To which field is this quotient isomorphic $\frac{\mathbb{Z}[x]}{\langle 2x-1, 5 \rangle}$?

Question

I need to show the quotient $\frac{\mathbb{Z}[x]}{\langle 2x-1, 5 \rangle}$ is a field and find to which field is this quotient isommorphic.

I think this: $\frac{\mathbb{Z}[x]}{\langle 2x-1, 5 \rangle}\cong \frac{\mathbb{Z}_5[x]}{\langle 2x-1 \rangle}$. The polynomial $2x-1$ is irreducible in $\mathbb{Z}_5[x]$ thus, the quotient $\frac{\mathbb{Z}_5[x]}{\langle 2x-1 \rangle}$ must to be a field, but I don´t know to which is isomorphic.

$(2x-1)=(x-2^{-1})$ in $\mathbb Z_5[x]$ and $R[x]/(x-a)\simeq R$ for every commutative ring $R$. — user26857, May 09 '21 at 06:25

score 2 · Answer 1 · answered May 09 '21 at 04:57

2

The ideal ${\langle 2x-1, 5 \rangle}$ is the kernel of the epimorphism from $\mathbb Z[x]$ to $\mathbb Z_5$ that sends $x$ to $3$ (and $1$ to $1$).

answered May 09 '21 at 04:57

David

19

3

You probably mean "surjective morphism". Ring epimorphisms don't need to be surjective! See here: https://math.stackexchange.com/questions/81123 – user26857 May 09 '21 at 06:18

Fuat Ray · Answer 2 · 2025-06-14T21:20:55.450

0

$\dfrac{\mathbb{Z}[x]}{( 2x-1, 5)}=\dfrac{\mathbb{Z}[x]/(5)}{(2x-1)}\cong \dfrac{\mathbb{Z}_5[x]}{(2x-1)}\cong \mathbb Z_5$

edited Jun 14 '25 at 21:20

answered Jun 14 '25 at 19:50

Fuat Ray

1,342

You are absolutely right. How we can see that $\mathbb Z[\frac12]/(5)\cong \mathbb Z_5$ – Fuat Ray Jun 14 '25 at 21:21
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How is the last isomorphism given explicitly? – Ted Shifrin Jun 14 '25 at 22:11
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$\mathbb Z_5[x]\to\mathbb Z_5$ by $p(x)\mapsto p(\overline 3)$ – Fuat Ray Jun 16 '25 at 20:35

To which field is this quotient isomorphic $\frac{\mathbb{Z}[x]}{\langle 2x-1, 5 \rangle}$?

2 Answers2