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Why there exists a surjective ring homomorphism $f:\mathbb Z[\sqrt2]\to\mathbb Z_p$ for $p=7$, but not for $p=5?$

My try: $\mathbb Z[\sqrt 2]=\frac{\mathbb Z[X]}{X^2-2}$ where $X^2-2 $ is prime ideal. Can we define canonical ring homomorphism for $p=7$ to show surjective? Thank you.

1 Answers 1

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There is no such homomorphism for $ p = 5 $, because this would imply that $ 2 $ is a quadratic residue modulo $ 5 $. (Why?)

For $ p = 7 $, there is an obvious map $ \mathbf Z[\sqrt{2}] \cong \mathbf Z[X]/(X^2 - 2) \to \mathbf F_7[X]/(X^2 - 2) \cong \mathbf F_7 \times \mathbf F_7 $ given by reducing modulo $ 7 $, and composing it with projection onto either factor gives the desired homomorphism $ \mathbf Z[\sqrt{2}] \to \mathbf Z/7 \mathbf Z $.

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    Strictly speaking, it implies that $2\cdot f(1)$ is a quadratic residue. You need a line or two to prove that we have $f(1) = 1$, or something similar. – 2017-02-20
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    @Arthur $ f(1)^2 = f(1) $, so $ f(1) = 0 $ or $ f(1) = 1 $. $ f $ is not surjective in the former case. – 2017-02-20
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    And there is the line or two I'm talking about. – 2017-02-20
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    can you plz mention that surjective ring homomorphism for later case? – 2017-02-20
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    @MatheMagic Reduce modulo $ 3 + \sqrt{2} $. – 2017-02-20
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    sorry..m little dumb to understand that. – 2017-02-20
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    @MatheMagic You have a quotient map $ \mathbf Z[\sqrt{2}] \to \mathbf Z[\sqrt{2}]/(3 + \sqrt{2}) $, and that last quotient is isomorphic to $ \mathbf Z/7 \mathbf Z $. This is what you get when you unpack the map I mentioned in my answer. – 2017-02-20