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IB Tripos - Find all ring homomorphisms...

Find all ring homomorphisms

\displaystyle\phi : \frac{\mathbb{Q}[X]}{(X^3-1)}\rightarrow \mathbb{C}

.

I've got an answer, it just seems a bit of a scrappy way to get to it and i was wondering if anyone can see a more elegant approach?

Immediately before this, I've been asked whether

\displaystyle\frac{\mathbb{Q}[X]}{(X^3-1)}

is an integral domain (it isn't), and so it seems natural to look at the elements in

\displaystyle\frac{\mathbb{Q}[X]}{(X^3-1)}

which give a counter-example:

(X-1 +(X^3-1))(1+X+X^2+(X^3-1))=0+(X^3-1)

So, if a ring homomorphism does exist, then we must have either

\phi(X-1 +(X^3-1))=0

\phi(1+X+X^2 +(X^3-1))=0

. A bit of messing around with this gives three possible homomorphisms:

\phi_1 (a + bX + cX^2 + (X^3-1))=a+b+c

\phi_2 (a + bX + cX^2 + (X^3-1))=a+b\omega+c\omega^2

\phi_3 (a + bX + cX^2 + (X^3-1))=a+b\omega^2+c\omega

where

\omega = e^{2\pi i/3}

I can see that

\displaystyle\frac{\mathbb{Q}[X]}{(X^3-1)}

is isomorphic to

\displaystyle\mathbb{Q}[\omega]

, but it's not very clear to me how to argue that the three homomorphisms above are the only ones that exist, without resorting to algebraic manipulation.

(edited 12 years ago)

Reply 1

Zhen Lin

Original post by Mark13

Find all ring homomorphisms

\displaystyle\phi : \frac{\mathbb{Q}[X]}{(X^3-1)}\rightarrow \mathbb{C}

.

I've got an answer, it just seems a bit of a scrappy way to get to it and i was wondering if anyone can see a more elegant approach?

It's a really easy question if you know what it "really" is. It's just the rings

\mathbb{Q}

and

\mathbb{Q}(\omega)

glued together, where

\omega

is a primitive 3rd root of unity. (N.B. It isn't

\mathbb{Q}(\omega)

, which is what you get if you quotient by

X^2 + X + 1

.)

It's "obviously" not an integral domain, because

X^3 - 1

is not irreducible. More concretely,

(X - 1)(X^2 + X + 1) = 0

but the two factors are non-zero.

Because your ring is a quotient of a polynomial ring, any homomorphism out of it is uniquely specified by the image of X, but it has to map to an element

x \in \mathbb{C}

such that

x^3 - 1 = 0

, so you only have 3 choices. (I feel this is a sufficient justification, although it's tapping into the fact that a polynomial ring over a field k is a free unital associative commutative k-algebra generated by a single element...)

If you really want to be clear about it, perhaps something like this: there is a canonical homomorphism

\pi : \mathbb{Q}[X] \to \mathbb{Q}[X]/J

with kernel J, namely the quotient map, so any homomorphism

\varphi : \mathbb{Q}[X]/J \to R

induces a homomorphism

\varphi \circ \pi : \mathbb{Q}[X] \to \mathbb{R}

with kernel containing J; but we know what the homomorphisms out of

\mathbb{Q}[X]

look like, and every homomorphism

\tilde{\varphi} : \mathbb{Q}[X] \to R

with kernel containing J factors through

\pi

and induces a unique homomorphism

\varphi : \mathbb{Q}[X] / J \to R

, so it suffices to look at homomorphisms

\mathbb{Q}[X] \to R

with kernel containing J. In this case

R = \mathbb{C}

is an algebraically closed field and

J = (X^3 - 1)

is a principal ideal so it's quite easy to classify those homomorphisms.

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IB Tripos - Find all ring homomorphisms...

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