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I was reading Advanced Modern Algebra by Rotman and I came across something that confused me.

On page 186 Proposition 3.117 (iii), he mentioned that $k[x] \subseteq K[x]$ where $K = k[x]/I$ with $I = (p(x))$ and $p(x)$ is a monic irreducible polynomial and $k$ is an arbitrary field. However, $k$ isn't even a subset of $k[x]/I$. Why is this the case?

Chirag Verma
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  • Did you read statement (i) in the proposition? "Therefore if $k'$ is identified with $k$, then $k$ is a subfield of $K$." – anon Nov 06 '19 at 01:29
  • Yes I did. I noticed that and the other line. I just didn't get the part about identifying $k'$ with $k$. My comment in the answer below gives my confusion – Chirag Verma Nov 06 '19 at 02:40

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The canonical projection $k[x] \to k[x]/I$ is injective on $k$ and so there is an isomorphic copy of $k$ in $k[x]/I$.

lhf
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    Thank you, I have noticed that. The only problem that I have with this is that the canonical projection $\phi: k \rightarrow k[x]/I$ results in a field that is isomorphic to $k$ and is a subfield of $k[x]/I$ but, it is not equal to $k$. Is it still ok to treat $k$ as being a subfield of $K$? – Chirag Verma Nov 06 '19 at 01:19
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    Yes, because for almost all practical purposes we can treat the isomorphic copy as actually equaling $k$ even though (as you note) they're technically two different objects. – Robert Shore Nov 06 '19 at 01:21
  • Sorry but, why can we treat them as the same? I understand that both operations in each field are similar (or preserved) and that there is a bijection between them. However, I just didn't understand why we can treat them as being exactly the same. (I am a self-learner so sorry if this is very basic. It is just one of the things I never understood) – Chirag Verma Nov 06 '19 at 01:34
  • @ChiragVerma, see https://math.stackexchange.com/a/2274891/589 and https://math.stackexchange.com/a/1553546/589 – lhf Nov 06 '19 at 10:21
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There is a canonical (and injective) homomorphism $k[x]\to K[y]$ (different variable names for clarity) sending $k\ni 1\mapsto 1+I\in K$ and $x\mapsto y$.

Hagen von Eitzen
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