Using field axioms to prove a set is not a field

[notnottrue]

Homework Statement

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

 

[notnottrue]

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