Math 542
Exit Cards Day 18
0. Is there any way to have a field of order something other than a prime or a power of a
prime? Answer: Nope! Any finite field will have a prime characteristic and hence a subfield
isomorphic to Zp for some prime p. So it will be a finite extension (and so an algebraic
extension) of Zp which will make it of order pn for some n.
1. I was a little confused about the existence of the finite field. First you claimed there was a
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field in which x p  x has pn roots. Later it seemed like you needed this field to have
characteristic p. Is this obvious?
Answer: Well we know that any finite field must be of prime characteristic and must have a
subfield isomorphic to Zq for some q prime. But we also know it will be a finite extension of
some degree n of Zq and so will have qn elements. This means that the q and p need to be
the same since this field has pn elements.
It kind of makes sense that a field with pn must be of characteristic p as opposed to some
other prime, but I guess it is not quite obvious.
2. If E is algebraic over F and  is in E but not in F do we know that B = {n: n is in N} is a
basis for E over F? That is, will any such  work or are there restrictions. Answer: No, this
will not work in general. For example Q( 6 2 ) is algebraic over Q and 2 is an element in
Q( 6 2 ) but not in Q. But the powers of 2 do not form a basis for Q( 6 2 ) since we can’t get
the sixth root of 2 this way.
We do know that we can use a generator of the multiplicative group (in the finite case) from
Corollary 33.6. (See notes from Day 18 Uniqueness Part 2).
3. Does their exist a field with any number of elements? Answer: No, finite fields must have pn
elements where p is prime. So for example there is no field with 6 elements.
4. If p(x) is a polynomial in F[x] and its of degree m greater than or equal to 1 and I is the
principle ideal generated by p(x) then is the dimension of F[x]/I equal to m? Why or why not?
Answer: I assume you mean “…as a vector space over F” In any case the answer is yes only if
the polynomial is irreducible. If it is not irreducible then you don’t even get a field. If it is
irreducible then it is a vector space of dimension m over F since the powers (from 0 to m-1) of
one of the roots of p(x) form a basis (as expressed in question 2 above).
5. What we were doing were simple extensions. What is a non-simple field extension? Answer:
It turns out that all the finite fields are simple extensions. However, R for example is not a
simple extension of Q. It takes a lot more than just throwing in one transcendental number and
closing the field up in order to get all the real numbers.
6. How are algebraic closures and topological/analytic closures related? How are they different?
Answer: They are the same in the sense that they are big enough to contain all of the original
set/space/field and all of something else. In the algebraic closure case the field is big enough to
contain all the roots of all of the polynomials. In the analytic/topological case it is big enough to
contain all the limits of all the Cauchy sequences.
7. Is it always true that if ker is {0} then  is 1-1? Answer: Yes if  is a homomorphism. 0
always maps to 0, so if  1-1 than that is the only thing that can go to 0 so the kernel is {0}. For
the other direction, suppose (a) = (b). Then (a) - (b) = 0 which means that (a - b) = 0
which means that a-b is in the kernel of . And so if the kernel is {0} then a – b = 0 so a = b.
8. If  is a root of x p  x how do we know that -1 is a root? Isn’t this needed to show this
splitting field is also a field? Answer: Yes you are correct! I completely left out this step (it is
fixed now on Day 18 notes). Also note that if I multiply -1 by itself pn times I get the inverse of
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p .
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(Since  p (-1) p = (-1) p = 1.) So since  p = , (-1) p is also the inverse of .
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Thus (-1) p = -1. So -1 is also a root of x p  x .
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