Using field axioms to prove a set is not a field

[notnottrue]

Homework Statement

Let F = {a + b$\sqrt[3]{2}$:a,b$\in$Q}.
Using the fact that $\sqrt[3]{2}$ is irrational, show that F is not a field.

[Hint: What is the inverse of $\sqrt[3]{2}$ under multiplication?]

Homework Equations

For a field,
For all c $\in$ F, there exists c^-1 $\in$ F s.t. c*c^-1 =1

The Attempt at a Solution

I am unsure how to relate the axioms to the set. Is c = (a +b$\sqrt[3]{2}$)?
Or show there is no b$\sqrt[3]{2}$*b^-1$\sqrt[3]{2}$=1?
Or b$\sqrt[3]{2}$*(b$\sqrt[3]{2}$)^-1=1?

 

[Hurkyl]

There are lots of ways to go about it, but if you're going to follow the hint, it specifically suggests showing that $\sqrt[3]{2}$ is an element that doesn't have an inverse.

 

[notnottrue]

$\sqrt[3]{2}$$\in$F with a=0 and b=1
So there must be c such that c*$\sqrt[3]{2}$=1
c=1/$\sqrt[3]{2}$$\notin$Q
Since $\sqrt[3]{2}$$\in$F, F must not be a field because it has no inverse under multiplication.
Is this sound?

 

[notnottrue]

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