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I'm a little bit confused by the notation "$K[t]$", what does it denote? I know that $K[X]$ is the polynomial ring over $K$, so the elements $P$ of $K[X]$ look like $P(X) = \sum_{n=0}^\infty k_n X^n$ with $k_n \in K$. Is $K[t]$ then just the image of the homomorphism $P \mapsto P(t)$ with $P \in K[X]$? Wouldn't that just be same as $K(t)$, which is the smallest field that contains both $K$ and $t$? What's the difference?

LinearAlgebruh
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    It would help to know specifically the context in which the notation occured, but one immediate difference between the image of the homomorphism and the field extension: $\frac1t$ is a member of the latter but not the former... – Steven Stadnicki May 20 '21 at 13:48
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    $K[t]$ and $K(t)$ are the same only if $t$ is algebraic over $K$ – Evariste May 20 '21 at 13:50
  • Oh, maybe my definition of $K(t)$ is wrong, I've read the wikipedia article and elements of $K(t)$ are actually quotients of such polynomials in $K[t]$ – LinearAlgebruh May 20 '21 at 13:50

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$K(t)$ is the smallest field that contains $K$ and $t$, but $K[t]$ is the smallest ring that contains $K$ and $t$. For example, $t^{-1}$ is in $K(t)$ but not in $K[t]$.

Arno Fehm
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    It might be useful to add that, if $t$ is transcendental over $K$, $K[t]$ is isomorphic to the ring of polynomials over $K$, and $K(t)$ is isomorphic to the field of rational fractions (in a single indeterminate). – Bernard May 20 '21 at 13:54
  • Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque May 20 '21 at 20:43