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I'm having a bit of trouble with the idea of an integral morphisms, and algebra homomorphisms for that matter. I'm wondering if the above is just "automatically" true. Does an algebra over a field always carry a "copy" of the field? I have seen that "if $m$ is a maximal ideal of $A$ then $A/m$ is a finite field extension" which makes me think yes, but then are trivial algebras allowed like "$A=\{0\}$"? thank you for your help, km very confused.

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As for your second question, note that any $k$-algebra $A$ has, by definition, a homomorphic image of $k$ in $Z(A)$, the center of $A$. However, if $A \neq 0$, then since non-trivial field homomorphisms are actually embeddings, therefore a copy of $k$ does indeed sit inside $A$.

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    $\begingroup$ Unless $A=0$ is the zero algebra. $\endgroup$ – Najib Idrissi Jun 20 '16 at 14:53
  • $\begingroup$ That's why I said non-trivial field homomorphisms. $\endgroup$ – Hmm. Jun 20 '16 at 14:54
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    $\begingroup$ Yes of course, I just wanted to clarify that it's possible for the unit $k \to A$ to actually be a trivial homomorphism (iff $A = 0$, in fact). $\endgroup$ – Najib Idrissi Jun 20 '16 at 15:00
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No, it is not true. For example $k$ and $k[x]$ are two finitely-generated $k$-algebras, but the inclusion $k \to k[x]$ is not integral ($x$ is not the root of any monic polynomial, pretty much by definition of the polynomial ring $k[x]$).

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