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$R$ is a commutative ring with $1$, prove that there exist epimorphism from $R[x]$ onto $R$.

I maybe able to show that R[x] onto R is a homomorphism but I'm not sure how to show that it is onto and how will I use the fact that $R$ is a commutative ring with $1$ to prove the statement? Please can someone help.

Ka Em
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    For rings, epimorphisms and surjective homomorphisms are not the same! – Fred Rohrer Apr 02 '20 at 14:56
  • I learnt in my class that epimorphism = homomorphism + onto... What is epimorphism then meant to be? – Ka Em Apr 02 '20 at 15:04
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    Epimorphism usually means that $g_1 \circ f = g_2 \circ f$ implies $g_1 = g_2$. So for sets, if you have a surjective function $f$ and $g_1(f(x)) = g_2(f(x))$ for all $x$ then $g_1(y) = g_2(y)$ for all $y$ (because $f$ is surjective). It would be nice if $g_1 \circ f = g_2 \circ f \implies g_1 = g_2$ was equivalent to $f$ being surjective but that's not always so. See https://en.wikipedia.org/wiki/Epimorphism for more info. – Trevor Gunn Apr 02 '20 at 15:28

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Consider the map

$$ev_0: R[X] \to R: P \mapsto P(0)$$

Clearly this is a ring homomorphism and this is surjective since $ev_0(R) = R$.

In fact, evaluating in every other ring element works equally well.

J. De Ro
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    See here: https://math.stackexchange.com/questions/1213572/why-is-commutativity-needed-for-polynomial-evaluation-to-be-a-ring-homomorphism – J. De Ro Apr 02 '20 at 15:13