Showing that R [ X , Y ] / ( X 2 + 1 , Y 2 + 1 ) ≅ C × C A problem in my textbook asks me to show that R [ X , Y ] / ( X 2 + 1 , Y 2 + 1 ) ≅ C × C . After I've tried solving this for a long time, I looked at the textbook's hint and it said that I should consider the homomorphism ϕ : R [ X , Y ] → C × C , ϕ ( f ) = ( f ( i , i ) , f ( i , − i ) ) and this would be a surjective ring homomorphism whose kernel is the ideal ( X 2 + 1 , Y 2 + 1 ), so I would be done.The fact that this is a ring homomorphism is pretty obvious to me. However, I am really struggling to show that this function is surjective and to find the kernel.I think that this all stems from the fact that I am not used to the arithmetic properties of polynomial rings in two indeterminates. Anyway, back to the problem. For the surjectivity I tried playing arround with a few polynomials and I hoped to kind of guess the right one, but no succes. Showing that ( X 2 + 1 , Y 2 + 1 ) ⊂ Ker ⁡ ϕ is really straightforward, but the other inclusion is really killing me. So, how should I go about doing this?

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Showing that       R    [  X  ,  Y  ]      /    (      X    2    +  1  ,      Y    2    +  1  )  ≅      C    ×      C  A problem in my textbook asks me to show that       R    [  X  ,  Y  ]      /    (      X    2    +  1  ,      Y    2    +  1  )  ≅      C    ×      C  . After I&#039;ve tried solving this for a long time, I looked at the textbook&#039;s hint and it said that I should consider the homomorphism   ϕ  :      R    [  X  ,  Y  ]  →      C    ×      C  ,   ϕ  (  f  )  =  (  f  (  i  ,  i  )  ,  f  (  i  ,  −  i  )  ) and this would be a surjective ring homomorphism whose kernel is the ideal   (      X    2    +  1  ,      Y    2    +  1  ), so I would be done.The fact that this is a ring homomorphism is pretty obvious to me. However, I am really struggling to show that this function is surjective and to find the kernel.I think that this all stems from the fact that I am not used to the arithmetic properties of polynomial rings in two indeterminates. Anyway, back to the problem. For the surjectivity I tried playing arround with a few polynomials and I hoped to kind of guess the right one, but no succes. Showing that   (      X    2    +  1  ,      Y    2    +  1  )  ⊂  Ker  ⁡  ϕ is really straightforward, but the other inclusion is really killing me. So, how should I go about doing this?

Patrick Arnold · Accepted Answer

Step 1Kernel: Doing polynomial division where you have two polynomials in multiple indeterminates is in general a tricky business where you first have to fix an ordering. In this case, we can kind of hack our way through.Let f(X,Y) be the polynomial we want to divide by       X    2    +  1 and       Y    2    +  1. We can write   f  (  X  ,  Y  )  =  (      X    2    +  1  )  (  .  .  .  )  +  (      Y    2    +  1  )  (  .  .  .  )  +  (      c    1    X  Y  +      c    2    X  +      c    3    Y  +      c    4    ).When we divide by       X    2    +  1, we are effectively saying replace       X    2   with -1 and likewise with       Y    2   and -1. So you can get a &quot;remainder&quot; where all the X terms have degree 1 and Y terms have degree 1.Step 2Now, evaluate at (i,i) and (i,-i) to see  0  =  f  (  i  ,  i  )  =  0  (  .  .  .  )  +  0  (  .  .  .  )  +      c    1    (  −  1  )  +      c    2    i  +      c    3    i  +      c    4    =  (  −      c    1    +      c    4    )  +  i  (      c    2    +      c    3    )  0  =  f  (  i  ,  −  i  )  =  0  (  .  .  .  )  +  0  (  .  .  .  )  +      c    1    (  1  )  +      c    2    i  −      c    3    i  +      c    4    =  (      c    1    +      c    4    )  +  i  (      c    2    −      c    3    )You can solve the simultaneous equations in       c    i   to see that all       c    i   have to be 0.Surjective: Let   (  a  +  b  i  ,  c  +  d  i  )  ∈            C        2  .We look at the &quot;remainder&quot; term from before   f  (  X  ,  Y  )  =      c    1    X  Y  +      c    2    X  +      c    3    Y  +      c    4  . We want   f  (  i  ,  i  )  =  a  +  b  i  ,  f  (  i  ,  −  i  )  =  c  +  d  i.You get   a  +  b  i  =  (  −      c    1    +      c    4    )  +  (      c    2    +      c    3    )  i and   c  +  d  i  =  (      c    1    +      c    4    )  +  i  (      c    2    −      c    3    ). You can again repeat the same thing as before and solve for the simultaneous equations.

Brenda Jordan · Accepted Answer

Step 1Question: &quot;So, how should I go about doing this?&quot;Answer:      R    [  X  ,  Y  ]      /    (      X    2    +  1  ,      Y    2    +  1  )  ≅      C        ⊗                  R                  C    ≅      R    {  1  ,      x    ¯    }      ⊗                  R                  C    ≅      C    ×      C  If       F    1    :=  x  +  y  ,      F    2    :=  x  −  y it follows  ϕ  (      F    1    )  =  (  i  ,  0  )  ,  ϕ  (      F    2    )  =  (  0  ,  i  )  ,  ϕ  (      F    1    2    )  =  (  −  1  ,  0  )  ,  ϕ  (      F    2    2    )  =  (  0  ,  −  1  )hence  ϕ  (  b      F    1    −  a      F    1    2    )  =  (  a  +  i  b  ,  0  )  ,  ϕ  (  −  a      F    2    2    +  b      F    2    )  =  (  0  ,  a  +  i  b  )hence the map is surjective. The quotient  A  :=      R    [  X  ,  Y  ]      /    (      X    2    +  1  ,      Y    2    +  1  )has a basis given as follows  A  :=      R    {  1  ,      x    ¯    ,      y    ¯    ,            x      y        ¯    }and you must use this basis to give an explicit isomorphism  ϕ  :  A  ≅      C    ×      C    .Step 2You find  ϕ  (  1  )  =  (  1  ,  1  )  ,  ϕ  (  x  )  =  (  i  ,  i  )  ,  ϕ  (  y  )  =  (  i  ,  −  i  )  ,  ϕ  (  x  y  )  =  (  −  1  ,  1  )  .You must check that the following vectors in       C    ×      C   are linearly independent:  (  1  ,  1  )  ,  (  i  ,  i  )  ,  (  i  ,  −  i  )  ,  (  −  1  ,  1  )  .

Showing that R[X,Y]/(X^2+1, Y^2+1) cong C times C

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