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Let $f:K \rightarrow K$ be a unital ring homomorphism ,i.e.it is a ring homomorphism which sends unity to unity. Here K is some field.

Then clearly it is injective as $\ker(f) =\{ 0 \}$.

I want to ask whether this is surjective also? Here $K$ is a field. If we take $K=R$, then what happens?

(2).Another question I wanted to ask that is $Fp$ adjoined uncountable many variables is a field with positive characteristic and uncountable cardinality?

Any help will be grateful. Thank you.

Derwal Meena
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  • "The clearly it injective, as $\ker f=0$." Are you claiming that every unital ring homomorphism is injective? That is not true. – Arturo Magidin May 09 '24 at 04:17
  • "If we take $K=R$"... what is $R$? Is it supposed to be $\mathbb{R}$? If $Fp$ supposed to be $\mathbb{F}_p$, the finite field of $p$ elements? When you say "adjoin", are you taking the polynomial ring, or its field of fractions? This needs clarity, and focus. Ask one question at a time, and make clear what you mean. – Arturo Magidin May 09 '24 at 04:19
  • It is surjective for $\mathbb{R}$, it does not have to be in general, for example $\mathbb{C}$. – freakish May 09 '24 at 05:34
  • Consider $f: \Bbb Z \times \Bbb Z \to \Bbb Z$ defined by $f(x, y)= x$. Then $f(1, 1)=1$ but $f$ clearly is not injective. – Robert Shore May 09 '24 at 05:35
  • R denote field of real numbers and Fp finite field of order p. – Derwal Meena May 09 '24 at 05:36
  • @freakish,can you give me some example? – Derwal Meena May 09 '24 at 05:37
  • @DerwalMeena read this: https://math.stackexchange.com/questions/4110689/proper-subfields-of-mathbbc-isomorphic-to-mathbbc – freakish May 09 '24 at 05:39
  • Please only ask one question per post. – Greg Martin May 09 '24 at 07:56

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