slight tangent there, back to the main point (sorry this post isn't extremely coherent but any reply to any part of the post is very appreciated and useful to me).

    Let's say F is a field with character 3 and 3^3 elements. Character 3 means any element multiplied by 3 will return zero.
    Let's list the elements of F in order of smallest to largest: {0 1 a b c d e f g h i j k l m n o p q r s t u v w x y}.(Can we even order the elements some smallest to largest in this case? I think so...)

    Now, i'm trying to figure out what elements the Z_3 sub field of F would contain. would it be {g p y} because those are the 9th, 18th and 27th elements? Well that can't be, it's a subfield so we need to include the same multiplicative and additive identities as the original field, right? So would the elements of Z_3 be {0, 1, and some other thing}?

    Anyway, if anyone can shed some light on my incoherent rambling that'd be appreciated. Thank you all for the great help you've been.
  2. jcsd
  3. Feb 26, 2015 #2
    False. Look at the field ##\mathbb{Q}##.


    If ##p## is a prime (like you intended), then yes. But don't make the mistake that this holds for general ##\mathbb{Z}_n##.

    No, we can't order elements in such a field. We can only order elements in ordered fields which can be proven to have characteristic 0.

    It would be {0, 1, 1+1}
  4. Feb 26, 2015 #3
    you're awesome
  5. Feb 26, 2015 #4
    That said, if we define ##\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}##, then we would actually have that ##\mathbb{Z}_0 = \mathbb{Z}##. And in that case, every field does contain a copy of ##\mathbb{Z}_n## where ##n## is either prime or zero. Somehow, this does not seem like a popular convention, but I think it is nice.
  6. Feb 26, 2015 #5
    Z_0 = Z? I understand this in regards to the character of Z_0 is 0 and so 1+1+1+.... will never equal zero, so it will be an infinite series and be equal to the integers. However, in terms of notation Z_n = Z/nZ, how do we make sense of this when n = 0? Z_0 = Z / (0*Z)? Wouldn't that break a the universe and send us all into the endless void or something?
  7. Feb 26, 2015 #6
    Well, ##0\mathbb{Z}## is just the zero ideal. So ##0\mathbb{Z} = \{0\}##. And thus we have ##\mathbb{Z}_0 = \mathbb{Z}/\{0\}## which is (by definition of the quotient space) isomorphic to ##\mathbb{Z}##.
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