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Show that every quadratic in $\mathbb F_{\,7}[t]$ has a zero in $\mathbb F_{\,7}(\sqrt{3})$.

Is somebody able to shine a light on how I'm meant to go about showing this?

Your help will be much appreciated!

Jessie
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    Hint: Can you show that every element of $\mathbb F_{,7}$ has a square root in the extended field? If so, then just apply the quadratic formula. – lulu Oct 06 '16 at 12:19
  • how do i go about doing that? I am at the end of a very long day (week!) and I don't think my brain is functioning anymore.... – Jessie Oct 06 '16 at 12:24
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    Just write down the square roots! We know $0^2=0,1^2=1,2^2=4,3^2=2\pmod 7$. If $\alpha^2=3$ then $(2\alpha)^2=4\times 3=12=5$ so the square root of $5$ is $2\alpha$. And so on. – lulu Oct 06 '16 at 12:26
  • Did you actually mean $;\Bbb F_7(\sqrt3);$ or indeed you meant $;\Bbb F_7t;$ ? For example, the last one isn't even a field... – DonAntonio Oct 06 '16 at 12:32
  • It was meant to be F$_7(\sqrt3)$ - sorry it was mistakenly edited that way and didn't notice it. I'll fix it now. – Jessie Oct 06 '16 at 12:36
  • Related material. Quite a bit of it actually :-) – Jyrki Lahtonen Oct 07 '16 at 10:35

2 Answers2

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Hints

  1. The polynomial $t^2-3$ has no root in $\mathbb{F}_7$

  2. The finite field with $49$ elements is unique up to isomorphism

Alternatively, show that also, $5$ and $6$ have square roots in $\mathbb{F}_7(\sqrt{3})$, which is sufficient for any quadratic polynomial in $\mathbb{F}_7[t]$ to have roots. For instance, $(2\sqrt{3})^2=\dots$

egreg
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  • Thanks egreg, I can see that t$^2$ - 3 has a root of $\sqrt3$ - but it says every quadratic, and so t$^2$ - 2 has a root of $\sqrt2$ which isn't in F$_7(\sqrt3)$ - or am I looking at it wrong? – Jessie Oct 06 '16 at 12:32
  • $\sqrt2=3$. @Jessie – Clayton Oct 06 '16 at 12:36
  • oh, coz it's mod 7! – Jessie Oct 06 '16 at 12:38
  • @Jessie For instance, $5$ is not a square in $\mathbb{F}7$; however, $\mathbb{F}{7}(\sqrt{5})$ has $49$ elements and so it is isomorphic to $\mathbb{F}_7(\sqrt{3})$. Hence $5$ has a square root in $\mathbb{F}_7(\sqrt{3})$, because it has one in an isomorphic field. – egreg Oct 06 '16 at 12:41
  • @egreg - what about $t^2$ - 6? $\sqrt6$ is not a square in F$_7$ - How do I show this one is in F$_7(\sqrt3)$ – Jessie Oct 06 '16 at 13:00
  • @Jessie Exactly in the same way. Another example: the polynomial $t^2-3t+1$ has no root in $\mathbb{F}_7$, so its splitting field has $49$ elements; therefore it is isomorphic to $\mathbb{F}_7(\sqrt{3})$. – egreg Oct 06 '16 at 13:07
  • We haven't studied splitting fields, so I think that's why I'm getting a bit lost! – Jessie Oct 06 '16 at 13:34
  • @Jessie You need to find $\sqrt{5}$ and $\sqrt{6}$. – egreg Oct 06 '16 at 16:03
  • @Jessie: You don't need to know anything about splitting fields. As for $\sqrt 6$, this is $\sqrt 2\cdot\sqrt 3$, and you already know what $\sqrt 2$ is. – TonyK Oct 07 '16 at 10:57
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To flesh out the comments:

The quadratic formula works perfectly well in finite fields, so long as the characteristic isn't $2$. It follows that the only obstruction to finding roots to quadratics is that we might not be able to extract the relevant square root. This, in turn, implies, that we'll be done if we can show that every element of $\mathbb F_{\,7}$ has a square root in $\mathbb F_{\,7}\left(\sqrt 3\right)$.

Let's just do this explicitly. Before adjoining $\sqrt 3$ the squares were $\{0^2=0,1^2=1,2^2=4,3^2=2\}$. Let $\alpha =\sqrt 3$. We need to find square roots of $5$ and $6$. General theory tells us that the product of a non-square and a square is a non-square, which leads us to consider $2\alpha$ and $3\alpha$. We compute: $$\left(2\alpha\right)^2=4\times 3=12=5\pmod 7\quad \&\quad \left(3\alpha\right)^2=9\times 3=27=6\pmod 7$$

Thus, every element of $\mathbb F_{\,7}$ has a square root in the extended field, and the argument is complete.

Note: it isn't difficult to extend this argument to larger fields. There, instead of writing out all the square roots, it is better to simply point out that you can get all the non-squares by multiplying a given non-square by the set of squares. All we did above was to make that explicit.

lulu
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