Are A[X, Y]/(XY) and A[X]⊕A[Y] Isomorphic?

[6c 6f 76 65]

Homework Statement

My textbook says that A[X, Y]/(XY) is a subring of A[X]⊕A[Y], but aren't they isomorphic? (A is any commutative ring)

Homework Equations

1st Ismorphism Theorem

The Attempt at a Solution

I can construct the map

φ: A[X, Y] → A[X]⊕A[Y]
f(X)+g(Y)+h(X, Y)*X*Y → f(X)+g(Y), this map is surjective (since the inverse of any f(X)+g(Y) is f(X)+g(Y)+(XY))

Ker(φ)=(XY), then by the 1st isomorphism theorem: A[X, Y]/(XY)≅A[X]⊕A[Y], am I stupid?

 

[fresh_42]

My textbook says that A[X, Y]/(XY) is a subring of A[X]⊕A[Y], but aren't they isomorphic?

Firstly, you are not stupid. Such things happen. (At least to most of us.)
Secondly, you have only one copy of $A$ in $A[X, Y]/(XY)$ whereas there are two in $A[X]⊕A[Y]: (0,A)$ and $(A,0)$.

 

[6c 6f 76 65]

Thank you for the reply! Sorry if I'm wasting your time, but I don't get what you mean by copy? (Is ⊕ the same as ×?)
And $A[X, Y]/(XY)$ means, as far as I know, all polynomials, with coefficients from A, of variable X and Y. Mod XY means, XY = 0, so shouldn't the new ring be functions of only X and functions of only Y? (That is $A[X]⊕A[Y]$ ?)

 

[fresh_42]

Thank you for the reply! Sorry if I'm wasting your time, ...

Never mind!

... but I don't get what you mean by copy? (Is ⊕ the same as ×?)

Well, it might depend on how the author uses it. I thought it's the same, as the circle around the plus suggests there is only {$0$} $=${$(0,0)$} as intersection. The authors of my textbook which I consulted to tell no nonsense use is it that way. (In the finite case of summation.)

Therefore $A[X] + A[Y] = A[X,Y]$ and $A[X] ⊕ A[Y] = A[X] × A[Y] =${$(f,g) | f ∈ A[X] ∧ g ∈ A[Y]$}.
With the embedding of $A[X,Y] / (XY)$, i.e. all polynomials over $A$ without any mixed $X$-$Y$-terms into $A[X] ⊕ A[Y]$ one has to choose whether $A$ is identified with $(A,0)$ or with $(0,A)$. So the image is either $A[X] ⊕ Y \cdot A[Y]$ or $X \cdot A[X] ⊕ A[Y]$.

At least this is my reading of the situation.

 

[Samy_A]

Never mind!

Well, it might depend on how the author uses it. I thought it's the same, as the circle around the plus suggests there is only {$0$} $=${$(0,0)$} as intersection. The authors of my textbook which I consulted to tell no nonsense use is it that way. (In the finite case of summation.)

Therefore $A[X] + A[Y] = A[X,Y]$ and $A[X] ⊕ A[Y] = A[X] × A[Y] =${$(f,g) | f ∈ A[X] ∧ g ∈ A[Y]$}.
With the embedding of $A[X,Y] / (XY)$, i.e. all polynomials over $A$ without any mixed $X$-$Y$-terms into $A[X] ⊕ A[Y]$ one has to choose whether $A$ is identified with $(A,0)$ or with $(0,A)$. So the image is either $A[X] ⊕ Y \cdot A[Y]$ or $X \cdot A[X] ⊕ A[Y]$.

I have a question about this embedding: does it map the multiplicative identity of $A[X,Y] / (XY)$ to the multiplicative identity of $A[X] ⊕ A[Y]$?

 

[fresh_42]

I have a question about this embedding: does it map the multiplicative identity of $A[X,Y] / (XY)$ to the multiplicative identity of $A[X] ⊕ A[Y]$?

You are totally right. I haven't thought about the unit because it wasn't required to have one. Unfortunately my negligence led to the lost of ring homomorphy, i.e. this isn't an embedding at all. Seems the way might go over tensor product and convolution which I tried to avoid. (And without units it's not yet clear to me.) Let me think about another way or do you know one which also works without 1?
Sorry to all for this mistake.

 

[Samy_A]

You are totally right. I haven't thought about the unit because it wasn't required to have one. Unfortunately my negligence led to the lost of ring homomorphy, i.e. this isn't an embedding at all. Seems the way might go over tensor product and convolution which I tried to avoid. (And without units it's not yet clear to me.) Let me think about another way or do you know one which also works without 1?
Sorry to all for this mistake.

No need to be sorry.

Would this work?

$xf(x)+yg(y)+c \mapsto (xf(x)+c,yg(y)+c)$, where $c$ is an element of the the ring A, and $f,g$ polynomials..

It looks weird, but it seems to add up.

 

[fresh_42]

It looks weird, but it seems to add up.

I agree. Brilliant idea to double the constant. I've been trapped in trying to split it.

EDIT: (Just as formal note with respect to the OP) The embedded ring always has the same constant c in both components, i.e. it is a proper subring. Elements (c,d) can't be reached. And it's not an ideal for the same reasons (multiplying with (r,0)).

 

[micromass]

Well, it might depend on how the author uses it. I thought it's the same, as the circle around the plus suggests there is only {$0$} $=${$(0,0)$} as intersection. The authors of my textbook which I consulted to tell no nonsense use is it that way. (In the finite case of summation.)

Well, there are actually two uses for $\oplus$, which are very related. In linear algebra, you can have subspaces $U$ and $V$ of a big space $W$. Then $U\oplus V$ simply denotes $U+V$ with the property that $U\cap U=\{0\}$. This is called the interior direct sum.
There is also an exterior direct sum, in the case of vector spaces, it is simply the cartesian product $U\times V$ with the pointswise operation.

These two operations are related because they are isomorphic vector spaces.

Anyway, here we are dealing with rings, in which case $R\oplus R'$ is simply the cartesian product with pointswise operations.