 3:49 AM
6 messages moved from 
:)
WOW!

9 hours later… 1:17 PM
@user21820 What was the book of which you were talking about?  @user21820 I think that they have implicitly taken $n\in\mathbb{N}\setminus \{0\}$ but forgot to say this explicitly. @user170039 You mean the degree. But as I said the error is stated twice.
Wait, you must have n > 1. Otherwise it is still wrong.
Anyway the second error cannot be put down to carelessness..
It explicitly says "prime field", which shows a conceptual misunderstanding. 1:32 PM
@user21820 Why $n>1$? Oh you are right. n > 0 will do.
My point is not that such webpages are totally useless, but that we cannot take anything at face value. @user21820 I think here they are talking about the length of the tuple and not of its elements (although I must say that even if this is the case, it is not easy (at least for me) to understand this from what is written). @user170039 It's fine if we consider (1) as an accidental error, because it is not a conceptually important one; we can consider constant polynomials as "trivial case". However (2) is not an accidental error. Nobody would explicitly say "prime field" unless they really meant it, as it is a technical term in algebra. 1:48 PM
There is also one more thing that I would like to mention. Your example of $\mathbb{C}(t)$ is wrong because $\mathbb{C}(t)$ is not even a field (what's the multiplicative inverse the polynomial $t$? in $\mathbb{C}(t)$). Do you wanted to mean that $\mathbb{C}$ is an algebraically closed field? @user170039 C(t) stands for the field of fractions of C[t]. This is standard notation in field theory, so I don't know why you can be so quick to say that I'm wrong.......
And that article is wrong so you linking to it is supremely ironic.
For this particular definition, Wikipedia is correct: Algebraically closed field
In abstract algebra, an algebraically closed field F contains a root for every non-constant polynomial in F[x], the ring of polynomials in the variable x with coefficients in F. == Examples == As an example, the field of real numbers is not algebraically closed, because the polynomial equation x2 + 1 = 0 has no solution in real numbers, even though all its coefficients (1 and 0) are real. The same argument proves that no subfield of the real field is algebraically closed; in particular, the field of rational numbers is not algebraically closed. Also, no finite field F is algebraically closed,...
Sorry I thought you linked the same article. 2:42 PM
4 messages moved from Philosophy of Mathematics