Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Which of the following statement(s) is / are true?
1. $ \mathbb{C} (x)$ is algebraically closed, where $x$ is an indeterminate.
2. An algebraically closed field must be of characteristic $0$.
3. If $E$ is an algebraically closed extension field of $F$, then $E$ is an algebraic extension of $F$.

can anyone help me to solve this problem.thanks for your time.

share|cite|improve this question

closed as off-topic by Bookend, Eric Wofsey, Zachary Selk, S.Panja-1729, Ivo Terek Oct 11 '15 at 2:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Bookend, Eric Wofsey, Zachary Selk, S.Panja-1729, Ivo Terek
If this question can be reworded to fit the rules in the help center, please edit the question.

  1. The question is whether all polynomials with coefficients in $\mathbf C(x)$ have a root. Hint: consider $f(T) = T^2-x$. In other words, does $x$ have a square root in $\mathbf C(x)$?

  2. That's a bit weird; why would that be true? Anyway, if you are familiar with the fact that every field has an algebraic closure, look at the algebraic closure of $\mathbf F_2$. What can you say about its characteristic?

  3. What do you think of $\mathbf C$ over $\mathbf Q$?

share|cite|improve this answer
About 2: an algebraically closed field is infinite and a field of characteristic 0 is infinite. It is not that weird to ask wether there is an inclusion. – Siméon Jan 14 '13 at 17:22
@Ju'x Ah thanks, that makes sense. – jathd Jan 14 '13 at 17:58
@Ju'x: inclusion is a homomorphism (injection), ie trivial kernel, ie characteristic is preserved. – Mark Bennet Jan 14 '13 at 19:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.